hcju/proofwikiqa huggingface dataset:
When $1_R$ denotes the unity of $R$, we have: {{begin-eqn}} {{eqn | l = 1_R       | r = \map \det {\mathbf I_n}       | c = Determinant of Unit Matrix }} {{eqn | r = \map \det {\mathbf A \mathbf B}       | c = by assumption }} {{eqn | r = \map \det {\mathbf A} \map \det {\mathbf B}       | c = Determinant of Matrix Product }} {{end-eqn}}  From Matrix is Invertible iff Determinant has Multiplicative Inverse, it follows that $\mathbf A$ and $\mathbf B$ are invertible. Then: {{begin-eqn}} {{eqn | l = \mathbf B       | r = \mathbf I_n \mathbf B       | c = Unit Matrix is Unity of Ring of Square Matrices }} {{eqn | r = \paren {\mathbf A^{-1} \mathbf A} \mathbf B       | c = {{Defof|Inverse Matrix}} }} {{eqn | r = \mathbf A^{-1} \paren {\mathbf A \mathbf B}       | c = Matrix Multiplication is Associative }} {{eqn | r = \mathbf A^{-1} \mathbf I_n       | c = by assumption }} {{eqn | r = \mathbf A^{-1}       | c = Unit Matrix is Unity of Ring of Square Matrices }} {{end-eqn}} {{qed}}
By definition, $\displaystyle \prod \O$ is the limit of the empty subcategory $\mathbf 0$ of $\mathbf C$. The result follows from Terminal Object as Limit. {{qed}} Category:Limits and Colimits Category:Category Theory lfph52fpugtjevxom96agpp4cyf6qnn
Let $X$ be a normed vector space. Denote by $B$ the closed unit ball in $X$. Let $X^*$ be the dual of $X$. Denote by $B^*$ the closed unit ball in $X^*$. Let $\map {\mathcal F} B = \closedint {-1} 1^B$ be the topological space of functions from $B$ to $\closedint {-1} 1$. By Tychonoff's Theorem, $\map {\mathcal F} B$ is compact with respect to the product topology. We define the restriction map: :$R: B^*\to \map {\mathcal F} B$ by $\map R \psi = \psi \restriction_B$. === Lemma 1 === $\map R {B^*}$ is a closed subset of $\map {\mathcal F} B$. === Proof === {{ProofWanted}} {{qed|lemma}} === Lemma 2 === $R$ is a homeomorphism from $B^*$ with the weak* topology to its image $\map R {B^*}$ seen as a subset of $\map {\mathcal F} B$ with the product topology. === Proof === {{ProofWanted}} {{qed|lemma}} Thus by lemma 2, $B^*$ in the weak* topology is homeomorphic with $\map R {B^*}$. This is a closed subset of $\map {\mathcal F} B$ (by lemma 1) and thus compact. {{qed}} {{Namedfor|Leonidas Alaoglu|cat = Alaoglu}}
We apply the Method of Truth Tables. $\begin{array}{|cccc||c|} \hline p & \land & \neg & p & \bot \\ \hline F & F & T & F & F \\ T & F & F & T & F \\ \hline \end{array}$ As can be seen by inspection, the truth value of the main connective, that is $\land$, is $F$ for each boolean interpretation for $p$. {{qed}}
Let $\mathbf a = \begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix}$, and $\mathbf b = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix}$. Then the dot product of $\mathbf a$ and $\mathbf a \times \mathbf b$ is: {{begin-eqn}} {{eqn | l = \mathbf a \cdot \left({\mathbf a \times \mathbf b}\right)       | r = a_1 \left({a_2 b_3 - a_3 b_2}\right) + a_2 \left({a_3 b_1 - a_1 b_3}\right) + a_3 \left({a_1 b_2 - a_2 b_1}\right)       | c = {{Defof|Dot Product}} and {{Defof|Vector Cross Product}} }} {{eqn | r = a_1 a_2 b_3 - a_1 a_3 b_2 + a_2 a_3 b_1 - a_1 a_2 b_3 + a_1 a_3 b_2 - a_2 a_3 b_1 }} {{eqn | r = 0 }} {{end-eqn}} Since the dot product is equal to zero, the vectors are orthogonal by definition. Similarly, $\mathbf b$ and $\mathbf a \times \mathbf b$ are orthogonal: {{begin-eqn}} {{eqn | l = \mathbf b \cdot \left({\mathbf a \times \mathbf b}\right)       | r = b_1 \left({a_2 b_3 - a_3 b_2}\right) + b_2 \left({a_3 b_1 - a_1 b_3}\right) + b_3 \left({a_1 b_2 - a_2 b_1}\right) }} {{eqn | r = a_2 b_1 b_3 - a_3 b_1 b_2 + a_3 b_1 b_2 - a_1 b_2 b_3 + a_1 b_2 b_3 - a_2 b_1 b_3 }} {{eqn | r = 0 }} {{end-eqn}} {{qed}}
