hcju/proofwikiqa huggingface dataset:
Using the One-Step Subgroup Test: : $(1): \quad e \in \set e \leadsto \set e \ne \O$ : $(2): \quad e \in \set e \leadsto e \circ e^{-1} = e \in \set e$ {{qed}}
First we note that the trivial group $\struct {\set e, \circ}$ is a subgroup of $G$. To show $\struct {\set e, \circ}$ is normal in $G$: :$\forall a \in G: a \circ e \circ a^{-1} = a \circ a^{-1} = e$ Hence the result. {{qed}}
First we note that $\struct {G, \circ}$ is a subgroup of itself. To show $\struct {G, \circ}$ is normal in $G$: :$\forall a, g \in G: a \circ g \circ a^{-1} \in G$ as $G$ is closed by definition. Hence the result. {{qed}}
The proof proceeds by Induction on Well-Formed Formulas of $P$. $P$ must be of the form: :$x \in y$ :$\left({Q \land R}\right)$ :$\neg Q$ or: :$\forall x: Q$ for some propositions $Q$ and $R$. === Atoms === Let $P$ be of the form $x \in y$. Then: :$A = \left\{ {x, y}\right\}$ By definition of standard structure: :$B \models P \iff \left({x \in y \land x \in B \land y \in B}\right)$ By definition of relativisation: :$\displaystyle P^B \iff x \in y$ If $A \subseteq B$, then: :$x \in B \land y \in B$ and the two statements are equivalent. === Inductive Step for $\neg$ === Let $P$ be of the form $\neg Q$ and that the statement holds for $Q$. Then the free variables in $Q$ are precisely those in $P$. {{begin-eqn}} {{eqn | l = A \subseteq B       | o = \implies       | r = \left({B \models Q \iff Q^B}\right)       | c = Inductive Hypothesis }} {{eqn | o = \implies       | r = \left({\neg B \models Q \iff \neg Q^B}\right)       | c =  }} {{eqn | o = \implies       | r = \left({B \models P \iff P^B}\right)       | c = Definition of $P$ }} {{end-eqn}} === Inductive Step for $\implies$ === Suppose $P$ is of the form $\left({Q \implies R}\right)$. Suppose, further, that the statement holds for $Q$ and $R$. The free variables of $Q$ and $R$ have to all be members of $A$, and thus are members of $B$ since $A \subseteq B$. {{begin-eqn}} {{eqn | l = A \subseteq B       | o = \implies       | r = \left({B \models Q \iff Q^B}\right) \land \left({B \models R \iff R^B}\right)       | c = Inductive Hypothesis }} {{eqn | o = \implies       | r = \left({\left({B \models Q \land B \models R}\right) \iff \left({Q^B \land R^B}\right)}\right)       | c =  }} {{eqn | o = \implies       | r = \left({B \models P \iff P^B}\right)       | c = Definitions of Standard Structure and Relativisation }} {{end-eqn}} === Inductive Step for $\forall x:$ === Let $P$ be of the form: :$\forall x: Q$ Let the statement hold for $Q$. Then the free variables of $Q$ are either in $A$ or $x$. Furthermore, $x \notin A$ because: :$x \in A \implies x$ is a free variable in $\forall x: Q$ which is a contradiction. {{begin-eqn}} {{eqn | l = A \subseteq B \land x \in B       | o = \implies       | r = \left({B \models Q \iff Q^B}\right)       | c = Inductive Hypothesis }} {{eqn | l = A \subseteq B       | o = \implies       | r = \left({\left({x \in B \implies B \models Q}\right) \iff \left({x \in B \implies Q^B}\right)}\right)       | c = Propositional logic manipulation }} {{eqn | o = \implies       | r = \left({\forall x \in B: B \models Q \iff \forall x \in B: Q^B}\right)       | c = Universal Generalization }} {{eqn | o = \implies       | r = \left({B \models P \iff P^B}\right)       | c = Definitions of Standard Structure and Relativisation }} {{end-eqn}} {{qed}}
By definition of bottom-up grammar, the well-formed formulas of $\mathcal F$ comprise: :letters of $\mathcal F$; :expressions resulting from rules of formation. Either case is dealt with by the assumptions on $\Phi$. Hence the result, from Proof by Cases. {{qed}} Category:Formal Systems 130mstbdlgv4psjvlwcp6wj7c3bb7t1
