Properties of Ordered Group/OG1

{{begin-axiom}} {{axiom|n = \operatorname{OG}1.1        |m = x \preceq y \iff x \circ z \preceq y \circ z }} {{axiom|n = \operatorname{OG}1.2        |m = x \preceq y \iff z \circ x \preceq z \circ y }} {{axiom|n = \operatorname{OG}1.1'        |m = x \prec y \iff x \circ z \prec y \circ z }} {{axiom|n = \operatorname{OG}1.2'        |m = x \prec y \iff z \circ x \prec z \circ y }} {{end-axiom}}


Relations Compatible with Operation Form Complete Distributive Lattice

Let $\left({S, \circ}\right)$ be an algebraic structure. Let $C$ be the set of relations on $S$ compatible with $\circ$. Then $\left({C, \cap,\cup,\subseteq}\right)$ is a complete distributive lattice.


Relations Compatible with Group Form Complete Boolean Algebra

Let $\left({S,\circ}\right)$ be a group. Let $C$ be the set of relations on $S$ which are compatible with $\circ$. Then $\left({C, \cap, \cup, \subseteq}\right)$ is a complete Boolean lattice.


Duality Principle (Order Theory)

Let $\Sigma$ be a statement about ordered sets (whether in natural or a formal language). Let $\Sigma^*$ be the dual statement of $\Sigma$.


Hypothetical Syllogism/Formulation 3

:$\vdash \paren {\paren {p \implies q} \land \paren {q \implies r} } \implies \paren {p \implies r}$


Duality Principle (Order Theory)/Global Duality

The following are equivalent: :$(1): \quad \Sigma$ is true for all ordered sets :$(2): \quad \Sigma^*$ is true for all ordered sets


Duality Principle (Order Theory)/Local Duality

Let $\struct {S, \preceq}$ be an ordered set, and let $\struct {S, \succeq}$ be its dual. Then the following are equivalent: :$(1): \quad \Sigma$ is true for $\struct {S, \preceq}$ :$(2): \quad \Sigma^*$ is true for $\struct {S, \succeq}$


Diagonal Complement Relation Compatible with Group Operation

Let $\left({G, \circ}\right)$ be a group. Let $\Delta_G$ be the diagonal relation on $G$. Then $\Delta_G^c = \complement_{G \times G} \Delta_G$ is a relation compatible with $\circ$. {{explain|Find another way to write the above, it's ugly}} In other words, $\ne$ is a relation compatible with $\circ$.


Reflexive Closure of Relation Compatible with Operation is Compatible

Let $\left({S, \circ}\right)$ be a magma. Let $\mathcal R$ be a relation on $S$ which is compatible with $\circ$. Let $\mathcal R^=$ be the reflexive closure of $\prec$. That is, $\mathcal R^=$ is defined as the union of $\mathcal R$ with the diagonal relation for $S$. Then $\mathcal R^=$ is compatible with $\circ$.


Reflexive Reduction of Relation Compatible with Group Operation is Compatible

Let $\struct {S, \circ}$ be a group. Let $\RR$ be a relation on $S$ which is compatible with $\circ$. Let $\RR^\ne$ be the reflexive reduction of $\RR$. Then $\RR^\ne$ is compatible with $\circ$.


Double Negation/Double Negation Introduction/Sequent Form/Formulation 2

: $\vdash p \implies \neg \neg p$


Dual of Dual Ordering

Let $\left({S, \preceq}\right)$ be an ordered set. Let $\left({S, \succeq}\right)$ be its dual. Then the dual of $\left({S, \succeq}\right)$ is again $\left({S, \preceq}\right)$.


Dual of Dual Statement (Order Theory)

Let $\Sigma$ be a statement about ordered sets. Let $\Sigma^*$ be its dual statement. Then $\Sigma$ is also the dual statement of $\Sigma^*$.


Double Negation/Double Negation Elimination/Sequent Form/Formulation 2

: $\vdash \neg \neg p \implies p$


The rule of '''double negation elimination''' is a valid deduction sequent in propositional logic. === Proof Rule === {{:Double Negation/Double Negation Elimination/Proof Rule}} === Sequent Form === {{:Double Negation/Double Negation Elimination/Sequent Form}}


Double Negation/Formulation 1

:$p \dashv \vdash \neg \neg p$


Double Negation/Formulation 1/Proof 2

:$p \dashv \vdash \neg \neg p$


Rule of Transposition/Formulation 1

A statement and its contrapositive have the same truth value: :$p \implies q \dashv \vdash \neg q \implies \neg p$ Its abbreviation in a tableau proof is $\textrm {TP}$.


Rule of Transposition/Formulation 1/Proof 1

:$p \implies q \dashv \vdash \neg q \implies \neg p$


Rule of Transposition/Formulation 1/Proof 2


Positive Infinity is Maximal

Let $\struct {\overline \R, \le}$ be the extended real numbers with the usual ordering. Then $+\infty$ is a maximal element of $\overline \R$.


Negative Infinity is Minimal

Let $\left({\overline \R, \le}\right)$ be the extended real numbers with the usual ordering. Then $-\infty$ is a minimal element of $\overline \R$.


Inversion Mapping Reverses Ordering in Ordered Group

{{begin-eqn}} {{eqn | n = \text{OG} 3       | l = x \preceq y       | o = \iff       | r = y^{-1} \preceq x^{-1} }} {{eqn | n = \text{OG}3'       | l = x \prec y       | o = \iff       | r = y^{-1} \prec x^{-1} }} {{end-eqn}}


Operating on Transitive Relationships Compatible with Operation

Let $\left({S, \circ}\right)$ be a magma. Let $\mathcal R$ be a transitive relation on $S$ which is compatible with $\circ$. Let $\mathcal R^=$ be the reflexive closure of $\mathcal R$. Let $x, y, z, w \in S$. Then the following implications hold: $(1)\quad$ If $x \mathrel{\mathcal R} y$ and $z \mathrel{\mathcal R} w$, then $x \circ z \mathrel{\mathcal R} y \circ w$. $(2)\quad$ If $x \mathrel{\mathcal R} y$ and $z \mathrel{\mathcal R^=} w$, then $x \circ z \mathrel{\mathcal R} y \circ w$. $(3)\quad$ If $x \mathrel{\mathcal R^=} y$ and $z \mathrel{\mathcal R} w$, then $x \circ z \mathrel{\mathcal R} y \circ w$. $(4)\quad$ If $x \mathrel{\mathcal R^=} y$ and $z \mathrel{\mathcal R^=} w$, then $x \circ z \mathrel{\mathcal R^=} y \circ w$.


Rule of Transposition/Formulation 2/Proof 1

: $\vdash \left({p \implies q}\right) \iff \left({\neg q \implies \neg p}\right)$


Rule of Transposition/Formulation 2/Forward Implication/Proof

:$\vdash \paren {p \implies q} \implies \paren {\neg q \implies \neg p}$


Rule of Transposition/Formulation 2/Reverse Implication/Proof

: $\vdash \left({\neg q \implies \neg p}\right) \implies \left({p \implies q}\right)$


Complex Modulus is Norm

The complex modulus is a norm on the set of complex numbers $\C$.


Set Difference of Relations Compatible with Group Operation is Compatible

Let $\struct {G, \circ}$ be a group. Let $\RR, \QQ$ be relations on $G$ which are compatible with $\circ$. Then the difference $\RR \setminus \QQ$ is compatible with $\circ$.


Transitive Relation Compatible with Semigroup Operation Relates Powers of Related Elements

Let $\left({S, \circ}\right)$ be a semigroup. Let $\mathcal R$ be a transitive relation on $S$ which is compatible with $\circ$. Let $x, y \in S$ such that $x \mathrel{\mathcal R} y$. Let $n \in \N_{>0}$ be a strictly positive integer. Then: : $x^n \mathrel{\mathcal R} y^n$ where $x^n$ is the $n$th power of $x$.


Reflexive Closure of Transitive Antisymmetric Relation is Ordering

Let $S$ be a set. Let $\prec$ be a transitive, antisymmetric relation on $S$. Let $\preceq$ be the reflexive closure of $\prec$. Then $\preceq$ is an ordering on $S$.


Reflexive Closure of Transitive Relation is Transitive

Let $S$ be a set. Let $\mathcal R$ be a transitive relation. Let $\mathcal R^=$ be the reflexive closure of $\mathcal R$. Then $\mathcal R^=$ is also transitive.


Positive Infinity is Greatest Element

Let $\left({\overline \R, \le}\right)$ be the extended real numbers with their usual ordering. Then $+\infty$ is the greatest element of $\overline \R$.


Negative Infinity is Smallest Element

Let $\left({\overline \R, \le}\right)$ be the extended real numbers with their usual ordering. Then $-\infty$ is the smallest element of $\overline \R$.


Properties of Relation Compatible with Group Operation/CRG1

:$x \mathrel {\mathcal R} y \iff x \circ z \mathrel {\mathcal R} y \circ z$ :$x \mathrel {\mathcal R} y \iff z \circ x \mathrel {\mathcal R} z \circ y$


Properties of Relation Compatible with Group Operation/CRG2

:$(1): \quad x \mathrel {\mathcal R} y \iff e \mathrel {\mathcal R} y \circ x^{-1}$ :$(2): \quad x \mathrel {\mathcal R} y \iff e \mathrel {\mathcal R} x^{-1} \circ y$ :$(3): \quad x \mathrel {\mathcal R} y \iff x \circ y^{-1} \mathrel {\mathcal R} e$ :$(4): \quad x \mathrel {\mathcal R} y \iff y^{-1} \circ x \mathrel {\mathcal R} e$


Properties of Relation Compatible with Group Operation/CRG4

:$x \mathrel{\mathcal R} e \iff e \mathrel{\mathcal R} x^{-1}$ :$e \mathrel{\mathcal R} x \iff x^{-1} \mathrel{\mathcal R} e$


Dual Pairs (Order Theory)

Let $\left({S, \preceq}\right)$ be an ordered set. Let $a, b \in S$, and let $T \subseteq S$. Then the following phrases about, and concepts pertaining to $\left({S, \preceq}\right)$ are dual to one another: :{| style="text-align:center" | $b \preceq a$ | width = "20px" | | $a \preceq b$ |- | $a$ succeeds $b$ | | $a$ precedes $b$ |- | $a$ strictly succeeds $b$ | | $a$ strictly precedes $b$ |- | $a$ is an upper bound for $T$ | | $a$ is a lower bound for $T$ |- | $a$ is a supremum for $T$ | | $a$ is an infimum for $T$ |- | $a$ is a maximal element of $T$ | | $a$ is a minimal element of $T$ |- | $a$ is the greatest element | | $a$ is the smallest element |- | the weak lower closure $a^\preceq$ of $a$ | | the weak upper closure $a^\succeq$ of $a$ |- | the strict lower closure $a^\prec$ of $a$ | | the strict upper closure $a^\succ$ of $a$ |- | the strict lower closure $T^\prec$ of $T$ | | the strict upper closure $T^\succ$ of $T$ |- | the join $a \vee b$ of $a$ and $b$ | | the meet $a \wedge b$ of $a$ and $b$ |- | $T$ is a lower set in $S$ | | $T$ is an upper set in $S$ |}


Conditional is not Left Self-Distributive/Formulation 1

While this holds: :$\paren {p \implies q} \implies r \vdash \paren {p \implies r} \implies \paren {q \implies r}$ its converse does not: :$\paren {p \implies r} \implies \paren {q \implies r} \not \vdash \paren {p \implies q} \implies r$


Cauchy-Riemann Equations

Let $D \subseteq \C$ be an open subset of the set of complex numbers $\C$. Let $f: D \to \C$ be a complex function on $D$. Let $u, v: \set {\tuple {x, y} \in \R^2: x + i y = z \in D} \to \R$ be two real-valued functions defined as: :$\map u {x, y} = \map \Re {\map f z}$ :$\map v {x, y} = \map \Im {\map f z}$ where: :$\map \Re {\map f z}$ denotes the real part of $\map f z$ :$\map \Im {\map f z}$ denotes the imaginary part of $\map f z$. Then $f$ is complex-differentiable in $D$ {{iff}}: :$u$ and $v$ are differentiable in their entire domain and: :The following two equations, known as the '''Cauchy-Riemann equations''', hold for the continuous partial derivatives of $u$ and $v$: :$(1): \quad \dfrac {\partial u} {\partial x} = \dfrac {\partial v} {\partial y}$ :$(2): \quad \dfrac {\partial u} {\partial y} = -\dfrac {\partial v} {\partial x}$ If the conditions are true, then for all $z \in D$: :$\map {f'} z = \map {\dfrac {\partial f} {\partial x} } z = -i \map {\dfrac {\partial f} {\partial y} } z$


Cauchy-Riemann Equations/Necessary Condition

Let $D \subseteq \C$ be an open subset of the set of complex numbers $\C$. Let $f: D \to \C$ be a complex function on $D$. Let $u, v: \left\{ {\left({x, y}\right) \in \R^2: x + i y = z \in D }\right\} \to \R$ be two real-valued functions defined as: :$u \left({x, y}\right) = \operatorname{Re} \left({f \left({z}\right) }\right)$ :$v \left({x, y}\right) = \operatorname{Im} \left({f \left({z}\right) }\right)$ where: : $\operatorname{Re} \left({f \left({z}\right)}\right)$ denotes the real part of $f \left({z}\right)$ : $\operatorname{Im} \left({f \left({z}\right)}\right) $ denotes the imaginary part of $f \left({z}\right)$. Then $f$ is complex-differentiable in $D$ {{iff}}: : $u$ and $v$ are differentiable in their entire domain and: : The following two equations, known as the '''Cauchy-Riemann equations''', hold for the partial derivatives of $u$ and $v$: :$(1): \quad \dfrac{\partial u}{\partial x} = \dfrac{\partial v}{\partial y}$ :$(2): \quad \dfrac{\partial u}{\partial y} = - \dfrac{\partial v}{\partial x}$ If the conditions are true, then for all $z \in D$: :$f' \left({z}\right) = \dfrac{\partial f}{\partial x} \left({z}\right) = -i \dfrac{\partial f}{\partial y} \left({z}\right)$