Let $\epsilon > 0$. Since $r$ converges to $0$, there exists an $N$ such that $n > N \implies r_n < \epsilon$. But then by Extended Transitivity, $n > N \implies d \left({p, x_n}\right) < \epsilon$. Thus $\left\langle{x_n}\right\rangle$ converges to $p$. {{qed}} Category:Metric Spaces Category:Limits of Sequences ttjzk3l7syew0kp9sci703bxd57re3f
Let $m \in S$ and $m < p$. Then $\left\langle{x_n}\right\rangle$ eventually succeeds $m$. Thus by Extended Transitivity, $\left\langle{y_n}\right\rangle$ eventually succeeds $m$. A similar argument using $\left\langle{z_n}\right\rangle$ proves the dual statement. Thus $\left\langle{y_n}\right\rangle$ is eventually in each ray containing $p$, so it converges to $p$. {{qed}} Category:Order Topology Category:Limits of Sequences skzmfnjs6c5wl2fah2fqawkhp58tt2k
Let $q \in S$ such that $q < p$. We will show that $F_2$ has an element which is a subset of $q^\ge$. {{explain|$q^\ge$ is the upper closure in what set?}} Since $F_1$ converges to $p$, it has an element: :$A \subseteq q^\ge$. Thus there is an element $k$ in $A$ and an element $M$ in $F_2$ such that all elements of $M$ succeed $k$. Then by Extended Transitivity: :$M \subseteq q^\ge$ A similar argument using $F_3$ proves the dual statement. Thus $F_2$ converges to $p$. {{explain}} {{qed}} Category:Filter Theory Category:Order Topology Category:Convergence ia5fwo0pwzub1pc2obn184j9ojlmykf
=== Necessary Condition === When $\theta$ denotes the angle between $\mathbf u$ and $\mathbf v$ measured in radians, we have: {{begin-eqn}} {{eqn | l = 0       | r = \mathbf u \cdot \mathbf v       | c = {{Defof|Orthogonal Vectors}} }} {{eqn | r = \left\Vert{\mathbf u}\right\Vert \left\Vert{\mathbf v}\right\Vert \cos \theta       | c = Cosine Formula for Dot Product }} {{end-eqn}} where $\left\Vert{\cdot}\right\Vert$ denotes the Euclidean norm. As $\mathbf u, \mathbf v$ are non-zero vectors, it follows from the norm axiom of positive definiteness that $\left\Vert{\mathbf u}\right\Vert \ne 0$, and $\left\Vert{\mathbf v}\right\Vert \ne 0$. Then $\cos \theta = 0$. It follows from Zeroes of Sine and Cosine that $\theta = \dfrac \pi 2$, given that $\theta \in \left[{0\,.\,.\,\pi}\right]$ by the convention of the definition of angle between vectors. Then $\mathbf u$ and $\mathbf v$ are perpendicular by the measurement of a right angle. {{qed|lemma}} === Sufficient Condition === If $\mathbf u$ and $\mathbf v$ are perpendicular, then the angle between them measures $\dfrac \pi 2$ in radians. Then $\cos \theta = 0$, and the calculations above show that $\mathbf u \cdot \mathbf v = 0$. Hence, $\mathbf u$ and $\mathbf v$ are orthogonal. {{qed}} Category:Linear Algebra Category:Analytic Geometry ojyl8da3qmaktopioezmvwcsytw6juf
This proof depends on the limit definition of the exponential function. Let: :$\displaystyle \map \exp x = \lim_{n \mathop \to \infty} \paren {1 + \dfrac x n}^n$ Fix $x_0 \in \R$. Consider $I := \closedint {x_0 - 1} {x_0 + 1}$. From Closed Bounded Subset of Real Numbers is Compact, $I$ is compact. From Exponential Sequence is Uniformly Convergent on Compact Sets: :$\paren {1 + \dfrac x n}^n$ is uniformly convergent on $I$. By the Uniform Limit Theorem: :$\displaystyle \lim_{n \mathop \to \infty} \paren {1 + \dfrac x n}^n = \exp$ is continuous on $I$. In particular, $\map \exp x$ is continuous at $x_0$. {{qed}}
