hcju/proofwikips huggingface dataset:
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
: $\vdash \neg \neg p \implies p$
Double Negation/Double Negation Elimination
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
:$p \implies q \dashv \vdash \neg q \implies \neg p$
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 $D$ : :$u$ 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 $D$ : : $u$ 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 $D$ : : $u$ 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
: $p \implies \left({q \implies r}\right) \vdash \left({p \implies q}\right) \implies \left({p \implies r}\right)$
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: :$a \preceq' b$  $a \vee b = b$ 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
: $p \implies \left({q \land r}\right) \vdash \left({p \implies q}\right) \land \left({p \implies r}\right)$
Implication is Left Distributive over Conjunction/Reverse Implication/Formulation 1
: $\left({p \implies q}\right) \land \left({p \implies r}\right) \vdash p \implies \left({q \land r}\right)$
Implication is Left Distributive over Conjunction/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
: $\left({p \implies r}\right) \land \left({q \implies r}\right) \vdash \left({p \lor q}\right) \implies r$