Cauchy-Riemann Equations/Sufficient Condition

Let $D \subseteq \C$ be an open subset of the set of complex numbers $\C$. Let $f: D \to \C$ be a complex function on $D$. Let $u, v: \left\{ {\left({x, y}\right) \in \R^2: x + i y = z \in D }\right\} \to \R$ be two real-valued functions defined as: :$u \left({x, y}\right) = \operatorname{Re} \left({f \left({z}\right) }\right)$ :$v \left({x, y}\right) = \operatorname{Im} \left({f \left({z}\right) }\right)$ where: : $\operatorname{Re} \left({f \left({z}\right)}\right)$ denotes the real part of $f \left({z}\right)$ : $\operatorname{Im} \left({f \left({z}\right)}\right) $ denotes the imaginary part of $f \left({z}\right)$. Let: :$u$ and $v$ be differentiable in their entire domain and: :The following two equations, known as the '''Cauchy-Riemann equations''', hold for the continuous partial derivatives of $u$ and $v$: ::$(1): \quad \dfrac{\partial u}{\partial x} = \dfrac{\partial v}{\partial y}$ ::$(2): \quad \dfrac{\partial u}{\partial y} = - \dfrac{\partial v}{\partial x}$ Then: :$f$ is complex-differentiable in $D$ and: :for all $z \in D$: ::$f' \left({z}\right) = \dfrac {\partial f} {\partial x} \left({z}\right) = -i \dfrac{\partial f}{\partial y} \left({z}\right)$


Self-Distributive Law for Conditional/Formulation 1/Proof 2

: $p \implies \left({q \implies r}\right) \dashv \vdash \left({p \implies q}\right) \implies \left({p \implies r}\right)$


Cauchy-Riemann Equations/Expression of Derivative

Let $D \subseteq \C$ be an open subset of the set of complex numbers $\C$. Let $f: D \to \C$ be a complex function on $D$. Let $u, v: \left\{ {\left({x, y}\right) \in \R^2: x + i y = z \in D }\right\} \to \R$ be two real-valued functions defined as: :$u \left({x, y}\right) = \operatorname{Re} \left({f \left({z}\right) }\right)$ :$v \left({x, y}\right) = \operatorname{Im} \left({f \left({z}\right) }\right)$ where: : $\operatorname{Re} \left({f \left({z}\right)}\right)$ denotes the real part of $f \left({z}\right)$ : $\operatorname{Im} \left({f \left({z}\right)}\right) $ denotes the imaginary part of $f \left({z}\right)$. Then $f$ is complex-differentiable in $D$ {{iff}}: : $u$ and $v$ are differentiable in their entire domain and: : The following two equations, known as the '''Cauchy-Riemann equations''', hold for the partial derivatives of $u$ and $v$: :$(1): \quad \dfrac {\partial u} {\partial x} = \dfrac {\partial v} {\partial y}$ :$(2): \quad \dfrac {\partial u} {\partial y} = - \dfrac {\partial v} {\partial x}$ If the conditions are true, then for all $z \in D$: :$f' \left({z}\right) = \dfrac{\partial f}{\partial x} \left({z}\right) = -i \dfrac{\partial f}{\partial y} \left({z}\right)$


Self-Distributive Law for Conditional/Forward Implication/Formulation 1/Proof

: $p \implies \left({q \implies r}\right) \vdash \left({p \implies q}\right) \implies \left({p \implies r}\right)$


Succeed is Dual to Precede

Let $\left({S, \preceq}\right)$ be an ordered set. Let $a, b \in S$. The following are dual statements: :$a$ succeeds $b$ :$a$ precedes $b$


Strictly Succeed is Dual to Strictly Precede

Let $\struct {S, \preceq}$ be an ordered set. Let $a, b \in S$. The following are dual statements: :$a$ strictly succeeds $b$ :$a$ strictly precedes $b$


Upper Bound is Dual to Lower Bound

Let $\left({S, \preceq}\right)$ be an ordered set. Let $a \in S$ and $T \subseteq S$. The following are dual statements: :$a$ is an upper bound for $T$ :$a$ is a lower bound for $T$


Supremum is Dual to Infimum

Let $\left({S, \preceq}\right)$ be an ordered set. Let $a \in S$ and $T \subseteq S$. The following are dual statements: :$a$ is a supremum for $T$ :$a$ is an infimum for $T$


Maximal Element is Dual to Minimal Element

Let $\struct {S, \preceq}$ be an ordered set. Let $T \subseteq S$, and $a \in T$. The following are dual statements: :$a$ is a maximal element of $T$ :$a$ is a minimal element of $T$


Greatest Element is Dual to Smallest Element

Let $\struct {S, \preceq}$ be an ordered set. Let $a \in S$. The following are dual statements: :$a$ is the greatest element of $S$ :$a$ is the smallest element of $S$


Lower Closure is Dual to Upper Closure

Let $\left({S, \preccurlyeq}\right)$ be an ordered set. Let $a, b \in S$. Let $T \subseteq S$ The following are pairs of dual statements: :$b \in a^\preccurlyeq$, the lower closure of $a$ :$b \in a^\succcurlyeq$, the upper closure of $a$ :$b \in T^\preccurlyeq$, the lower closure of $T$ :$b \in T^\succcurlyeq$, the upper closure of $T$


Strict Lower Closure is Dual to Strict Upper Closure

Let $\left({S, \preceq}\right)$ be an ordered set. Let $a, b \in S$. The following are dual statements: :$b \in a^\prec$, the strict lower closure of $a$ :$b \in a^\succ$, the strict upper closure of $a$


Join is Dual to Meet

Let $\left({S, \preceq}\right)$ be an ordered set. Let $a, b, c \in S$. The following are dual statements: :$c = a \vee b$, the join of $a$ and $b$ :$c = a \wedge b$ the meet of $a$ and $b$


Self-Distributive Law for Conditional/Forward Implication/Formulation 1


Self-Distributive Law for Conditional/Reverse Implication/Formulation 1

:$\paren {p \implies q} \implies \paren {p \implies r} \vdash p \implies \paren {q \implies r}$


Square Matrix with Duplicate Columns has Zero Determinant

If two columns of a square matrix over a commutative ring $\struct {R, +, \circ}$ are identical, then its determinant is zero.


Properties of Relation Compatible with Group Operation/CRG3

:$x \mathrel{\mathcal R} y \iff y^{-1} \mathrel{\mathcal R} x^{-1}$


Ordering Induced by Join Semilattice

Let $\struct {S, \vee, \preceq}$ be a join semilattice. By Join Semilattice is Semilattice, $\struct {S, \vee}$ is a semilattice. By Semilattice Induces Ordering, $\struct {S, \vee}$ induces an ordering $\preceq'$ on $S$, by: :$a \preceq' b$ {{iff}} $a \vee b = b$ for all $a, b \in S$. The ordering $\preceq'$ coincides with the original ordering $\preceq$.


Combination Theorem for Complex Derivatives

Let $D$ be an open subset of the set of complex numbers $\C$. Let $f, g: D \to \C$ be complex-differentiable functions on $D$ Let $z \in D$. Let $w, c \in \C$ be arbitrary complex numbers. Then the following results hold:


Combination Theorem for Complex Derivatives/Sum Rule

:$\left({f + g}\right)' \left({z}\right) = f' \left({z}\right) + g' \left({z}\right)$


Combination Theorem for Complex Derivatives/Multiple Rule

:$\left({w f}\right)' \left({z}\right) = w f' \left({z}\right)$


Join Absorbs Meet

Let $\struct {S, \wedge, \preceq}$ be a meet semilattice. Let $\vee$ denote join. Then $\vee$ absorbs $\wedge$. That is, for all $a, b \in S$: :$a \vee \paren {a \wedge b} = a$


Meet Absorbs Join

Let $\left({S, \vee, \preceq}\right)$ be a join semilattice. Let $\wedge$ denote meet. Then $\wedge$ absorbs $\vee$. That is, for all $a, b \in S$: :$a \wedge \left({a \vee b}\right) = a$


Rule of Exportation/Reverse Implication/Formulation 2/Proof 1

:$\vdash \paren {p \implies \paren {q \implies r} } \implies \paren {\paren {p \land q} \implies r}$


Rule of Exportation/Forward Implication/Formulation 1

: $\left ({p \land q}\right) \implies r \vdash p \implies \left ({q \implies r}\right)$


Rule of Exportation/Reverse Implication/Formulation 1

:$p \implies \paren {q \implies r} \vdash \paren {p \land q} \implies r$


Rule of Commutation/Conjunction/Formulation 1/Proof 2

:$p \land q \dashv \vdash q \land p$


Rule of Commutation/Conjunction/Formulation 1

:$p \land q \dashv \vdash q \land p$


Rule of Commutation/Disjunction/Formulation 1

:$p \lor q \dashv \vdash q \lor p$


Rule of Commutation/Conjunction/Formulation 2

:$\vdash \paren {p \land q} \iff \paren {q \land p}$


Rule of Commutation/Disjunction/Formulation 2

:$\vdash \paren {p \lor q} \iff \paren {q \lor p}$


Combination Theorem for Complex Derivatives/Product Rule

:$\left({f g}\right)' \left({z}\right) = f' \left({z}\right) g \left({z}\right) + f \left({z}\right) g' \left({z}\right)$


Dedekind Completion is Unique up to Isomorphism

Let $S$ be an ordered set. Let $\struct {X, f}$ and $\struct {Y, g}$ be Dedekind completions of $S$. Then there exists a unique order isomorphism $\phi: X \to Y$ such that $\phi \circ f = g$.


Combination Theorem for Complex Derivatives/Combined Sum Rule

Let $\dfrac \d {\d z} \left({c f + w g}\right)$ denote the derivative of $c f + w g$. Then: :$\dfrac \d {\d z} \left({c f + w g}\right) \left({z}\right) = c \dfrac \d {\d z} f \left({z}\right) + w \dfrac \d {\d z} g \left({z}\right)$


Reflexive Reduction of Transitive Antisymmetric Relation is Strict Ordering

Let $S$ be a set. Let $\mathcal R$ be a transitive, antisymmetric relation on $S$. Let $\mathcal R^\ne$ denote the reflexive reduction of $\mathcal R$. Then $\mathcal R^\ne$ is a strict ordering.


Reflexive Reduction of Antisymmetric Relation is Asymmetric

Let $S$ be a set. Let $\mathcal R$ be an antisymmetric relation on $S$. Let $\mathcal R^\ne$ be the reflexive reduction of $\mathcal R$. Then $\mathcal R^\ne$ is asymmetric.


Properties of Ordered Group

Let $\left({G, \circ, \preceq}\right)$ be an ordered group with identity $e$. Let $x,y,z,w \in G$. Then the following hold:


Reductio ad Absurdum/Variant 1

: $\neg p \implies \bot \vdash p$


Reductio ad Absurdum/Variant 2

:$\neg p \implies \paren {q \land \neg q} \vdash p$


Proof by Contradiction/Variant 1

:$\left({p \vdash \left({q \land \neg q}\right)}\right) \vdash \neg p$


Proof by Contradiction/Variant 2

==== Formulation 1 ==== {{:Proof by Contradiction/Variant 2/Formulation 1}} ==== Formulation 2 ==== {{:Proof by Contradiction/Variant 2/Formulation 2}}


Proof by Contradiction/Variant 3

==== Formulation 1 ==== {{:Proof by Contradiction/Variant 3/Formulation 1}} ==== Formulation 2 ==== {{:Proof by Contradiction/Variant 3/Formulation 2}}


Equivalence of Definitions of Topological Group

Let $\left({G, \odot}\right)$ be a group. On its underlying set $G$, let $\left({G, \tau}\right)$ be a topological space. {{TFAE|def = Topological Group}}


Implication is Left Distributive over Conjunction/Forward Implication/Formulation 1/Proof

: $p \implies \left({q \land r}\right) \vdash \left({p \implies q}\right) \land \left({p \implies r}\right)$


Implication is Left Distributive over Conjunction/Reverse Implication/Formulation 1/Proof

: $\left({p \implies q}\right) \land \left({p \implies r}\right) \vdash p \implies \left({q \land r}\right)$


Implication is Left Distributive over Conjunction/Forward Implication/Formulation 1


Implication is Left Distributive over Conjunction/Reverse Implication/Formulation 1


Implication is Left Distributive over Conjunction/Formulation 1/Proof 1

: $p \implies \left({q \land r}\right) \dashv \vdash \left({p \implies q}\right) \land \left({p \implies r}\right)$


Implication is Left Distributive over Conjunction/Forward Implication/Formulation 2/Proof

: $\vdash \left({p \implies \left({q \land r}\right)}\right) \implies \left({\left({p \implies q}\right) \land \left({p \implies r}\right)}\right)$


Implication is Left Distributive over Conjunction/Reverse Implication/Formulation 2/Proof

: $\vdash \left({\left({p \implies q}\right) \land \left({p \implies r}\right)}\right) \implies \left({p \implies \left({q \land r}\right)}\right)$


Implication is Left Distributive over Conjunction/Formulation 2/Proof 1

: $\vdash \left({p \implies \left({q \land r}\right)}\right) \iff \left({\left({p \implies q}\right) \land \left({p \implies r}\right)}\right)$


Criterion for Ring with Unity to be Topological Ring

Let $\struct {R, +, \circ}$ be a ring with unity. Let $\tau$ be a topology over $R$. Suppose that $+$ and $\circ$ are $\tau$-continuous mappings. Then $\struct {R, +, \circ, \tau}$ is a topological ring.


Union of Relations is Relation

Let $S$ and $T$ be sets. Let $\mathcal F$ be a family of relations from $S$ to $T$. Let $\displaystyle \mathcal R = \bigcup \mathcal F$, the union of all the elements of $\mathcal F$.   Then $\mathcal R$ is a relation from $S$ to $T$. {{expand|Binary case}}


Combination Theorem for Complex Derivatives/Quotient Rule

For all $z \in D$ with $\map g z \ne 0$: :$\map {\paren {\dfrac f g}'} z = \dfrac {\map {f'} z \map g z - \map f z \map {g'} z} {\paren {\map g z}^2}$


Complex-Differentiable Function is Continuous

Let $f: D \to \C$ be a complex function, where $D \subseteq \C$ is an open set. Suppose that $f$ is complex-differentiable at $z \in D$. Then $f$ is continuous at $z$.


Proof by Cases/Formulation 1/Forward Implication/Proof 1

: $\left({p \implies r}\right) \land \left({q \implies r}\right) \vdash \left({p \lor q}\right) \implies r$


Closed Ball in Euclidean Space is Compact

Let $x \in \R_n$ be a point in the Euclidean space $\R^n$. Let $\epsilon \in \R_{>0}$. Then the closed $\epsilon$-ball $\map {B_\epsilon^-} x$ is compact.