This proof depends on the definition of the exponential function as the function inverse of the natural logarithm. From Logarithm is Strictly Increasing, $\ln$ is strictly monotone on $\R_{>0}$. From Natural Logarithm Function is Continuous, $\ln$ is continuous on $\R_{>0}$ Thus, from the Continuous Inverse Theorem, $\exp := \ln^{-1}$ is continuous. {{qed}}
This proof depends on the differential equation definition of the exponential function. The result follows from Differentiable Function is Continuous. {{qed}}
This proof depends on the differential equation definition of the exponential function. The result follows from Complex-Differentiable Function is Continuous. {{qed}} Category:Exponential Function is Continuous 445lwmth2kcf596w3ms1rmazvjrik3d
Suppose that $\RR$ is not antireflexive. Then there is some $x \in G$ such that $x \mathrel \RR x$. Let $y \in G$. Then by the definition of compatibility: :$\paren {x \circ \paren {x^{-1} \circ y} } \mathrel \RR \paren {x \circ \paren {x^{-1} \circ y} }$ By {{GroupAxiom|1}} and {{GroupAxiom|3}}: :$y \mathrel \RR y$ Since this holds for arbitrary $y$, $\RR$ is reflexive. Thus $\mathcal R$ is either reflexive or antireflexive. {{qed}} {{explain|While the above demonstrates that if such a relation is not antireflexive it must be reflexive, it does not approach the task of demonstrating that such a relation can actually *be* antireflexive.}} Category:Compatible Relations Category:Group Theory ii85i0461niyz1id47zcu28phlmvv0h
{{improve|Reword so as to be less wordy. Add pictures so as to be easier to visualize.}} By the definition of the order topology, the sets of open rays in $\left({S, \preceq}\right)$ and $\left({A, \preceq}\right)$ form sub-bases for $\tau$ and $\upsilon$, respectively. By Sub-Basis for Topological Subspace, we need only show that the set of open rays in $\left({S, \preceq}\right)$ induces a subbase for $\upsilon$. Specifically, we will show that each open ray in $A$ is the intersection of $A$ with an open ray in $S$, and we will show that the intersection of any open ray in $S$ with $A$ is either an open ray in $A$, the empty set, or $A$. By Dual Pairs (Order Theory), strict upper closure is dual to strict lower closure, so by the duality principle we need show this only for the upward-pointing rays. For each $p \in S$, let $p^{\succ S}$ be the strict upper closure of $p$ in $S$. For each $p \in A$, let $p^{\succ A}$ be the strict upper closure of $p$ in $A$. Let $p \in A$. Then by Strict Upper Closure in Restricted Ordering: :$p^{\succ A} = A \cap p^{\succ S}$ Let $q \in S$. If $q \in A$, then $A \cap q^{\succ S} = q^{\succ A}$, so $A \cap q^{\succ S}$ is an [[Definition:Open Ray|open ray] in $A$. Suppose instead that $q \notin A$. If $A \cap q^{\succ S} = \varnothing$, the proof is complete. Otherwise, $A \cap q^{\succ S}$ must have some element $x$. Suppose for the sake of contradiction that there is a $y \in A \setminus \left({ A \cap q^{\succ S} }\right) = A \cap q^{\preceq S}$. Then since $q \notin A$, it follows that $y \prec q$. Thus $y \prec q \prec x$, $y, x \in A$, and $q \notin A$, contradicting the fact that $A$ is convex. Thus $A \cap q^{\succ S} = A$. {{qed}} Category:Order Topology Category:Topological Subspaces 27m938ooaqvwwhgxxqyy2t2qkuhvld9