Isometric Image of Cauchy Sequence is Cauchy Sequence

Let $\struct {S_1, d_1}$ and $\struct {S_2, d_2}$ be metric spaces. Let $f: S_1 \to S_2$ be an isometry. Let $\sequence {x_n}$ be a Cauchy sequence in $S_1$. Let $\sequence {y_n} = \sequence {\map f {x_n} }$ be the image of $\sequence {x_n}$ under $f$. Then $\sequence {y_n}$ is a Cauchy sequence.


Union of Subsets is Subset

Let $S_1$, $S_2$, and $T$ be sets. Let $S_1$ and $S_2$ both be subsets of $T$. Then: :$S_1 \cup S_2 \subseteq T$ That is: :$\paren {S_1 \subseteq T} \land \paren {S_2 \subseteq T} \implies \paren {S_1 \cup S_2} \subseteq T$


Proof by Cases/Formulation 1

: $\left({p \implies r}\right) \land \left({q \implies r}\right) \dashv \vdash \left({p \lor q}\right) \implies r$


Distance between Closed Sets in Euclidean Space

Let $S, T \subseteq \R^n$ be closed, non-empty subsets of the real Euclidean space $R^n$. Suppose that $S$ is bounded, and $S$ and $T$ are disjoint. Then there exists $x \in S$ and $y \in T$ such that: :$\map d {x, y} = \map d {S, T} > 0$ where: :$d$ denotes the Euclidean metric :$\map d {S, T}$ is the distance between $S$ and $T$.


Complement in Distributive Lattice is Unique

Let $\left({S, \vee, \wedge, \preceq}\right)$ be a bounded distributive lattice. Then every $a \in S$ admits at most one complement.


Equivalence of Definitions of Top

{{TFAE|def = Top (Lattice Theory)|view = $\top$ (top)|context = Lattice Theory}} Let $\left({S, \vee, \wedge, \preceq}\right)$ be a lattice.


Equivalence of Definitions of Bottom

Let $\left({S, \vee, \wedge, \preceq}\right)$ be a lattice. Let $\bot$ be a bottom of $\left({S, \vee, \wedge, \preceq}\right)$. {{TFAE|def = Bottom (Lattice Theory)|view = Bottom|context = Lattice Theory}}


Top is Unique

Let $\struct {S, \vee, \wedge, \preceq}$ be a lattice. Then $S$ has at most one top.


Isometry is Homeomorphism of Induced Topologies

Let $\struct {S_1, d_1}$ and $\struct {S_2, d_2}$ be metric spaces or pseudometric spaces. Let $f: S_1 \to S_2$ be an isometry from $\struct {S_1, d_1}$ to $\struct {S_2, d_2}$. Let $\tau_1$ and $\tau_2$ be the topologies induced on $S_1$ and $S_2$ by the metrics $d_1$ and $d_2$, respectively. Then $f$ is a homeomorphism from $\struct {S_1, \tau_1}$ to $\struct {S_2, \tau_2}$.


Equivalence of Definitions of Distributive Lattice

{{TFAE|def = Distributive Lattice}} Let $\left({S, \vee, \wedge, \preceq}\right)$ be a lattice. {{begin-axiom}} {{axiom | n = 1         | q = \forall x, y, z \in S         | m = x \wedge \left({y \vee z}\right) = \left({x \wedge y}\right) \vee \left({x \wedge z}\right) }} {{axiom | n = 1'         | q = \forall x, y, z \in S         | m = \left({x \vee y}\right) \wedge z = \left({x \wedge z}\right) \vee \left({y \wedge z}\right) }} {{axiom | n = 2         | q = \forall x, y, z \in S         | m = x \vee \left({y \wedge z}\right) = \left({x \vee y}\right) \wedge \left({x \vee z}\right) }} {{axiom | n = 2'         | q = \forall x, y, z \in S         | m = \left({x \wedge y}\right) \vee z = \left({x \vee z}\right) \wedge \left({x \vee y}\right) }} {{end-axiom}}


Distance-Preserving Surjection is Isometry of Metric Spaces

Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be metric spaces. Let $\phi: M_1 \to M_2$ be a surjective distance-preserving mapping. That is: :$\forall a, b \in M_1: d_1 \tuple {a, b} = d_2 \tuple {\map \phi a, \map \phi b}$ Then $\phi$ is an isometry.


Existence of Positive Root of Positive Real Number

Let $x \in \R$ be a real number such that $x > 0$. Let $n \in \Z$ be an integer such that $n \ne 0$. Then there exists a $y \in \R: y \ge 0$ such that $y^n = x$.


Proof by Cases/Formulation 2

:$\vdash \paren {\paren {p \implies r} \land \paren {q \implies r} } \iff \paren {\paren {p \lor q} \implies r}$


Proof by Cases/Formulation 2/Forward Implication

: $\vdash \left({\left({p \implies r}\right) \land \left({q \implies r}\right)}\right) \implies \left({\left({p \lor q}\right) \implies r}\right)$


Proof by Cases/Formulation 2/Reverse Implication

: $\vdash \left({\left({p \lor q}\right) \implies r}\right) \implies \left({\left({p \implies r}\right) \land \left({q \implies r}\right)}\right)$


Set is Subset of Union/Set of Sets

Let $\mathbb S$ be a set of sets. Then: : $\displaystyle \forall T \in \mathbb S: T \subseteq \bigcup \mathbb S$


Proof by Cases/Formulation 3

: $\vdash \left({\left({p \lor q}\right) \land \left({p \implies r}\right) \land \left({q \implies r}\right)}\right) \implies r$


Properties of Relation Compatible with Group Operation

Let $\left({G,\circ}\right)$ be a group with identity element $e$. Let $\mathcal R$ be a endorelation on $G$ which is compatible with $\circ$. Let $x,y,z \in G$. Then the following hold:


Existence of Positive Root of Positive Real Number/Positive Exponent

Let $x \in \R$ be a real number such that $x > 0$. Let $n \in \Z$ be an integer such that $n > 0$. Then there exists a $y \in \R: y \ge 0$ such that $y^n = x$.


Existence of Positive Root of Positive Real Number/Negative Exponent

Let $x \in \R$ be a real number such that $x > 0$. Let $n \in \Z$ be an integer such that $n < 0$. Then there exists a $y \in \R: y \ge 0$ such that $y^n = x$.


Power of Ring Negative

Let $\struct {R, +, \circ}$ be a ring. Let $n \in \N_{>0}$ be a strictly positive integer. Let $x \in R$. Then: :If $n$ is even: :::$\map {\circ^n} {-x} = \map {\circ^n} x$ :If $n$ is odd: :::$\map {\circ^n} {-x} = -\map {\circ^n} x$


Paving Lemma

Let $S$ be an open subset of the Euclidean space $\R^m$ or the set of complex numbers $\C$. Let $\gamma: \closedint a b \to S$ be a path in $S$. Then there exists $K \in \R_{>0}$ such that: :For all $\epsilon \in \openint 0 K$, there exists a normal subdivision $\set {x_0, x_1, \ldots, x_{n-1}, x_n}$ of the closed interval $\closedint a b$ such that: ::$\displaystyle \bigcup_{i \mathop = 0}^n \map {B_\epsilon} {\map \gamma {x_i} } \subseteq S$ :and for all $i \in \set {0, 1, \ldots, n - 1}$: ::$\map \gamma {\closedint {x_i} {x_{i + 1} } } \subseteq \map {B_\epsilon} {\map \gamma {x_i} }$ Here, $\map {B_\epsilon} {\map \gamma {x_i} }$ denotes the open ball of $\map \gamma {x_i}$ with radius $\epsilon$.


Proof by Cases/Formulation 1/Reverse Implication

: $\left({p \lor q}\right) \implies r \vdash \left({p \implies r}\right) \land \left({q \implies r}\right)$


Praeclarum Theorema/Formulation 1/Proof 2

:$\paren {p \implies q} \land \paren {r \implies s} \vdash \paren {p \land r} \implies \paren {q \land s}$


Praeclarum Theorema/Formulation 1

: $\left({p \implies q}\right) \land \left({r \implies s}\right) \vdash \left({p \land r}\right) \implies \left({q \land s}\right)$


Praeclarum Theorema/Formulation 2

: $\vdash \left({\left({p \implies q}\right) \land \left({r \implies s}\right)}\right) \implies \left({\left({p \land r}\right) \implies \left({q \land s}\right)}\right)$


Strictly Positive Integer Power Function Strictly Succeeds Each Element

Let $\struct {R, +, \circ, \le}$ be an ordered ring with unity. Let $\struct {R, \le}$ be a directed set with no upper bound. Let $n \in \N_{>0}$. Let $f: R \to R$ be defined by: :$\forall x \in R: \map f x = \circ^n x$ Then the image of $f$ has elements strictly succeeding each elements of $R$.


Constructive Dilemma/Formulation 1/Proof 2

: $p \implies q, r \implies s \vdash p \lor r \implies q \lor s$


Constructive Dilemma/Formulation 1


Constructive Dilemma/Formulation 2

: $\vdash \paren {\paren {p \lor r} \land \paren {p \implies q} \land \paren {r \implies s} } \implies \paren {q \lor s}$


Clavius's Law/Formulation 1

:$\neg p \implies p \vdash p$


Equivalent Matrices may not be Similar

If two square matrices of order $n > 1$ over a ring with unity $R$ are equivalent, they are not necessarily similar.


Odd Power Function is Strictly Increasing/General Result

Let $\struct {R, +, \circ, \le}$ be a totally ordered ring. Let $n$ be an odd positive integer. Let $f: R \to R$ be the mapping defined by: :$\map f x = \map {\circ^n} x$ Then $f$ is strictly increasing on $R$.


Number of Bijective Restrictions

Let $f: S \to T$ be a surjection. Let $B$ be the set of all bijective restrictions of $f$. Then the cardinality of $B$ is: :$\displaystyle \card {\prod_{i \mathop \in I} \family {S / \mathcal R_f}_i}$ where $S / \mathcal R_f$ denotes the quotient set of the induced equivalence of $f$ indexed by $I$.


Number of Injective Restrictions

Let $f: S \to T$ be a mapping. Let $Q$ be the set of all injective restrictions of $f$. Then the cardinality of $Q$ is: :$\displaystyle \card {\prod_{i \mathop \in I} \prod_{j \mathop \in J_i} \family {\family {\powerset {S / \mathcal R_f} }_i}_j}$ where: :$\mathcal P$ denotes power set :$S / \mathcal R_f$ denotes quotient set of the induced equivalence of $f$.


Box Topology on Finite Product Space is Tychonoff Topology

Let $n \in \N$. For all $k \in \set {1, \ldots, n}$, let $T_k = \struct {X_k, \tau_k}$ be topological spaces. Let $\displaystyle X = \prod_{k \mathop = 1}^n X_k$ be the cartesian product of $X_1, \ldots, X_n$. Then the box topology and the Tychonoff topology on $X$ are identical.


Strictly Positive Integer Power Function is Unbounded Above/General Case

Let $\struct {R, +, \circ, \le}$ be a totally ordered ring with unity. Suppose that $R$ has no upper bound. Let $n \in \N_{>0}$. Let $f: R \to R$ be defined by: :$\map f x = \circ^n x$ Then the image of $f$ is unbounded above in $R$.


Strictly Positive Integer Power Function is Unbounded Above

Let $\R$ be the real numbers with the usual ordering. Let $n \in \N_{>0}$. Let $f: \R \to \R$ be defined by: :$\map f x = x^n$ Then $f$ is unbounded above.


Ring Product preserves Inequalities on Positive Elements

Let $\struct {R, +, \circ, \le}$ be an ordered ring. Let $x, y, z, w \in R$. Let $0 < x < y$ and $0 < z < w$. Then: :$0 < z \circ x < w \circ y$


Power Function is Strictly Increasing on Positive Elements

Let $\struct {R, +, \circ, \le}$ be an ordered ring. Let $x, y \in R$. Let $n \in \N_{>0}$ be a strictly positive integer. Let $0 < x < y$. Then: :$0 < \map {\circ^n} x < \map {\circ^n} y$


Biconditional is Associative/Formulation 1

: $p \iff \paren {q \iff r} \dashv \vdash \paren {p \iff q} \iff r$


Biconditional is Associative/Formulation 2

:$\vdash \paren {p \iff \paren {q \iff r} } \iff \paren {\paren {p \iff q} \iff r}$


Open Ball is Convex Set

Let $V$ be a normed vector space with norm $\left\Vert{\cdot}\right\Vert$ over $\R$ or $\C$. An open ball in the metric induced by $\left\Vert{\cdot}\right\Vert$ is a convex set.


Pseudometric Space is Metric Space iff Kolmogorov

Let $M = \struct {S, d}$ be a pseudometric space. Let $T = \struct {S, \tau}$ be the topological space over $S$ induced by $d$. Then $M$ is a metric space {{iff}} $T$ is a $T_0$ (Kolmogorov} space.


Sequence Converges to Point Relative to Metric iff it Converges Relative to Induced Topology

Let $M = \struct {S, d}$ be a metric space or a pseudometric space. Let $T = \struct {S, \tau}$ be the topological space induced by $d$. Let $\sequence {x_n}$ be a infinite sequence in $S$. Let $l \in S$. Then $\sequence {x_n}$ converges to $l$ relative to $d$ {{iff}} $\sequence {x_n}$ converges to $l$ relative to $\tau$.


Isometry Preserves Sequence Convergence

Let $M_1 = \struct {S_1, d_1}$ and $M_2 = \struct {S_2, d_2}$ both be metric spaces or pseudometric spaces. Let $\phi: S_1 \to S_2$ be an isometry. Let $\sequence {x_n}$ be an infinite sequence in $S_1$. Suppose that $\sequence {x_n}$ converges to a point $p \in S_1$. Then $\sequence {\map \phi {x_n}}$ converges to $\map \phi p$.


Connected Domain is Connected by Staircase Contours

Let $D \subseteq \C$ be an open set. Then $D$ is a connected domain {{iff}}: :for all $z, w \in \C$, there exists a staircase contour in $D$ with start point $z$ and end point $w$.


Dedekind Completeness is Self-Dual

Let $\struct {S, \preceq}$ be an ordered set. Then $\struct {S, \preceq}$ is Dedekind complete {{iff}} every non-empty subset of $S$ that is bounded below admits an infimum in $S$. That is, an ordered set is Dedekind complete {{iff}} its dual is Dedekind complete.


Biconditional is Transitive/Formulation 1

:$p \iff q, q \iff r \vdash p \iff r$


Biconditional is Transitive/Formulation 2

: $\vdash \left({\left({p \iff q}\right) \land \left({q \iff r}\right)}\right) \implies \left({p \iff r}\right)$


Biconditional as Disjunction of Conjunctions/Formulation 1/Reverse Implication

: $\left({p \land q}\right) \lor \left({\neg p \land \neg q}\right) \vdash p \iff q$


Biconditional as Disjunction of Conjunctions/Formulation 1

: $p \iff q \dashv \vdash \paren {p \land q} \lor \paren {\neg p \land \neg q}$


Biconditional as Disjunction of Conjunctions/Formulation 2

:$\vdash \paren {p \iff q} \iff \paren {\paren {p \land q} \lor \paren {\neg p \land \neg q} }$


==== Formulation 1 ==== {{:Biconditional as Disjunction of Conjunctions/Formulation 1}} ==== Formulation 2 ==== {{:Biconditional as Disjunction of Conjunctions/Formulation 2}}


Biconditional Equivalent to Biconditional of Negations/Formulation 1/Reverse Implication

:$\neg p \iff \neg q \vdash p \iff q$


Biconditional Equivalent to Biconditional of Negations/Formulation 1

: $p \iff q \dashv \vdash \neg p \iff \neg q$


Biconditional Equivalent to Biconditional of Negations/Formulation 2

: $\vdash \left({p \iff q}\right) \iff \left({\neg p \iff \neg q}\right)$


==== Formulation 1 ==== {{:Biconditional Equivalent to Biconditional of Negations/Formulation 1}} ==== Formulation 2 ==== {{:Biconditional Equivalent to Biconditional of Negations/Formulation 2}}


Biconditional iff Disjunction implies Conjunction/Formulation 1/Forward Implication

: $p \iff q \vdash \left({p \lor q}\right) \implies \left({p \land q}\right)$


Biconditional iff Disjunction implies Conjunction/Formulation 1/Reverse Implication

: $\left({p \lor q}\right) \implies \left({p \land q}\right) \vdash p \iff q$


Biconditional iff Disjunction implies Conjunction/Formulation 1

:$p \iff q \dashv \vdash \paren {p \lor q} \implies \paren {p \land q}$


Biconditional iff Disjunction implies Conjunction/Formulation 2

: $\vdash \left({p \iff q}\right) \iff \left({\left({p \lor q}\right) \implies \left({p \land q}\right)}\right)$


==== Formulation 1 ==== {{:Biconditional iff Disjunction implies Conjunction/Formulation 1}} ==== Formulation 2 ==== {{:Biconditional iff Disjunction implies Conjunction/Formulation 2}}


Biconditional is Transitive/Formulation 1/Proof 1

: $p \iff q, q \iff r \vdash p \iff r$


Order of Squares in Totally Ordered Ring without Proper Zero Divisors

Let $\struct {R, +, \circ, \le}$ be a totally ordered ring without proper zero divisors whose zero is $0_R$. Let $x, y \in R$ be positive, that is, $0_R \le x, y$. Then $x \le y \iff x \circ x \le y \circ y$. That is, the square mapping is an order embedding of $\struct {R_{\ge 0}, \le}$ into itself. When $R$ is one of the standard sets of numbers $\Z, \Q, \R$, then this translates into: :If $x, y$ are positive, then $x \le y \iff x^2 \le y^2$.


Order of Squares in Ordered Field

Let $\struct {R, +, \circ, \le}$ be an ordered field whose zero is $0_R$ and whose unity is $1_R$. Suppose that $\forall a \in R: 0 < a \implies 0 < a^{-1}$. Let $x, y \in \struct {R, +, \circ, \le}$ such that $0_R \le x, y$. Then $x \le y \iff x \circ x \le y \circ y$. That is, the square function is an order embedding of $\struct {R_{\ge 0}, \le}$ into itself. When $R$ is one of the standard fields of numbers $\Q$ and $\R$, then this translates into: :If $x, y$ are positive then $x \le y \iff x^2 \le y^2$.


Law of Identity/Formulation 1/Proof 1

: $p \vdash p$


Law of Identity/Formulation 1/Proof 2

: $p \vdash p$


Law of Identity/Formulation 2

Every proposition entails itself: :$\vdash p \implies p$


Vandermonde Determinant/Alternative Formulations

The '''Vandermonde determinant of order $n$''' is the determinant defined as follows: :$V_n = \begin{vmatrix}      1 &    x_1 &  x_1^2 & \cdots & x_1^{n - 2} & x_1^{n - 1} \\      1 &    x_2 &  x_2^2 & \cdots & x_2^{n - 2} & x_2^{n - 1} \\ \vdots & \vdots & \vdots & \ddots &      \vdots &      \vdots \\      1 &    x_n &  x_n^2 & \cdots & x_n^{n - 2} & x_n^{n - 1} \end{vmatrix}$ Its value is given by: :$\displaystyle V_n = \prod_{1 \mathop \le i \mathop < j \mathop \le n} \paren {x_j - x_i}$


Neighborhood Condition for Coarser Topology

Let $S$ be a set. Let $\tau_1$ and $\tau_2$ be two topologies on $S$. Suppose that for all $z \in S$ and for all open neighborhoods $N_z$ of $z$ with respect to $\tau_1$, there exists $U \in \tau_2$ such that $U \subseteq N_z$. Then $\tau_1$ is coarser than $\tau_2$.


Absorption Laws (Logic)/Conjunction Absorbs Disjunction/Proof 2

:$p \land \left({p \lor q}\right) \dashv \vdash p$


Absorption Laws (Logic)/Conjunction Absorbs Disjunction

:$p \land \paren {p \lor q} \dashv \vdash p$


Absorption Laws (Logic)/Disjunction Absorbs Conjunction/Proof 2

:$p \lor \left ({p \land q}\right) \dashv \vdash p$


Absorption Laws (Logic)/Disjunction Absorbs Conjunction


Extended Transitivity

Let $S$ be a set. Let $\RR$ be a transitive relation on $S$. Let $\RR^=$ be the reflexive closure of $\RR$. Let $a, b, c \in S$. Then: {{begin-eqn}} {{eqn | n = 1       | l = \paren {a \mathrel \RR b} \land \paren {b \mathrel \RR c}       | o = \implies       | r = a \mathrel \RR c }} {{eqn | n = 2       | l = \paren {a \mathrel \RR b} \land \paren {b \mathrel {\RR^=} c}       | o = \implies       | r = a \mathrel \RR c }} {{eqn | n = 3       | l = \paren {a \mathrel {\RR^=} b} \land \paren {b \mathrel \RR c}       | o = \implies       | r = a \mathrel \RR c }} {{eqn | n = 4       | l = \paren {a \mathrel {\RR^=} b} \land \paren {b \mathrel {\RR^=} c}       | o = \implies       | r = a \mathrel {\RR^=} c }} {{end-eqn}}


Rule of Idempotence/Conjunction

{{:Rule of Idempotence/Conjunction/Formulation 1}}


Rule of Idempotence/Disjunction

{{:Rule of Idempotence/Disjunction/Formulation 1}}


Rule of Idempotence/Disjunction/Formulation 1/Proof

: $p \dashv \vdash p \lor p$


Ordering Cycle implies Equality

Let $\left({S,\preceq}\right)$ be an ordered set. Let $x_1$, $x_2$, and $x_3$ be elements of $S$. Suppose that {{begin-eqn}} {{eqn|l = x_1      |o = \preceq      |r = x_2 }} {{eqn|l = x_2      |o = \preceq      |r = x_3 }} {{eqn|l = x_3      |o = \preceq      |r = x_1 }} {{end-eqn}} Then $x_1 = x_2 = x_3$.


Euclidean Topology is Tychonoff Topology

Let $T_1 = \left({\R, \tau_1}\right)$ be the topological space such that $\tau_1$ is the Euclidean topology on $\R$. Let $T_n = \left({\R^n, \tau_n}\right)$ be the topological space such that $\tau_n$ is the Tychonoff topology on the cartesian product $\displaystyle \R_n = \prod_{i \mathop = 1}^n \R$. Then the Euclidean topology on $\R^n$ and the Tychonoff topology on $\R^n$ are the same.


Modus Ponendo Tollens/Variant/Formulation 1/Proof

:$\neg \paren {p \land q} \dashv \vdash p \implies \neg q$


Modus Ponendo Tollens/Variant/Formulation 1/Forward Implication

: $\neg \left({p \land q}\right) \vdash p \implies \neg q$


Modus Ponendo Tollens/Variant/Formulation 1/Reverse Implication

:$p \implies \neg q \vdash \neg \paren {p \land q}$


Modus Ponendo Tollens/Variant/Formulation 1

: $\neg \left({p \land q}\right) \dashv \vdash p \implies \neg q$


Modus Ponendo Tollens/Variant/Formulation 2

: $\vdash \left({\neg \left({p \land q}\right)}\right) \iff \left({p \implies \neg q}\right)$


Implication Equivalent to Negation of Conjunction with Negative/Formulation 1/Reverse Implication

:$\neg \left({p \land \neg q}\right) \vdash p \implies q$


Implication Equivalent to Negation of Conjunction with Negative/Formulation 1/Forward Implication

:$p \implies q \vdash \neg \left({p \land \neg q}\right)$


Implication Equivalent to Negation of Conjunction with Negative/Formulation 1

:$p \implies q \dashv \vdash \neg \paren {p \land \neg q}$


==== Formulation 1 ==== {{:Implication Equivalent to Negation of Conjunction with Negative/Formulation 1}} ==== Formulation 2 ==== {{:Implication Equivalent to Negation of Conjunction with Negative/Formulation 2}}


Zero Derivative implies Constant Complex Function

Let $D \subseteq \C$ be a connected domain of $\C$. Let $f: D \to \C$ be a complex-differentiable function. For all $z \in D$, let $\map {f'} z = 0$. Then $f$ is constant on $D$.


Relation between P-Product Metric and Chebyshev Distance on Real Vector Space

For $n \in \N$, let $\R^n$ be a Euclidean space. Let $p \in \R_{\ge 1}$. Let $d_p$ be the $p$-product metric on $\R^n$.  Let $d_\infty$ be the Chebyshev distance on $\R^n$. Then :$\forall x, y \in \R^n: \map {d_\infty} {x, y} \le \map {d_p} {x, y} \le n^{1/p} \map {d_\infty} {x, y}$


P-Product Metrics on Real Vector Space are Topologically Equivalent/Inequality for General Case

For $n \in \N$, let $\R^n$ be a real vector Space. Let $r, t \in \R_{\ge 1}$. Let $d_r$ and $d_t$ be $p$-product metrics on $\R^n$. Then $d_r$ and $d_t$ are topologically equivalent.


De Morgan's Laws (Logic)/Disjunction of Negations/Formulation 1/Forward Implication

: $\neg p \lor \neg q \vdash \neg \left({p \land q}\right)$


De Morgan's Laws (Logic)/Disjunction of Negations/Formulation 1/Reverse Implication

: $\neg \left({p \land q}\right) \vdash \neg p \lor \neg q$


De Morgan's Laws (Logic)/Disjunction of Negations/Formulation 1

:$\neg p \lor \neg q \dashv \vdash \neg \paren {p \land q}$


De Morgan's Laws (Logic)/Disjunction of Negations/Formulation 2

:$\vdash \paren {\neg p \lor \neg q} \iff \paren {\neg \paren {p \land q} }$


De Morgan's Laws (Logic)/Conjunction of Negations/Formulation 1/Forward Implication

:$\neg p \land \neg q \vdash \neg \paren {p \lor q}$


De Morgan's Laws (Logic)/Conjunction of Negations/Formulation 1/Reverse Implication

:$\neg \paren {p \lor q} \vdash \neg p \land \neg q$


{{:De Morgan's Laws (Logic)/Conjunction of Negations/Formulation 1}}


De Morgan's Laws (Logic)/Conjunction

{{:De Morgan's Laws (Logic)/Conjunction/Formulation 1}}


De Morgan's Laws (Logic)/Conjunction/Formulation 1/Forward Implication

: $p \land q \vdash \neg \left({\neg p \lor \neg q}\right)$


De Morgan's Laws (Logic)/Conjunction/Formulation 1/Reverse Implication

: $\neg \left({\neg p \lor \neg q}\right) \vdash p \land q$


De Morgan's Laws (Logic)/Conjunction/Formulation 1/Proof

: $p \land q \dashv \vdash \neg \left({\neg p \lor \neg q}\right)$


De Morgan's Laws (Logic)/Disjunction

{{:De Morgan's Laws (Logic)/Disjunction/Formulation 1}}


=== Disjunction is Left Distributive over Conjunction === {{:Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive}} === Disjunction is Right Distributive over Conjunction === {{:Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive}}


=== Conjunction is Left Distributive over Disjunction === {{:Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive}} === Conjunction is Right Distributive over Disjunction === {{:Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive}}


Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 1/Proof

:$p \land \left({q \lor r}\right) \dashv \vdash \left({p \land q}\right) \lor \left({p \land r}\right)$


Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 2

:$\vdash \paren {p \land \paren {q \lor r} } \iff \paren {\paren {p \land q} \lor \paren {p \land r} }$


==== Formulation 1 ==== {{:Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 1}} ==== Formulation 2 ==== {{:Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 2}}


==== Formulation 1 ==== {{:Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive/Formulation 1}} ==== Formulation 2 ==== {{:Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive/Formulation 2}}


Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive/Formulation 1

:$\paren {q \lor r} \land p \dashv \vdash \paren {q \land p} \lor \paren {r \land p}$


Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive/Formulation 2

:$\vdash \paren {\paren {q \lor r} \land p} \iff \paren {\paren {q \land p} \lor \paren {r \land p} }$


==== Formulation 1 ==== {{:Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 1}} ==== Formulation 2 ==== {{:Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 2}}


==== Formulation 1 ==== {{:Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive/Formulation 1}} ==== Formulation 2 ==== {{:Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive/Formulation 2}}


Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 2

:$\vdash \paren {p \lor \paren {q \land r} } \iff \paren {\paren {p \lor q} \land \paren {p \lor r} }$


Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive/Formulation 1

:$\paren {q \land r} \lor p \dashv \vdash \paren {q \lor p} \land \paren {r \lor p}$


Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive/Formulation 2

:$\vdash \paren {\paren {q \land r} \lor p} \iff \paren {\paren {q \lor p} \land \paren {r \lor p} }$


Union of Subsets is Subset/Set of Sets

Let $T$ be a set. Let $\mathbb S$ be a set of sets. Suppose that for each $S \in \mathbb S$, $S \subseteq T$. Then: :$\displaystyle \bigcup \mathbb S \subseteq T$


Union is Smallest Superset/Set of Sets

Let $T$ be a set. Let $\mathbb S$ be a set of sets. Then: :$\displaystyle \paren {\forall X \in \mathbb S: X \subseteq T} \iff \bigcup \mathbb S \subseteq T$


Intersection of Relations is Relation

Let $S$ and $T$ be sets. Let $\mathcal F$ be a family of relations from $S$ to $T$. Let $\displaystyle \mathcal R = \bigcap \mathcal F$, the intersection of all the elements of $\mathcal F$.   Then $\mathcal R$ is a relation from $S$ to $T$. {{Expand|Binary case}}


Intersection of Subsets is Subset/Set of Sets

Let $T$ be a set. Let $\mathbb S$ be a non-empty set of sets. Suppose that for each $S \in \mathbb S$: :$S \subseteq T$ Then: :$\bigcap \mathbb S \subseteq T$


Equivalence of Definitions of Initial Topology

Let $X$ be a set. Let $I$ be an indexing set. Let $\family {\struct{Y_i, \tau_i}}_{i \mathop \in I}$ be an indexed family of topological spaces indexed by $I$. Let $\family {f_i: X \to Y_i}_{i \mathop \in I}$ be an indexed family of mappings indexed by $I$. {{TFAE|def = Initial Topology}}


Radius of Convergence from Limit of Sequence/Complex Case

Let $\xi \in \C$ be a complex number. Let $\displaystyle S \paren z = \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n$ be a (complex) power series about $\xi$. Let the sequence $\sequence {\cmod {\dfrac {a_{n + 1} } {a_n} } }_{n \mathop \in \N}$ converges. Then $R$ is given by: :$\displaystyle \dfrac 1 R = \lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n + 1} } {a_n} }$ If: :$\displaystyle \lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n + 1} } {a_n} } = 0$ then the radius of convergence is infinite, and $S \paren z$ is absolutely convergent for all $z \in \C$.