From Condition for Edge to be Bridge, $e$ is a bridge iff $e$ does not lie on any circuit. Since $T$ is a tree, there are no circuits in $T$. The result follows. {{qed}} Category:Tree Theory d6vql0l2gsxkaymoedoq8ja7dk2y05u
Let $B$ be the basis for $\tau$ generated by $K$. By Basis for Topological Subspace, $B$ generates a basis, $B\,'$, for $\tau'$. We will show that $K'$ generates $B\,'$. Let $V' \in B\,'$. Then for some $V \in B$: :$V' = S \cap V$ By the definition of a sub-basis, there is a finite subset $K_V$ of $K$ such that: :$\displaystyle V = \bigcap K_V$ Let $K_V' = \left\{{S \cap U: U \in K_V }\right\}$. Then: :$\displaystyle V' = S \cap V = S \cap \bigcap K_V$ so: :$\displaystyle V' = \bigcap_{P \mathop \in K_V} \left({S \cap P}\right) = \bigcap K_V'$ Thus $B\,'$ is generated by $K'$. {{qed}} Category:Topological Bases Category:Topological Subspaces cwtvjt3c1l9iu0xs6z0bd8w714lclmo
We apply the Method of Truth Tables. $\begin{array}{|c|ccccc|} \hline p & (p & \implies & q) & \implies & q\\ \hline F & F & T & F & F & F \\ F & F & T & T & T & T \\ T & T & F & F & T & F \\ T & T & T & T & T & T \\ \hline \end{array}$ As can be seen by inspection, when $p$ is true, the value of the main connective is also true. {{qed}}
We apply the Method of Truth Tables. $\begin{array}{|c|c|ccccc|} \hline p & \implies & ((p & \implies & q) & \implies & q)\\ \hline F & T & F & T & F & F & F \\ F & T & F & T & T & T & T \\ T & T & T & F & F & T & F \\ T & T & T & T & T & T & T \\ \hline \end{array}$ As can be seen by inspection, the main connective is true throughout. {{qed}}
{{BeginTableau|\vdash \left({p \implies q}\right) \implies \left({\left({p \land r}\right) \implies \left ({q \land r}\right)}\right)}} {{Assumption|1|p \implies q}} {{SequentIntro|2|1|\left({p \land r}\right) \implies \left ({q \land r}\right)|1   |Factor Principles: Conjunction on Right: Formulation 1}} {{Implication|3|1|\left({p \implies q}\right) \implies \left({\left({p \land r}\right) \implies \left ({q \land r}\right)}\right)|1|2}} {{EndTableau}} {{qed}}
Let $p_1 ,p_2 \in U$. Then for $i \in \left\{ {1, 2}\right\}$, $p_i = p + s_i t_i \mathbf v + \left({1 - s_i}\right) t_i \mathbf w$ for some $s_i \in \left({0\,.\,.\,1}\right) , t_i \in \R_{>0}$. WLOG assume that $t_1 \le t_2$. Suppose that $q \in \R^2$ lies on the line segment joining $p_1$ and $p_2$, so: {{begin-eqn}} {{eqn |l= q       |r= p + s_1 t_1 \mathbf v + \left({1 - s_1}\right) t_1 \mathbf w + s \left({ p + s_2 t_2 \mathbf v + \left({1 - s_2}\right) t_2 \mathbf w - p - s_1 t_1 \mathbf v - \left({1 - s_1}\right) t_1 \mathbf w }\right)       |c= for some $s \in \left({0\,.\,.\,1}\right)$ }} {{eqn |r= p + \left({ \left({1 - s}\right) s_1 t_1 + s s_2 t_2}\right) \mathbf v +  \left({ \left({1 - s}\right) \left({1 - s_1}\right) t_1 + s \left({1 - s_2}\right) t_2}\right) \mathbf w }} {{eqn |r= p + \dfrac{\left({1 - s}\right) s_1 t_1 + s s_2 t_2}{r}r \mathbf v + \dfrac{t_1 + st_2 - st_1 - \left({1 - s}\right) s_1 t_1 - s s_2 t_2}{r} r \mathbf w       |c= where $r = t_1 + s \left({t_2 - t_1}\right)$ }} {{eqn |r= p + \dfrac{\left({1 - s}\right) s_1 t_1 + s s_2 t_2}{r}r \mathbf v + \left({ 1 - \dfrac{\left({1 - s}\right) s_1 t_1 + s s_2 t_2}{r} }\right) r \mathbf w }} {{end-eqn}} As $t_1 \le t_2$, it follows that $r \in \R_{>0}$. We have $\dfrac{ \left({1 - s}\right) s_1 t_1 + s s_2 t_2}{r}> 0$, and: {{begin-eqn}} {{eqn |l= 1 - \dfrac{\left({1 - s}\right) s_1 t_1 + s s_2 t_2}{r}       |r= \dfrac{ \left({1 - s}\right) \left({1 - s_1}\right) t_1 + s \left({1 - s_2}\right) t_2}{r} }} {{eqn |o= >       |r= 0 }} {{end-eqn}} It follows that $\dfrac{ \left({1 - s}\right) s_1 t_1 + s s_2 t_2}{r} \in \left({0\,.\,.\,1}\right)$. Then $q \in U$. By definition of convex set, it follows that $U$ is convex. {{qed}} Category:Vector Spaces 9bmr4r3gkhrw6nlhcxeeovr0al1t8ls