Radius of Convergence from Limit of Sequence/Real Case

Let $\xi \in \R$ be a real number. Let $\displaystyle S \paren x = \sum_{n \mathop = 0}^\infty a_n \paren {x - \xi}^n$ be a power series about $\xi$. Then the radius of convergence $R$ of $S \paren x$ is given by: : $\displaystyle \frac 1 R = \lim_{n \mathop \to \infty} \size {\frac {a_{n + 1} } {a_n} }$ if this limit exists and is nonzero. If : $\displaystyle \lim_{n \mathop \to \infty} \size {\frac {a_{n + 1} } {a_n} } = 0$ then the radius of convergence is infinite and therefore the interval of convergence is $\R$.


Existence of Radius of Convergence of Complex Power Series

Let $\xi \in \C$. Let $\displaystyle S \paren z = \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n $ be a (complex) power series about $\xi$. Then there exists a radius of convergence $R \in \overline \R$ of $S \paren z$.


Conjunction Equivalent to Negation of Implication of Negative/Formulation 1

:$p \land q \dashv \vdash \neg \left({p \implies \neg q}\right)$


Conjunction Equivalent to Negation of Implication of Negative/Formulation 2

:$\vdash \paren {p \land q} \iff \paren {\neg \paren {p \implies \neg q} }$


Conjunction Equivalent to Negation of Implication of Negative/Formulation 1/Forward Implication

:$p \land q \vdash \neg \left({p \implies \neg q}\right)$


Conjunction Equivalent to Negation of Implication of Negative/Formulation 1/Reverse Implication

:$\neg \left({p \implies \neg q}\right) \vdash p \land q$


Rule of Material Implication/Formulation 1/Forward Implication

:$p \implies q \vdash \neg p \lor q$


Rule of Material Implication/Formulation 1

:$p \implies q \dashv \vdash \neg p \lor q$


Rule of Material Implication/Formulation 1/Reverse Implication

:$\neg p \lor q \vdash p \implies q$


Rule of Material Implication/Formulation 1/Proof


Rule of Material Implication/Formulation 2

:$\vdash \paren {p \implies q} \iff \paren {\neg p \lor q}$


Modus Tollendo Ponens/Variant/Formulation 1

:$p \lor q \dashv \vdash \neg p \implies q$


Modus Tollendo Ponens/Sequent Form

==== Case 1 ==== {{:Modus Tollendo Ponens/Sequent Form/Case 1}} ==== Case 2 ==== {{:Modus Tollendo Ponens/Sequent Form/Case 2}}


Modus Tollendo Ponens/Variant/Formulation 1/Proof


Radius of Convergence of Power Series over Factorial/Complex Case

Let $\xi \in \C$ be a complex number. Let $\displaystyle \map f z = \sum_{n \mathop = 0}^\infty \dfrac {\paren {z - \xi}^n} {n!}$. Then $\map f z$ converges absolutely for all $z \in \C$. That is, the radius of convergence of the power series $\displaystyle \sum_{n \mathop = 0}^\infty \frac {\paren {z - \xi}^n} {n!}$ is infinite.


Radius of Convergence of Power Series over Factorial/Real Case

Let $\xi \in \R$ be a real number. Let $\displaystyle f \left({x}\right) = \sum_{n \mathop = 0}^\infty \frac {\left({x - \xi}\right)^n} {n!}$. Then $f \left({x}\right)$ converges for all $x \in \R$. That is, the interval of convergence of the power series $\displaystyle \sum_{n \mathop = 0}^\infty \frac {\left({x - \xi}\right)^n} {n!}$ is $\R$.


Sum of Logarithms/Natural Logarithm

Let $x, y \in \R$ be strictly positive real numbers. Then: :$\ln x + \ln y = \map \ln {x y}$ where $\ln$ denotes the natural logarithm.


Sum of Logarithms/General Logarithm

Let $x, y, b \in \R$ be strictly positive real numbers such that $b > 1$. Then: :$\log_b x + \log_b y = \map {\log_b} {x y}$ where $\log_b$ denotes the logarithm to base $b$.


Rule of Simplification/Sequent Form/Formulation 1/Proof

:$(1): \quad p \land q \vdash p$ :$(2): \quad p \land q \vdash q$


Rule of Simplification/Sequent Form

The Rule of Simplification can be symbolised by the sequents:


Rule of Conjunction/Sequent Form

The Rule of Conjunction can be symbolised in sequent form as follows:


Radius of Convergence of Derivative of Complex Power Series

Let $\xi \in \C$. For all $z \in \C$, define the power series: : $\displaystyle S \paren z = \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n$ and: : $\displaystyle S' \paren z = \sum_{n \mathop = 1}^\infty n a_n \paren {z - \xi}^{n - 1}$ Let $R$ be the radius of convergence of $S \paren z$, and let $R'$ be the radius of convergence of $S' \paren z$. Then $R =R'$.


Rule of Addition/Sequent Form

The Rule of Addition can be symbolised by the sequents: :$(1): \quad p \vdash p \lor q$ :$(2): \quad q \vdash p \lor q$


Rule of Addition/Sequent Form/Proof 2

:$(1): \quad p \vdash p \lor q$ :$(2): \quad q \vdash p \lor q$


Rule of Addition/Sequent Form/Proof 1


Rule of Addition/Sequent Form/Form 2/Proof 1

:$q \vdash p \lor q$


Proof by Cases/Sequent Form

Proof by Cases can be symbolised by the sequent: :$p \lor q, \paren {p \vdash r}, \paren {q \vdash r} \vdash r$


Modus Ponendo Ponens/Sequent Form

:$p \implies q, p \vdash q$


Logarithm of Power/Natural Logarithm

Let $x \in \R$ be a strictly positive real number. Let $a \in \R$ be a real number such that $a > 1$. Let $r \in \R$ be any real number. Let $\ln x$ be the natural logarithm of $x$. Then: :$\map \ln {x^r} = r \ln x$


Logarithm of Power/General Logarithm

Let $x \in \R$ be a strictly positive real number. Let $a \in \R$ be a real number such that $a > 1$. Let $r \in \R$ be any real number. Let $\log_a x$ be the logarithm to the base $a$ of $x$. Then: :$\map {\log_a} {x^r} = r \log_a x$


Rule of Implication/Sequent Form

The Rule of Implication can be symbolised by the sequent: :$\left({p \vdash q}\right) \vdash p \implies q$


Rule of Implication/Sequent Form/Proof 2

:$\left({p \vdash q}\right) \vdash p \implies q$


Sum of Absolutely Convergent Series

Let $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ and $\displaystyle \sum_{n \mathop = 1}^\infty b_n$ be two real or complex series that are absolutely convergent. Then the series $\displaystyle \sum_{n \mathop = 1}^\infty \paren {a_n + b_n}$ is absolutely convergent, and: :$\displaystyle \sum_{n \mathop = 1}^\infty \paren {a_n + b_n} = \displaystyle \sum_{n \mathop = 1}^\infty a_n + \sum_{n \mathop = 1}^\infty b_n$


Rule of Implication/Sequent Form/Proof 1


Rule of Simplification/Sequent Form/Formulation 1


Rule of Simplification/Sequent Form/Formulation 2

:$(1): \quad \vdash p \land q \implies p$ :$(2): \quad \vdash p \land q \implies q$


Rule of Simplification/Sequent Form/Formulation 2/Proof 1/Form 1

:$\vdash p \land q \implies p$


Rule of Simplification/Sequent Form/Formulation 2/Proof 1/Form 2

:$\vdash p \land q \implies q$


Complement of Complement in Uniquely Complemented Lattice

Let $\left({S, \wedge, \vee, \preceq}\right)$ be a uniquely complemented lattice. For each $x \in S$, let $\neg x$ be the complement of $x$. Then for each $x \in S$: :$\neg \neg x = x$


Complement of Bottom/Boolean Algebra

Let $\left({S, \vee, \wedge, \neg}\right)$ be a Boolean algebra. Then $\neg \bot = \top$.


Complement of Bottom/Bounded Lattice

Let $\left({S, \vee, \wedge, \preceq}\right)$ be a bounded lattice. Then the bottom $\bot$ has a unique complement, namely $\top$, top.


Complement of Top/Boolean Algebra

Let $\struct {S, \vee, \wedge, \neg}$ be a Boolean algebra. Then $\neg \top = \bot$.


Duality Principle (Boolean Algebras)

Let $\struct {S, \vee, \wedge}$ be a Boolean algebra. Then any theorem in $\struct {S, \vee, \wedge}$ remains valid if both $\vee$ and $\wedge$ are interchanged, and also $\bot$ and $\top$ are interchanged throughout the whole theorem.


Identities of Boolean Algebra also Zeroes

Let $\struct {S, \vee, \wedge, \neg}$ be a Boolean algebra, defined as in Definition 1. Let the identity for $\vee$ be $\bot$ and the identity for $\wedge$ be $\top$. Then: :$(1): \quad \forall x \in S: x \vee \top = \top$ :$(2): \quad \forall x \in S: x \wedge \bot = \bot$ That is, $\bot$ is a zero element for $\wedge$, and $\top$ is a zero element for $\vee$.


Complement of Complement (Boolean Algebras)

Let $\left({S, \vee, \wedge, \neg}\right)$ be a Boolean algebra. Then for all $a \in S$: :$\neg (\neg a) = a$


De Morgan's Laws (Boolean Algebras)

:$\neg \paren {a \vee b} = \neg a \wedge \neg b$ :$\neg \paren {a \wedge b} = \neg a \vee \neg b$


Complement in Boolean Algebra is Unique

Let $\left({S, \vee, \wedge}\right)$ be a Boolean algebra. Then for all $a \in S$, there is a unique $b \in S$ such that: :$a \wedge b = \bot, a \vee b = \top$ i.e., a valid choice for $\neg a$ as in axiom $(BA \ 4)$ for Boolean algebras.


Cancellation of Join in Boolean Algebra

Let $\left({S, \vee, \wedge, \neg}\right)$ be a Boolean algebra. Let $a, b, c \in S$, and suppose that: {{begin-eqn}} {{eqn | l = a \vee c       | r = b \vee c }} {{eqn | l = a \vee \neg c       | r = b \vee \neg c }} {{end-eqn}} Then $a = b$.


Cancellation of Meet in Boolean Algebra

Let $\struct {S, \vee, \wedge, \neg}$ be a Boolean algebra. Let $a, b, c \in S$. Let: {{begin-eqn}} {{eqn | l = a \wedge c       | r = b \wedge c }} {{eqn | l = a \wedge \neg c       | r = b \wedge \neg c }} {{end-eqn}} Then: : $a = b$


Operations of Boolean Algebra are Associative

Let $\struct {S, \vee, \wedge, \neg}$ be a Boolean algebra, defined as in Definition 1. Then: :$\forall a, b, c \in S: \paren {a \wedge b} \wedge c = a \wedge \paren {b \wedge c}$ :$\forall a, b, c \in S: \paren {a \vee b} \vee c = a \vee \paren {b \vee c}$ That is, both $\vee$ and $\wedge$ are associative operations.


Absorption Laws (Boolean Algebras)

Let $\struct {S, \vee, \wedge}$ be a Boolean algebra, defined as in Definition 1. Then for all $a, b \in S$: :$a = a \vee \paren {a \wedge b}$ :$a = a \wedge \paren {a \vee b}$ That is, $\vee$ absorbs $\wedge$, and $\wedge$ absorbs $\vee$.


Difference of Absolutely Convergent Series

Let $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ and $\displaystyle \sum_{n \mathop = 1}^\infty b_n$ be two real or complex series that are absolutely convergent. Then the series $\displaystyle \sum_{n \mathop = 1}^\infty \paren {a_n - b_n}$ is absolutely convergent, and: :$\displaystyle \sum_{n \mathop = 1}^\infty \paren {a_n - b_n} = \sum_{n \mathop = 1}^\infty a_n - \sum_{n \mathop = 1}^\infty b_n$


True Statement is implied by Every Statement/Formulation 1

:$p \vdash q \implies p$


True Statement is implied by Every Statement/Formulation 2

:$\vdash q \implies \paren {p \implies q}$


==== Formulation 1 ==== {{:True Statement is implied by Every Statement/Formulation 1}} ==== Formulation 2 ==== {{:True Statement is implied by Every Statement/Formulation 2}}


False Statement implies Every Statement

==== Formulation 1 ==== {{:False Statement implies Every Statement/Formulation 1}} ==== Formulation 2 ==== {{:False Statement implies Every Statement/Formulation 2}}


False Statement implies Every Statement/Formulation 1

: $\neg p \vdash p \implies q$


False Statement implies Every Statement/Formulation 1/Proof 2

:$\neg p \vdash p \implies q$


False Statement implies Every Statement/Formulation 2

:$\vdash \neg p \implies \paren {p \implies q}$


Modus Tollendo Ponens/Sequent Form/Case 1

:$p \lor q, \neg p \vdash q$


Modus Tollendo Ponens/Sequent Form/Case 2

:$p \lor q, \neg q \vdash p$


Two is Boolean Algebra

Let $\mathbf 2$ denote two. Then $\mathbf 2$ is a Boolean algebra.


Two-Valued Functions form Boolean Algebra

Let $\mathbf 2$ be the Boolean algebra two, and let $X$ be a set. Let $\mathbf 2^X$ be the set of all mappings $p: X \to \mathbf 2$. Define the operations $\vee$, $\wedge$ and $\neg$ on $\mathbf 2^X$ in pointwise fashion thus: :$\vee: \mathbf 2^X \times \mathbf 2^X \to \mathbf 2^X, \left({p \vee q}\right) (x) := p (x) \vee q (x)$ :$\wedge: \mathbf 2^X \times \mathbf 2^X \to \mathbf 2^X, \left({p \wedge q}\right) (x) := p (x) \wedge q (x)$ :$\neg: \mathbf 2^X \to \mathbf 2^X, \left({\neg p}\right) (x) := \neg p (x)$ Furthermore, write $\bot$ and $\top$ for the constant mappings with these values, viz: :$\bot: X \to \mathbf 2, \bot (x) := \bot$ :$\top: X \to \mathbf 2, \top (x) := \top$ Then $\left({\mathbf 2^X, \vee, \wedge, \neg}\right)$ is a Boolean algebra, with $\bot$ and $\top$ as identities for $\vee$ and $\wedge$, respectively.


Power Set is Boolean Ring

Let $S$ be a set, and let $\mathcal P \left({S}\right)$ be its power set. Denote with $*$ and $\cap$ symmetric difference and intersection, respectively. Then $\left({S, *, \cap}\right)$ is a Boolean ring.


Power Set and Two-Valued Functions are Isomorphic Boolean Rings

Let $S$ be a set. Let $\mathbf 2$ be the Boolean ring two. Let $\mathcal P \left({S}\right)$ be the power set of $S$; by Power Set is Boolean Ring, it is a Boolean ring. Let $\mathbf 2^S$ be the set of all mappings $f: S \to \mathbf 2$; by Two-Valued Functions form Boolean Ring, it is also a Boolean ring. Let $\chi_{\left({\cdot}\right)}: \mathcal P \left({S}\right) \to \mathbf 2^S$ be the characteristic function operation. Then $\chi_{\left({\cdot}\right)}$ is a ring isomorphism.


Peirce's Law/Formulation 1

: $\left({p \implies q}\right) \implies p \vdash p$


Peirce's Law/Formulation 2/Proof 2

: $\vdash \left({\left({p \implies q}\right) \implies p}\right) \implies p$


Peirce's Law/Strong Form/Formulation 1

:$\paren {\paren {p \implies q} \implies p} \dashv \vdash p$


==== Formulation 1 ==== {{:Conjunction with Negative Equivalent to Negation of Implication/Formulation 1}} ==== Formulation 2 ==== {{:Conjunction with Negative Equivalent to Negation of Implication/Formulation 2}}


Conjunction with Negative Equivalent to Negation of Implication/Formulation 1

:$p \land \neg q \dashv \vdash \neg \left({p \implies q}\right)$


Conjunction with Negative Equivalent to Negation of Implication/Formulation 1/Forward Implication

:$p \land \neg q \vdash \neg \paren {p \implies q}$


Conjunction with Negative Equivalent to Negation of Implication/Formulation 1/Reverse Implication

:$\neg \left({p \implies q}\right) \vdash p \land \neg q$


Conjunction with Negative Equivalent to Negation of Implication/Formulation 1/Proof


Conjunction with Negative Equivalent to Negation of Implication/Formulation 2

:$\vdash \paren {p \land \neg q} \iff \paren {\neg \paren {p \implies q} }$


Reflexive Circular Relation is Equivalence

Let $\mathcal R \subseteq S \times S$ be a reflexive and circular relation in $S$. Then $\mathcal R$ is an equivalence relation.


Inverse of Algebraic Structure Isomorphism is Isomorphism/General Result

Let $\phi: \left({S, \circ_1, \circ_2, \ldots, \circ_n}\right)$ and $\left({T, *_1, *_2, \ldots, *_n}\right)$ be algebraic structures. Let $\phi: \left({S, \circ_1, \circ_2, \ldots, \circ_n}\right) \to \left({T, *_1, *_2, \ldots, *_n}\right)$ be a mapping. Then: : $\phi: \left({S, \circ_1, \circ_2, \ldots, \circ_n}\right) \to \left({T, *_1, *_2, \ldots, *_n}\right)$ is an isomorphism {{iff}}: : $\phi^{-1}: \left({T, *_1, *_2, \ldots, *_n}\right) \to \left({S, \circ_1, \circ_2, \ldots, \circ_n}\right)$ is also an isomorphism.


Fundamental Theorem of Calculus for Complex Riemann Integrals

Let $\closedint a b$ be a closed real interval. Let $F, f: \closedint a b \to \C$ be complex functions. Suppose that $F$ is a primitive of $f$. Then the complex Riemann integral of $f$ satisfies: :$\displaystyle \int_a^b \map f t \rd t = \map F b - \map F a$


Fundamental Theorem of Calculus for Contour Integrals

Let $F, f: D \to \C$ be complex functions, where $D$ is a connected domain. Let $C$ be a contour that is a concatenation of the directed smooth curves $C_1, \ldots, C_n$. Let $C_k$ be parameterized by the smooth path $\gamma_k: \closedint {a_k} {b_k} \to D$ for all $k \in \set {1, \ldots, n}$. Suppose that $F$ is a primitive of $f$. If $C$ has start point $z$ and end point $w$, then: :$\displaystyle \int_C \map f z \rd z = \map F w - \map F z$ If $C$ is a closed contour, then: :$\displaystyle \oint_C \map f z \rd z = 0$


Power Series is Taylor Series

Let $\displaystyle f \left({z}\right) = \sum_{n \mathop = 0}^\infty a_n \left({z - \xi}\right)^n$ be a complex power series about $\xi \in \C$. Let $R$ be the radius of convergence of $f$. Then, $f$ is of differentiability class $C^\infty$. For all $n \in \N$: :$a_n = \dfrac{f^{\left({n}\right) } \left({\xi}\right) }{ n! }$ Hence, $f$ is equal to its Taylor series expansion about $\xi$: :$\displaystyle \forall z \in \C, \left\vert{z - \xi}\right\vert < R: \quad f \left({z}\right) = \sum_{n \mathop = 0}^\infty \dfrac{\left({z - \xi}\right)^n}{n!} f^{\left({n}\right) } \left({\xi}\right)$


Image of Complex Exponential Function

The image of the complex exponential function is $\C \setminus \left\{ {0}\right\}$.


Reciprocal of Complex Exponential

:$\dfrac 1 {\map \exp z} = \map \exp {-z}$


Power Function Strictly Preserves Ordering in Ordered Group

Let $n \in \N_{>0}$ be a strictly positive integer.


Period of Complex Exponential Function

:$\map \exp {z + 2 k \pi i} = \map \exp z$


Zero Staircase Integral Condition for Primitive

Let $f: D \to \C$ be a continuous complex function, where $D$ is a connected domain. Let $z_0 \in D$. Suppose that $\displaystyle \oint_C \map f z \rd z = 0$ for all closed staircase contours $C$ in $D$. Then $f$ has a primitive $F: D \to \C$ defined by: :$\displaystyle \map F w = \int_{C_w} \map f z \rd z$ where $C_w$ is any staircase contour in $D$ with start point $z_0$ and end point $w$.


Derivative of Complex Polynomial

Let $a_n \in \C$ for $n \in \left\{ {0, 1, \ldots, N}\right\}$, where $N \in \N$. Let $f: \C \to \C$ be a complex polynomial defined by $\displaystyle f \left({z}\right) = \sum_{n \mathop = 0}^N a_n z^n$. Then $f$ is complex differentiable and its derivative is: :$\displaystyle f' \left({z}\right) = \sum_{n \mathop = 1}^N n a_n z^{n-1}$


Nested Sphere Theorem

Let $M = \left({A, d}\right)$ be a complete metric space. Let $\left\langle{S_n}\right\rangle$ be a sequence of closed balls in $M$ defined by: : $S_n = B^-_{\rho_n} \left({x_n}\right)$ where $\rho_n \to 0$ as $n \to \infty$ and: : $S_1 \supseteq S_2 \supseteq \cdots \supseteq S_n \supseteq \cdots$ Then there exists $x \in A$ such that: :$\displaystyle \bigcap_{n \mathop = 1}^\infty S_n = \left\{{x}\right\}$


Principle of Dilemma/Formulation 1/Forward Implication

: $\left({p \implies q}\right) \land \left({\neg p \implies q}\right) \vdash q$


Principle of Dilemma/Formulation 1/Forward Implication/Proof 3


Principle of Dilemma/Formulation 1/Reverse Implication

: $q \vdash \left({p \implies q}\right) \land \left({\neg p \implies q}\right)$


Principle of Dilemma/Formulation 2/Reverse Implication

:$\vdash q \implies \paren {\paren {p \implies q} \land \paren {\neg p \implies q} }$


Properties of Complex Exponential Function

Let $z \in \C$ be a complex number. Let $\exp z$ be the exponential of $z$. Then:


Open Domain is Connected iff it is Path-Connected

Let $D \subseteq \C$ be a open subset of the set of complex numbers. Then $D$ is connected {{iff}} $D$ is path-connected.


Proof by Contradiction/Variant 2/Formulation 1

:$p \implies q, p \implies \neg q \vdash \neg p$


Proof by Contradiction/Variant 2/Formulation 2

:$\vdash \paren {\paren {p \implies q} \land \paren {p \implies \neg q} } \implies \neg p$


Proof by Contradiction/Variant 3/Formulation 1

:$p \implies \neg p \vdash \neg p$


Proof by Contradiction/Variant 3/Formulation 2

:$\vdash \paren {p \implies \neg p} \implies \neg p$


Primitive of Function on Connected Domain

Let $f: D \to \C$ be a continuous complex function, where $D$ is a connected domain. Then the following three conditions are equivalent: :$(1): \quad$ $f$ has a primitive. :$(2): \quad$ For any two contours $C_1, C_2$ in $D$ with identical start points $z_1 \in D$ and end points $z_2 \in D$, we have: ::$\displaystyle \int_{C_1} \map f z \rd z = \int_{C_2} \map f z \rd z$ :$(3): \quad$ For all closed contours $C$ in $D$, we have: ::$\displaystyle \oint_C \map f z \rd z = 0$ If the conditions hold, we can choose any $z_0 \in D$ and define a primitive $F: D \to \C$ of $f$ by: :$\displaystyle \map F w = \int_{C_w} \map f z \rd z$ where $C_w$ is any contour in $D$ with start point $z_0$ and end point $w$.


Mind the Gap

Let $\left({S, \preccurlyeq}\right)$ be a totally ordered set. Let $a, b \in S$ with $a \prec b$. Suppose that: :$\left\{ {x \in S: a \prec x \prec b}\right\} = \varnothing$ Then: : $a^\succ = b^\succcurlyeq$ and : $b^\prec = a^\preccurlyeq$ where: : $a^\succ$ is the strict upper closure of $a$ : $b^\succcurlyeq$ is the weak upper closure of $b$ : $b^\prec$ is the strict lower closure of $b$ : $a^\preccurlyeq$ is the weak lower closure of $a$.


Proof by Contradiction/Sequent Form

:$\paren {p \vdash \bot} \vdash \neg p$


Left or Right Inverse of Matrix is Inverse

Let $\mathbf A, \mathbf B$ be square matrices of order $n$ over a commutative ring with unity $\left({R, +, \circ}\right)$. Suppose that: : $\mathbf A \mathbf B = \mathbf I_n$ where $\mathbf I_n$ is the unit matrix of order $n$. Then $\mathbf A$ and $\mathbf B$ are invertible matrices, and furthermore: :$\mathbf B = \mathbf A^{-1}$ where $\mathbf A^{-1}$ is the inverse of $\mathbf A$.


Banach-Alaoglu Theorem

Let $X$ be a separable normed space. Then the closed unit sphere in its dual $X^*$ is weak* sequentially compact.


Empty Product is Terminal Object

Let $\mathbf C$ be a metacategory. Suppose $\mathbf C$ admits a product $\displaystyle \prod \O$ for the empty set. Then $\displaystyle \prod \O$ is a terminal object of $\mathbf C$.


Unary Product for Object is Itself

Let $\mathbf C$ be a metacategory. Let $C$ be an object of $\mathbf C$. Then $\displaystyle \prod \set C = C$, where $\displaystyle \prod$ denotes product.


Vector Cross Product is Orthogonal to Factors

Let $\mathbf a$ and $\mathbf b$ be vectors in the Euclidean space $\R^3$. Let $\mathbf a \times \mathbf b$ denote the vector cross product. Then: :$(1): \quad$ $\mathbf a$ and $\mathbf a \times \mathbf b$ are orthogonal. :$(2): \quad$ $\mathbf b$ and $\mathbf a \times \mathbf b$ are orthogonal.


Squeeze Theorem/Sequences/Metric Spaces

Let $M = \left({S, d}\right)$ be a metric space or pseudometric space. Let $\left\langle{x_n}\right\rangle$ be a sequence in $S$. Let $p \in S$.  Let $\left\langle{r_n}\right\rangle$ be a sequence in $\R_{\ge 0}$. Let $\left\langle{r_n}\right\rangle$ converge to $0$. For each $n$, let $d \left({p, x_n}\right) \le r_n$. Then $\left\langle{x_n}\right\rangle$ converges to $p$.


Squeeze Theorem/Sequences/Linearly Ordered Space

Let $\left({S, \le, \tau}\right)$ be a linearly ordered space. Let $\left\langle{x_n}\right\rangle$, $\left\langle{y_n}\right\rangle$, and $\left\langle{z_n}\right\rangle$ be sequences in $S$. Let $p \in S$. Let $\left\langle{x_n}\right\rangle$ and $\left\langle{z_n}\right\rangle$ both converge to $p$ For each $n$, let $x_n \le y_n \le z_n$. Then $\left\langle{y_n}\right\rangle$ converges to $p$.