{{BeginTableau|\vdash \paren {p \implies q} \implies \paren {\paren {r \land p} \implies \paren {r \land q} } }} {{Assumption|1|p \implies q}} {{SequentIntro|2|1|\paren {\paren {r \land p} \implies \paren {r \land q} }|1   |Factor Principles: Conjunction on Left: Formulation 1}} {{Implication|3|1|\paren {p \implies q} \implies \paren {\paren {r \land p} \implies \paren {r \land q} }|1|2}} {{EndTableau}} {{qed}}
Denote the triangle as $\triangle$, and the interior of the boundary of $\triangle$ as $\Int \triangle$. From Boundary of Polygon is Jordan Curve, it follows that the boundary of $\triangle$ is equal to the image of a Jordan curve, so $\Int \triangle$ is well-defined. Denote the vertices of $\triangle$ as $A_1, A_2, A_3$. For $i \in \set {1, 2, 3}$, put $j = i \bmod 3 + 1$, $k = \paren {i + 1} \bmod 3 + 1$, and: :$U_i = \set {A_i + s t \paren {A_j - A_i} + \paren {1 - s} t \paren {A_k - A_i} : s \in \openint 0 1, t \in \R_{>0} }$ Suppose that the angle $\angle A_i$ between is $A_j - A_i$ and $A_k - A_i$ is non-convex. As $\angle A_i$ is an internal angle in $\triangle$, it follows from definition of polygon that $\angle A_i$ cannot be zero or straight. Then $\angle A_i$ is larger than a straight angle, which is impossible by Sum of Angles of Triangle Equals Two Right Angles. It follows that $\angle A_i$ is convex. From Characterization of Interior of Triangle, it follows that: :$\displaystyle \Int \triangle = \bigcap_{i \mathop = 1}^3 U_i$ From Interior of Convex Angle is Convex Set, it follows for $i \in \set {1, 2, 3}$ that $U_i$ is a convex set. The result now follows from Intersection of Convex Sets is Convex Set (Vector Spaces). {{qed}} Category:Vector Spaces 62yculhuqp577pbsv8offiogbpmwcf4
Let $\mathbf a = \begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix}$, and $\mathbf b = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix}$. Then: {{begin-eqn}} {{eqn | l = \left\Vert{ \mathbf a \times \mathbf b }\right\Vert^2       | r = \left({ \mathbf a \times \mathbf b }\right) \cdot \left({ \mathbf a \times \mathbf b }\right)       | c = {{Defof|Euclidean Norm}} }} {{eqn | r = \begin{bmatrix} a_2 b_3 - a_3 b_2 \\ a_3 b_1 - a_1 b_3 \\ a_1 b_2 - a_2 b_1 \end{bmatrix} \cdot \begin{bmatrix} a_2 b_3 - a_3 b_2 \\ a_3 b_1 - a_1 b_3 \\ a_1 b_2 - a_2 b_1 \end{bmatrix}       | c = {{Defof|Vector Cross Product}} }} {{eqn | r = a_2^2 b_3^2 + a_3^2 b_2^2 - 2a_2 a_3 b_2 b_3 + a_3^2 b_1^2 + a_1^2 b_3^2 - 2a_1 a_3 b_1 b_3 + a_1^2 b_2^2 + a_2^2 b_1^2 - 2a_1 a_2 b_1 b_2       | c = {{Defof|Dot Product}} }} {{eqn | r = \left({a_1^2 + a_2^2 + a_3^2}\right) \left({b_1^2 + b_2^2 + b_3^2}\right) - \left({a_1 b_1 + a_2 b_2 + a_3 b_3}\right)^2       | c = by algebraic manipulations }} {{eqn | r = \left\Vert{\mathbf a}\right\Vert^2 \left\Vert{\mathbf b}\right\Vert^2 - \left({\mathbf a \cdot \mathbf b}\right)^2 }} {{end-eqn}} This proves $(1)$. {{qed|lemma}} If $\mathbf a$ or $\mathbf b$ is the zero vector, then $\left\Vert{\mathbf a}\right\Vert = 0$, or $\left\Vert{\mathbf b}\right\Vert = 0$ by the positive definiteness norm axiom. By calculation, it follows that $\mathbf a \times \mathbf b$ is also the zero vector, so $\left\Vert{\mathbf a \times \mathbf b}\right\Vert = 0$. Hence, equality $(2)$ holds. If both $\mathbf a$ or $\mathbf b$ are non-zero vectors, we continue the calculations from the first section: {{begin-eqn}} {{eqn | l = \left\Vert{ \mathbf a \times \mathbf b }\right\Vert^2       | r = \left\Vert{\mathbf a}\right\Vert^2 \left\Vert{\mathbf b}\right\Vert^2 - \left({\mathbf a \cdot \mathbf b}\right)^2 }} {{eqn | r = \left\Vert{\mathbf a}\right\Vert^2 \left\Vert{\mathbf b}\right\Vert^2 - \left\Vert{\mathbf a}\right\Vert^2 \left\Vert{\mathbf b}\right\Vert^2 \cos^2 \theta        | c = Cosine Formula for Dot Product }} {{eqn | r = \left\Vert{\mathbf a}\right\Vert^2 \left\Vert{\mathbf b}\right\Vert^2 \left({1 - \cos^2 \theta}\right) }} {{eqn | r = \left\Vert{\mathbf a}\right\Vert^2 \left\Vert{\mathbf b}\right\Vert^2 \sin^2 \theta       | c = Sum of Squares of Sine and Cosine }} {{end-eqn}} Equality $(2)$ now follows after taking the square root of both sides of the equality. This is possible as Square of Real Number is Non-Negative. {{qed}}
By Subset Relation is Compatible with Subset Product, $\subseteq$ is compatible with $\circ_\PP$. By Subset Relation is Transitive, $\subseteq$ is transitive. Thus the theorem holds by Operating on Transitive Relationships Compatible with Operation. {{qed}} Category:Subset Products Category:Compatible Relations n5123gorw863wboo6opdwomth1tut18
This follows from Subset Relation is Compatible with Subset Product and the definition of the subset product with a singleton. {{qed}} Category:Subset Products hwov202p0rmiee3qlizogofinxbtbx2
=== Definition 1 is Equivalent to Definition 2 === {{:Equivalence of Definitions of Normal Subset/1 iff 2}} === Definition 2 Implies Definition 3 === {{:Equivalence of Definitions of Normal Subset/2 implies 3}} === Definition 3 is Equivalent to Definition 4 === {{:Equivalence of Definitions of Normal Subset/3 iff 4}} === Definitions 3 and 4 imply Definition 2 === {{:Equivalence of Definitions of Normal Subset/3 and 4 imply 2}} === Definition 3 is Equivalent to Definition 5 === {{:Equivalence of Definitions of Normal Subset/3 iff 5}} {{ProofWanted|Include equivalence to definitions 6 and 7}} Category:Group Theory Category:Normal Subsets 4h2rm7ld8cz7o0zhts6ac7jyx5zelyj
Let $e$ be the identity of $G$. First note that: :$(4): \quad \left({\forall g \in G: g \circ S \circ g^{-1} = S}\right) \iff \left({\forall g \in G: g^{-1} \circ S \circ g = S}\right)$ which is shown by, for example, setting $h := g^{-1}$ and substituting. === Necessary Condition === Suppose that $S$ satisfies $(1)$. Then: {{begin-eqn}} {{eqn | lo= \forall g \in G:       | l = g \circ S       | r = S \circ g       | c = $(1)$ }} {{eqn | ll= \implies       | l = \left({g \circ S}\right) \circ g^{-1}       | r = \left({S \circ g}\right) \circ g^{-1}       | c =  }} {{eqn | ll= \implies       | l = g \circ S \circ g^{-1}       | r = S \circ \left({g \circ g^{-1} }\right)       | c = Subset Product within Semigroup is Associative: Corollary }} {{eqn | ll= \implies       | l = g \circ S \circ g^{-1}       | r = S \circ e       | c = {{Defof|Inverse Element}} }} {{eqn | ll= \implies       | l = g \circ S \circ g^{-1}       | r = S       | c = Subset Product by Identity Singleton }} {{eqn | lo= \forall g \in G:       | l = g^{-1} \circ S \circ g       | r = S       | c = $(4)$ }} {{end-eqn}} {{qed|lemma}} === Sufficient Condition === Let $S$ be a subset of $G$ such that: : $\forall g \in G: g \circ S \circ g^{-1} = S$ or: : $\forall g \in G: g^{-1} \circ S \circ g = S$ By $(4)$, as long as one of these statements holds, the other one holds as well. Then: {{begin-eqn}} {{eqn | lo= \forall g \in G:       | l = g \circ S \circ g^{-1}       | r = S       | c =  }} {{eqn | ll= \implies       | l = \left({g \circ S \circ g^{-1} }\right) \circ g       | r = S \circ g       | c =  }} {{eqn | ll= \implies       | l = \left({g \circ S}\right) \circ \left({g^{-1} \circ g}\right)       | r = S \circ g       | c = Subset Product within Semigroup is Associative: Corollary }} {{eqn | ll= \implies       | l = \left({g \circ S}\right) \circ e       | r = S \circ g       | c = {{Defof|Inverse Element}} }} {{eqn | ll= \implies       | l = g \circ S       | r = S \circ g       | c = Subset Product by Identity Singleton }} {{end-eqn}} {{qed}}
We have that: :$\left({\forall g \in G: g \circ S \circ g^{-1} = S}\right) \iff \left({\forall g \in G: g^{-1} \circ S \circ g = S}\right)$ The result follows by definition of set equality. {{qed}}
By Subset Relation is Compatible with Subset Product, the subset relation $\subseteq$ is compatible with $\circ_{\mathcal P}$. From Inverse of Subset Relation is Superset, the inverse of $\subseteq$ is $\supseteq$. The result follows from Inverse of Relation Compatible with Operation is Compatible. {{qed}} Category:Compatible Relations 0ld0w22afe9z6m2t6oiestk41kajtld
First note that: :$(5): \quad \left({\forall g \in G: g \circ S \circ g^{-1} \subseteq S}\right) \iff \left({\forall g \in G: g^{-1} \circ S \circ g \subseteq S}\right)$ :$(6): \quad \left({\forall g \in G: S \subseteq g \circ S \circ g^{-1}}\right) \iff \left({\forall g \in G: S \subseteq g^{-1} \circ S \circ g}\right)$ which is shown by, for example, setting $h := g^{-1}$ and substituting. Therefore: : conditions $(1)$ and $(2)$ are equivalent and: : conditions $(3)$ and $(4)$ are equivalent. It remains to be shown that condition $(1)$ is equivalent to condition $(3)$. Suppose that $(1)$ holds. Then: {{begin-eqn}} {{eqn | lo= \forall g \in G:       | l = g \circ S \circ g^{-1}       | o = \subseteq       | r = S }} {{eqn | l = g^{-1} \circ \left({ g \circ S \circ g^{-1} }\right)       | o = \subseteq       | r = g^{-1} \circ S       | c = Subset Relation is Compatible with Subset Product/Corollary 2 }} {{eqn | l = S \circ g^{-1}       | o = \subseteq       | r = g^{-1} \circ S       | c = Subset Product within Semigroup is Associative/Corollary and the definition of inverse }} {{eqn | l = \left({ S \circ g^{-1} }\right) \circ g       | o = \subseteq       | r = \left({ g^{-1} \circ S }\right) \circ g       | c = Subset Relation is Compatible with Subset Product/Corollary 2 }} {{eqn | l = S       | o = \subseteq       | r = g^{-1} \circ S \circ g       | c = Subset Product within Semigroup is Associative/Corollary and the definition of inverse }} {{end-eqn}} Thus condition $(1)$ implies condition $(3)$. The exact same argument, substituting $\supseteq$ for $\subseteq$ and using Superset Relation is Compatible with Subset Product instead of Subset Relation is Compatible with Subset Product proves that $(3)$ implies $(1)$. {{explain|the necessary corollary is currently only available in subset form, but these arguments are all the same.  Alternatively, the exact same method can be used to move the gs from the right to the left as was used to move them from the left to the right.}} {{qed}}