Non-Zero Vectors Orthogonal iff Perpendicular

Let $\mathbf u$, $\mathbf v$ be non-zero vectors in the Euclidean space $\R^n$. Then $\mathbf u$ and $\mathbf v$ are orthogonal {{iff}} they are perpendicular.


Exponential Function is Continuous/Complex

:$\forall z_0 \in \C: \displaystyle \lim_{z \mathop \to z_0} \exp z = \exp z_0$


Order Topology on Convex Subset is Subspace Topology

Let $\left({S, \preceq,\tau}\right)$ be a linearly ordered space. Let $A \subseteq S$ be a convex set in $S$. Let $\upsilon$ be the order topology on $A$. Let $\tau'$ be the $\tau$-relative subspace topology on $A$. Then $\upsilon = \tau'$.


Exponential Function is Continuous/Real Numbers

The real exponential function is continuous. That is: :$\forall x_0 \in \R: \displaystyle \lim_{x \mathop \to x_0} \ \exp x = \exp x_0$


Edge of Tree is Bridge

Let $T$ be a tree. Let $e$ be an edge of $T$. Then $e$ is a bridge of $T$.


Sub-Basis for Topological Subspace

Let $\left({X,\tau}\right)$ be a topological space. Let $K$ be a sub-basis for $\tau$. Let $\left({S, \tau'}\right)$ be a subspace of $\left({X, \tau}\right)$. Let $K' = \left\{{ U \cap S: U \in K }\right\}$. That is, $K'$ consists of the $\tau'$-open sets in $S$ corresponding to elements of $K$. Then $K'$ is a sub-basis for $\tau'$.


Modus Ponendo Ponens/Variant 1

:  $p \vdash \left({p \implies q}\right) \implies q$


Modus Ponendo Ponens/Variant 1/Proof 2

:$p \vdash \left({p \implies q}\right) \implies q$


Modus Ponendo Ponens/Variant 2

:  $\vdash p \implies \left({\left({p \implies q}\right) \implies q}\right)$


Modus Ponendo Ponens/Variant 2/Proof 2

:$\vdash p \implies \paren {\paren {p \implies q} \implies q}$


Rule of Commutation/Conjunction

Conjunction is commutative: === Formulation 1 === {{:Rule of Commutation/Conjunction/Formulation 1}} === Formulation 2 === {{:Rule of Commutation/Conjunction/Formulation 2}}


Rule of Commutation/Disjunction

Disjunction is commutative: === Formulation 1 === {{:Rule of Commutation/Disjunction/Formulation 1}} === Formulation 2 === {{:Rule of Commutation/Disjunction/Formulation 2}}


Factor Principles/Conjunction on Right/Formulation 1

: $p \implies q \vdash \left({p \land r}\right) \implies \left ({q \land r}\right)$


Factor Principles/Conjunction on Right/Formulation 2

: $\vdash \left({p \implies q}\right) \implies \left({\left({p \land r}\right) \implies \left ({q \land r}\right)}\right)$


Interior of Convex Angle is Convex Set

Let $\mathbf v, \mathbf w$ be two non-zero vectors in $\R^2$, and let $p$ be a point in $\R^2$. Suppose that the angle between $\mathbf v$ and $\mathbf w$ is a convex angle. Then the set  :$U = \left\{ {p + st \mathbf v + \left({1-s}\right) t \mathbf w : s \in \left({0\,.\,.\,1}\right) , t \in \R_{>0} }\right\}$ is a convex set. {{expand|It'd be really nice to have a picture of $U$ to support intuition and connect with the page title}}


Factor Principles/Conjunction on Right

==== Formulation 1 ==== {{:Factor Principles/Conjunction on Right/Formulation 1}} ==== Formulation 2 ==== {{:Factor Principles/Conjunction on Right/Formulation 2}}


Factor Principles/Conjunction on Left

==== Formulation 1 ==== {{:Factor Principles/Conjunction on Left/Formulation 1}} ==== Formulation 2 ==== {{:Factor Principles/Conjunction on Left/Formulation 2}}


Factor Principles/Conjunction on Left/Formulation 1

: $p \implies q \vdash \left({r \land p}\right) \implies \left ({r \land q}\right)$


Factor Principles/Conjunction on Left/Formulation 2

:$\vdash \paren {p \implies q} \implies \paren {\paren {r \land p} \implies \paren {r \land q} }$


Triangle is Convex Set

The interior of a triangle embedded in $\R^2$ is a convex set.


Subset Relation is Compatible with Subset Product

Let $\left({S,\circ}\right)$ be a magma. Let $\mathcal P \left({S}\right)$ be the power set of $S$. Let $\circ_\mathcal P$ be the operation induced on $\mathcal P \left({S}\right)$ by $\circ$. Then the subset relation $\subseteq$ is compatible with $\circ_\mathcal P$.


Subset Relation is Compatible with Subset Product/Corollary 1

Let $A, B, C, D \in \powerset S$. Let $A \subseteq B$ and $C \subseteq D$. Then: :$A \circ_\PP C \subseteq B \circ_\PP D$


Subset Relation is Compatible with Subset Product/Corollary 2

Let $A,B \in \mathcal P \left({S}\right)$, the power set of $S$. {{improve|No need to use power set. $A \subseteq B \subseteq S$ sufficient.}} Let $A \subseteq B$. Let $x \in S$. Then: :$x \circ A \subseteq x \circ B$ :$A \circ x \subseteq B \circ x$


Equivalence of Definitions of Normal Subset/1 iff 2

Let $\left({G, \circ}\right)$ be a group. Let $S$ be a subset of $G$. Then Normal Subset/Definition 1 is equivalent to Normal Subset/Definition 2. That is, the following three statements are equivalent: :$(1)\quad \forall g \in G: g \circ S = S \circ g$ :$(2)\quad \forall g \in G: g \circ S \circ g^{-1} = S$ :$(3)\quad \forall g \in G: g^{-1} \circ S \circ g = S$


Subset Product with Identity

Let $\left({S, \circ}\right)$ be a magma. Let $\left({S, \circ}\right)$ have an identity element $e$. Then $e \circ S = S \circ e = S$, where $\circ$ is understood to be the subset product with singleton.


Equivalence of Definitions of Normal Subset/2 implies 3

Let $\left({G, \circ}\right)$ be a group. Let $S \subseteq G$. Let $S$ be a normal subset of $G$ by Definition 2. Then $S$ is a normal subset of $G$ by Definition 3. That is, if: :$\forall g \in G: g \circ S \circ g^{-1} = S$ or: :$\forall g \in G: g^{-1} \circ S \circ g = S$ then: :$\forall g \in G: g \circ S \circ g^{-1} \subseteq S$ and: :$\forall g \in G: g^{-1} \circ S \circ g \subseteq S$


Superset Relation is Compatible with Subset Product

Let $\left({S,\circ}\right)$ be a magma. Let $\circ_{\mathcal P}$ be the subset product on $\mathcal P \left({S}\right)$, the power set of $S$. Then the superset relation $\supseteq$ is compatible with $\circ_{\mathcal P}$.


Equivalence of Definitions of Normal Subset/3 iff 4

Let $\left({G,\circ}\right)$ be a group. Let $S \subseteq G$. Then: : $S$ is a normal subset of $G$ by Definition 3 iff: : $S$ is a normal subset of $G$ by Definition 4. That is, the following conditions are equivalent: : $(1)\quad \forall g \in G: g \circ S \circ g^{-1} \subseteq S$ : $(2)\quad \forall g \in G: g^{-1} \circ S \circ g \subseteq S$ : $(3)\quad \forall g \in G: S \subseteq g \circ S \circ g^{-1}$ : $(4)\quad \forall g \in G: S \subseteq g^{-1} \circ S \circ g$


Equivalence of Definitions of Normal Subset/3 and 4 imply 2

Let $\left({G,\circ}\right)$ be a group. Let $S \subseteq G$. Let $S$ be a normal subset of $G$ by Definition 3 and Definition 4. Then $S$ is a normal subset of $G$ by Definition 2.


Equivalence of Definitions of Normal Subset/3 iff 5

Let $\struct {G, \circ}$ be a group. Let $S \subseteq G$. Then: : $S$ is a normal subset of $G$ by Definition 3 {{iff}}: : $S$ is a normal subset of $G$ by Definition 5.


Equivalence of Definitions of Normal Subgroup

Let $G$ be a group. Let $N$ be a subgroup of $G$. {{TFAE|def = Normal Subgroup}}


Subgroup is Normal iff Normal Subset

Let $\left({G, \circ}\right)$ be a group. Let $N$ be a subgroup of $G$. Then $N$ is normal in $G$ (by definition 1) {{iff}} it is a normal subset of $G$.


Rule of Explosion/Sequent Form

:$\bot \vdash \phi$


Law of Excluded Middle/Sequent Form

The Law of Excluded Middle can be symbolised by the sequent: :$\vdash p \lor \neg p$


Power Set of Monoid under Induced Operation is Monoid

Let $\left({G, \circ}\right)$ be a monoid with identity $e$. Let $\left({\mathcal P \left({G}\right), \circ_\mathcal P}\right)$ be the algebraic structure consisting of the power set of $G$ and the operation induced on $\mathcal P \left({S}\right)$ by $\circ$. Then $\left({\mathcal P \left({G}\right), \circ_\mathcal P}\right)$ is a monoid with identity $\left\{{e}\right\}$.


Principle of Non-Contradiction/Sequent Form/Formulation 2

:$\vdash \neg \paren {p \land \neg p}$


Power Set of Semigroup under Induced Operation is Semigroup

Let $\left({S, \circ}\right)$ be a semigroup. Let $\left({\mathcal P \left({S}\right), \circ_\mathcal P}\right)$ be the algebraic structure consisting of the power set of $S$ and the operation induced on $\mathcal P \left({S}\right)$ by $\circ$. Then $\left({\mathcal P \left({S}\right), \circ_\mathcal P}\right)$ is a semigroup.


Power Set of Group under Induced Operation is Monoid

Let $\left({G, \circ}\right)$ be a group with identity $e$. Let $\left({\mathcal P \left({G}\right), \circ_\mathcal P}\right)$ be the algebraic structure consisting of the power set of $G$ and the operation induced on $\mathcal P \left({G}\right)$ by $\circ$. Then $\left({\mathcal P \left({G}\right), \circ_\mathcal P}\right)$ is a monoid with identity $\left\{{e}\right\}$.


Power Set of Magma under Induced Operation is Magma

Let $\left({S, \circ}\right)$ be a magma. Let $\left({\mathcal P \left({S}\right), \circ_\mathcal P}\right)$ be the algebraic structure consisting of the power set of $S$ and the operation induced on $\mathcal P \left({S}\right)$ by $\circ$. Then $\left({\mathcal P \left({S}\right), \circ_\mathcal P}\right)$ is a magma.


Element in Left Coset iff Product with Inverse in Subgroup

Let $y H$ denote the left coset of $H$ by $y$. Then: :$x \in y H \iff x^{-1} y \in H$


Element in Right Coset iff Product with Inverse in Subgroup

Let $H \circ y$ denote the right coset of $H$ by $y$. Then: :$x \in H y \iff x y^{-1} \in H$


Inverse of Inverse of Subset of Group

Let $\struct {G, \circ}$ be a group. Let $X \subseteq G$. Then: : $\paren {X^{-1} }^{-1} = X$. where $X^{-1}$ denotes the inverse of $X$.


Exclusive Or is Commutative

: $p \oplus q \dashv \vdash q \oplus p$


Exclusive Or is Associative

Exclusive or is associative: : $p \oplus \left({q \oplus r}\right) \dashv \vdash \left({p \oplus q}\right) \oplus r$


Exclusive Or with Itself

Exclusive or destroys copies of itself: : $p \oplus p \dashv \vdash \bot$


Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Proof 1

: $\neg \left ({p \iff q}\right) \dashv \vdash \left({\neg p \land q}\right) \lor \left({p \land \neg q}\right)$


Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Proof 2


Non-Equivalence as Disjunction of Conjunctions/Formulation 1


Exclusive Or is Negation of Biconditional

Exclusive or is equivalent to the negation of the biconditional: :$p \oplus q \dashv \vdash \neg \paren {p \iff q}$


Exclusive Or as Disjunction of Conjunctions

:$p \oplus q \dashv \vdash \paren {\neg p \land q} \lor \paren {p \land \neg q}$


Non-Equivalence as Disjunction of Negated Implications/Proof 1

: $\neg \left ({p \iff q}\right) \dashv \vdash \neg \left({p \implies q}\right) \lor \neg \left({q \implies p}\right)$


Non-Equivalence as Disjunction of Negated Implications/Proof 2


Non-Equivalence as Disjunction of Negated Implications


Conjunction of Disjunction with Negation is Conjunction with Negation

:$\left({p \lor q}\right) \land \neg q \dashv \vdash p \land \neg q$


Non-Equivalence as Conjunction of Disjunction with Negation of Conjunction/Proof 1

: $\neg \left ({p \iff q}\right) \dashv \vdash \left({p \lor q} \right) \land \neg \left({p \land q}\right)$


Non-Equivalence as Conjunction of Disjunction with Negation of Conjunction/Proof 2


Non-Equivalence as Conjunction of Disjunction with Negation of Conjunction

: $\neg \left ({p \iff q}\right) \dashv \vdash \left({p \lor q} \right) \land \neg \left({p \land q}\right)$ That is, negation of the biconditional means the same thing as '''either-or but not both''', that is, exclusive or.


Inversion Mapping is Involution

Let $G$ be a group, and let $\iota: G \to G$ be the inversion mapping. Then $\iota$ is an involution. That is: :$\forall g \in G: \map \iota {\map \iota g} = g$


Rule of Association/Conjunction

Conjunction is associative: === Formulation 1 === {{:Rule of Association/Conjunction/Formulation 1}} === Formulation 2 === {{:Rule of Association/Conjunction/Formulation 2}}


Rule of Association/Disjunction

Disjunction is associative: === Formulation 1 === {{:Rule of Association/Disjunction/Formulation 1}} === Formulation 2 === {{:Rule of Association/Disjunction/Formulation 2}}


Rule of Association/Conjunction/Formulation 1

:$p \land \left({q \land r}\right) \dashv \vdash \left({p \land q}\right) \land r$


Rule of Association/Disjunction/Formulation 1

:$p \lor \paren {q \lor r} \dashv \vdash \paren {p \lor q} \lor r$


Rule of Association/Disjunction/Formulation 1/Proof 2

:$p \lor \left({q \lor r}\right) \dashv \vdash \left({p \lor q}\right) \lor r$


Rule of Association/Conjunction/Formulation 2

:$\vdash \paren {p \land \paren {q \land r} } \iff \paren {\paren {p \land q} \land r}$


Rule of Association/Disjunction/Formulation 2

:$\vdash \paren {p \lor \paren {q \lor r} } \iff \paren {\paren {p \lor q} \lor r}$


Principle of Commutation/Forward Implication/Formulation 1

: $p \implies \left({q \implies r}\right) \vdash q \implies \left({p \implies r}\right)$


Principle of Commutation/Reverse Implication/Formulation 1

:$q \implies \paren {p \implies r} \vdash p \implies \paren {q \implies r}$


Principle of Commutation/Reverse Implication/Formulation 1/Proof

: $q \implies \left({p \implies r}\right) \vdash p \implies \left({q \implies r}\right)$


Principle of Commutation/Formulation 1/Proof 1

: $p \implies \left({q \implies r}\right) \dashv \vdash q \implies \left({p \implies r}\right)$


Principle of Commutation/Formulation 1/Proof 2


Principle of Commutation/Formulation 2

:$\vdash \paren {p \implies \paren {q \implies r} } \iff \paren {q \implies \paren {p \implies r} }$


Linearly Ordered Space is Connected iff Linear Continuum

Let $T = \struct {S, \preceq, \tau}$ be a linearly ordered space. Then $S$ is a connected space {{iff}} it is a linear continuum.


Rule of Transposition/Variant 1/Formulation 2/Forward Implication

: $\vdash \left({p \implies \neg q}\right) \implies \left({q \implies \neg p}\right)$


Boundary of Polygon is Jordan Curve

Let $P$ be a polygon embedded in $\R^2$. Then there exists a Jordan curve $\gamma: \closedint 0 1 \to \R^2$ such that the image of $\gamma$ is equal to the boundary $\partial P$ of $P$.


Rule of Explosion/Variant 1

:$\vdash p \implies \paren {\neg p \implies q}$


Negation of Conditional implies Antecedent

:$\vdash \neg \paren {p \implies q} \implies p$


Negation of Conditional implies Negation of Consequent

:$\vdash \neg \left({p \implies q}\right) \implies \neg q$


Compactness from Basis

Let $\struct {X, \tau}$ be a topological space. Let $B$ be a basis for $\tau$. Then the following propositions are equivalent: {{begin-axiom}} {{axiom | n = 1         | t = $\struct {X, \tau}$ is compact. }} {{axiom | n = 2         | t = Every open cover of $X$ by elements of $B$ has a finite subcover. }} {{end-axiom}}


Implication Equivalent to Negation of Conjunction with Negative/Formulation 2/Reverse Implication

:$\vdash \left({\neg \left({p \land \neg q}\right)}\right) \implies \left({p \implies q}\right)$


Biconditional Elimination

The rule of '''biconditional  elimination''' is a valid deduction sequent in propositional logic.


Biconditional Introduction/Sequent Form

:$p \implies q, q \implies p \vdash p \iff q$


Biconditional Elimination/Sequent Form/Proof 1/Form 1

:$p \iff q \vdash p \implies q$


Biconditional Elimination/Sequent Form/Proof 1/Form 2

:$p \iff q \vdash q \implies p$


Biconditional Elimination/Sequent Form/Proof 1

:$(1): \quad p \iff q \vdash p \implies q$ :$(2): \quad p \iff q \vdash q \implies p$


Biconditional Elimination/Sequent Form/Proof 2


Biconditional is Commutative/Formulation 1/Proof 1

: $p \iff q \dashv \vdash q \iff p$


Rule of Material Equivalence/Formulation 1

:$p \iff q \dashv \vdash \paren {p \implies q} \land \paren {q \implies p}$


Equivalences are Interderivable/Proof 1

If two propositional formulas are interderivable, they are equivalent: : $\left ({p \dashv \vdash q}\right) \dashv \vdash \left ({p \iff q}\right)$


Equivalences are Interderivable/Forward Implication

: $\left ({p \dashv \vdash q}\right) \vdash \left ({p \iff q}\right)$


Compact Subspace of Linearly Ordered Space/Reverse Implication

Let $\struct {X, \preceq, \tau}$ be a linearly ordered space. Let $Y \subseteq X$ be a non-empty subset of $X$. Let the following hold: :$(1): \quad$ For every non-empty $S \subseteq Y$, $S$ has a supremum and an infimum in $X$. :$(2): \quad$ For every non-empty $S \subseteq Y$: $\sup S, \inf S \in Y$. Then $Y$ is a compact subspace of $\struct {X, \tau}$.


Modus Ponendo Tollens/Sequent Form

==== Case 1 ==== {{:Modus Ponendo Tollens/Sequent Form/Case 1}} ==== Case 2 ==== {{:Modus Ponendo Tollens/Sequent Form/Case 2}}


Modus Ponendo Tollens/Sequent Form/Case 1

:$\neg \paren {p \land q}, p \vdash \neg q$


Modus Ponendo Tollens/Sequent Form/Case 2

:$\neg \left({p \land q}\right), q \vdash \neg p$


Union of Subsets is Subset/Subset of Power Set

Let $S$ and $T$ be sets. Let $\powerset S$ be the power set of $S$. Let $\mathbb S$ be a subset of $\powerset S$. Then: :$\displaystyle \paren {\forall X \in \mathbb S: X \subseteq T} \implies \bigcup \mathbb S \subseteq T$


Conjunction with Negative Equivalent to Negation of Implication/Formulation 2/Forward Implication

:$\vdash \paren {p \land \neg q} \implies \paren {\neg \paren {p \implies q} }$


Neighborhood Sub-Basis Criterion for Filter Convergence

Let $\left({X, \tau}\right)$ be a topological space. Let $\mathcal F$ be a filter on $X$. Let $p \in X$. Then $\mathcal F$ converges to $p$ iff $\mathcal F$ contains as a subset a neighborhood sub-basis at $x$.


Rule of Material Implication/Formulation 2/Forward Implication

: $\vdash \left({p \implies q}\right) \implies \left({\neg p \lor q}\right)$


Implication is Left Distributive over Conjunction/Formulation 1/Proof 2

:$p \implies \left({q \land r}\right) \dashv \vdash \left({p \implies q}\right) \land \left({p \implies r}\right)$


Compact Subspace of Linearly Ordered Space

Let $\struct {X, \preceq, \tau}$ be a linearly ordered space. Let $Y \subseteq X$ be a non-empty subset of $X$. Then $Y$ is a compact subspace of $\struct {X, \tau}$ {{iff}} both of the following hold: :$(1): \quad$ For every non-empty $S \subseteq Y$, $S$ has a supremum and an infimum in $X$. :$(2): \quad$ For every non-empty $S \subseteq Y$: $\sup S, \inf S \in Y$.


Order Topology is Hausdorff

Let $\struct {X, \preceq, \tau}$ be a linearly ordered space. Then $\struct {X, \tau}$ is a Hausdorff space.


Subset of Linearly Ordered Space which is Order-Complete and Closed but not Compact

Let $X = \left[{0 \,.\,.\, 1}\right) \cup \left({2 \,.\,.\, 3}\right) \cup \left\{{4}\right\}$. Let $\preceq$ be the ordering on $X$ induced by the usual ordering of the real numbers. Let $\tau$ be the $\preceq$ order topology on $X$. Let $Y = \left[{0 \,.\,.\, 1}\right) \cup \left\{{4}\right\}$. Let $\tau'$ be the $\tau$-relative subspace topology on $Y$. Then: :$\left({Y, \preceq}\right)$ is a complete lattice :$Y$ is closed in $X$ but: :$\left({Y, \tau'}\right)$ is not compact.


Rule of Explosion/Variant 2

:$\vdash \paren {p \land \neg p} \implies q$


Jordan Curve Theorem

Let $\gamma: \closedint 0 1 \to \R^2$ be a Jordan curve. Let $\Img \gamma$ denote the image of $\gamma$. Then $\R^2 \setminus \Img \gamma$ is a union of two disjoint connected components. Both components are open in $\R^2$, and both components have $\Img \gamma$ as their boundary. One component is bounded, and is called the interior of $\gamma$. The other component is unbounded, and is called the exterior of $\gamma$.


Heine-Borel Theorem/Dedekind Complete Space

Let $T = \left({X, \preceq, \tau}\right)$ be a Dedekind-complete linearly ordered space. Let $Y$ be a non-empty subset of $X$. Then $Y$ is compact {{iff}} $Y$ is closed and bounded in $T$.


Heine-Borel iff Dedekind Complete

Let $\struct {X, \preceq, \tau}$ be a linearly ordered space. Then $X$ is Dedekind complete {{iff}} every closed, bounded subset of $X$ is compact.


Jordan Polygon Theorem

Let $P$ be a polygon embedded in $\R^2$. Denote the boundary of $P$ as $\partial P$. Then, $\R^2 \setminus \partial P$ is a union of two connected components. Both components are open in $\R^2$. One component is bounded, and is called the interior of $P$. The other component is unbounded, and is called the exterior of $P$.


Jordan Polygon Interior and Exterior Criterion

Let $P$ be a polygon embedded in $\R^2$. Let $q \in \R^2 \setminus \partial P$, where $\partial P$ denotes the boundary of $P$. Let $\mathbf v \in \R^2 \setminus \left\{ {\mathbf 0}\right\}$ be a non-zero vector, and let $\mathcal L = \left\{ {q + s \mathbf v: s \in \R_{\ge 0} }\right\}$ be a ray with start point $q$. Let $N \left({q}\right) \in \N$ be the number of crossings between $\mathcal L$ and $\partial P$. Then: :$(1): \quad$ $q \in \operatorname{Int} \left({P}\right)$, iff $N \left({q}\right) \equiv 1 \pmod 2$ :$(2): \quad$ $q \in \operatorname{Ext} \left({P}\right)$, iff $N \left({q}\right) \equiv 0 \pmod 2$ Here, $\operatorname{Int} \left({P}\right)$ and $\operatorname{Ext} \left({P}\right)$ denote the interior and exterior of $\partial P$, when $\partial P$ is considered as a Jordan curve.


Closed Set in Linearly Ordered Space

Let $\struct {X, \preceq, \tau}$ be a linearly ordered space. Let $C$ be a subset of $X$. Then $C$ is closed in $X$ {{iff}} for all non-empty subsets $S$ of $C$: :If $s \in X$ is a supremum or infimum of $S$ in $X$, then $s \in C$.


Factor Principles/Disjunction on Left/Formulation 2

: $\vdash \left({p \implies q}\right) \implies \left({\left({r \lor p}\right) \implies \left ({r \lor q}\right)}\right)$


Factor Principles/Disjunction on Right/Formulation 2

: $\vdash \left({p \implies q}\right) \implies \left({\left({p \lor r}\right) \implies \left ({q \lor r}\right)}\right)$


Factor Principles/Disjunction on Right/Formulation 1

:$p \implies q \vdash \paren {p \lor r} \implies \paren {q \lor r}$


Factor Principles/Disjunction on Left/Formulation 1

: $p \implies q \vdash \left({r \lor p}\right) \implies \left ({r \lor q}\right)$


Disjunction of Implications

:$\vdash \paren {p \implies q} \lor \paren {q \implies r}$


Principle of Composition/Formulation 1/Forward Implication

:$\paren {p \implies r} \lor \paren {q \implies r} \vdash \paren {p \land q} \implies r$


Principle of Composition/Formulation 1/Reverse Implication

:$\paren {p \land q} \implies r \vdash \paren {p \implies r} \lor \paren {q \implies r}$


Principle of Composition/Formulation 2

:$\left({\left({p \implies r}\right) \lor  \left({q \implies r}\right)}\right) \iff \left({\left({p \land q}\right) \implies r}\right)$


Inversion Mapping on Topological Group is Homeomorphism

Let $T = \struct {G, \circ, \tau}$ be a topological group. Let $\phi: G \to G$ be the inversion mapping of $T$. Then $\phi$ is a homeomorphism.


De Morgan's Laws (Logic)/Conjunction of Negations/Formulation 2/Forward Implication

: $\left({\neg p \land \neg q}\right) \implies \left({\neg \left({p \lor q}\right)}\right)$


De Morgan's Laws (Logic)/Conjunction of Negations/Formulation 2/Reverse Implication

: $\left({\neg \left({p \lor q}\right)}\right) \implies \left({\neg p \land \neg q}\right)$


Equivalence of Definitions of Generalized Ordered Space/Definition 2 implies Definition 1

Let $\struct {S, \preceq, \tau}$ be a generalized ordered space by Definition 2: {{:Definition:Generalized Ordered Space/Definition 2}} Then $\struct {S, \preceq, \tau}$ is a generalized ordered space by Definition 1: {{:Definition:Generalized Ordered Space/Definition 1}}


Inverse Image of Convex Set under Monotone Mapping is Convex

Let $\left({X, \le}\right)$ and $\left({Y, \preceq}\right)$ be ordered sets. Let $f: X \to Y$ be a monotone mapping. Let $C$ be a convex subset of $Y$. Then $f^{-1} \left[{C}\right]$ is convex in $X$.


Conjunction with Law of Excluded Middle

:$\vdash p \iff \paren {p \land q} \lor \paren {p \land \neg q}$


Proof by Cases with Contradiction

:$\vdash p \iff \left({p \lor q}\right) \land \left({p \lor \neg q}\right)$


Ray is Convex

Let $\left({S, \preceq}\right)$ be an ordered set. Let $I$ be a ray, either open or closed. Then $I$ is convex in $S$.


Intersection of Convex Sets is Convex Set (Order Theory)

Let $\left({S, \preceq}\right)$ be an ordered set. Let $\mathcal C$ be a set of convex sets in $S$. Then $\displaystyle \bigcap \mathcal C$ is convex.


Interval of Ordered Set is Convex

Let $\struct {S, \preceq}$ be a totally ordered set. Let $I$ be an interval: be it open, closed, or half-open in $S$. Then $I$ is convex in $S$.


Upper and Lower Closures are Convex

Let $\left({S, \preceq}\right)$ be an ordered set. Let $a \in S$. Then $a^\succeq$, $a^\succ$, $a^\preceq$, and $a^\prec$ are convex in $S$.