By the Axiom of Extension: : $a = b \implies \left({a \in x \implies b \in x}\right)$ Equality is Symmetric, so also by the Axiom of Extension: : $a = b \implies \left({b \in x \implies a \in x}\right)$ {{qed}}
$y$ is a limit ordinal and thus is nonzero, by definition. $x$ and $y$ are both nonzero. So by Ordinals have No Zero Divisors: :$\left({ x \times y }\right) \ne 0$ So by definition of limit ordinal: : $\left({ x \times y }\right) \in K_{II} \lor \exists z \in \operatorname{On}: \left({ x \times y }\right) = z^+$ Suppose that $\left({ x \times y }\right) = z^+$ for some ordinal $z$. {{begin-eqn}} {{eqn | l = \left({ x \times y }\right)       | r = \bigcup_{w \mathop \in y} \left({ x \times w }\right)       | c = Definition of Ordinal Multiplication }} {{end-eqn}} It follows that: {{begin-eqn}} {{eqn | l = z \in \bigcup_{w \mathop \in y} \left({ x \times w }\right)       | o = \implies       | r = \exists w \in y: z \in \left({ x \times w }\right)       | c = Definition of Set Union }} {{eqn | o = \implies       | r = \exists w \in y: z^+ \in \left({ x \times w }\right)^+       | c = Successor is Less than Successor }} {{eqn | l = \left({ x \times w }\right)^+       | r = \left({ x \times w }\right) + 1       | c = Ordinal Addition by One }} {{eqn | o = \implies       | r = \exists w \in y: z^+ \in \left({ \left({ x \times w }\right) + 1 }\right)       | c =  }} {{end-eqn}} But by Subset is Right Compatible with Ordinal Addition: : $\left({x \times w}\right) + 1 \subseteq \left({x \times w}\right) + x$ Therefore: {{begin-eqn}} {{eqn | ll = \exists w \in y:       | l = z^+        | o = \in       | r = \left({ x \times w }\right) + x       | c =  }} {{eqn | r = \left({ x \times w^+ }\right)       | c = Definition of Ordinal Multiplication }} {{end-eqn}} But by Successor in Limit Ordinal: : $w^+ \in y$ Therefore: :$z^+ \in \left({ x \times y }\right)$ contradicting the fact that $z^+ = \left({x \times y}\right)$. Thus: :$z^+ \ne \left({x \times y}\right)$ and: :$\left({x \times y}\right) \in K_{II}$ {{qed}}
{{begin-eqn}} {{eqn | l = z < \left({ x \times y }\right)       | o = \iff       | r = z < \bigcup_{w \mathop \in y} \left({ x \times w }\right)       | c = Definition of Ordinal Multiplication }} {{eqn | o = \iff       | r = \exists w \in y: z < \left({x \times w}\right)       | c = Definition of Set Union }} {{end-eqn}} {{qed}}
{{NotZFC}} Suppose that :$\forall x: \left({x \subseteq A \implies x \in A}\right)$ {{MissingLinks|to results, although most are up only for sets}} Consider $\Bbb U \setminus A$. Aiming for a contradiction, suppose that: : $\Bbb U \setminus A \ne \varnothing$ Then by Axiom of Foundation (Strong Form), we have that: :$\exists x \notin A: \left({x \cap \left({\Bbb U \setminus A}\right)}\right) = \varnothing$ But: {{begin-eqn}} {{eqn | l = \left({x \cap \left({\Bbb U \setminus A}\right)}\right)       | r = \left({\left({x \cap \Bbb U}\right) \setminus A}\right)       | c =  }} {{eqn | r = \left({x \setminus A}\right)       | c =  }} {{end-eqn}} Thus: :$x \setminus A = \varnothing$ So from Set Difference with Superset is Empty Set: :$x \subseteq A$ Thus by hypothesis: :$x \in A$ contradicting the fact that $x \notin A$. Therefore we can conclude that :$\left({\Bbb U \setminus A}\right) = \varnothing$ and so from Set Difference with Superset is Empty Set: :$\Bbb U \subseteq A$ Furthermore, by definition of universe: :$A \subseteq \Bbb U$ so by definition of set equality: :$A = \Bbb U$ {{qed}}
{{NotZFC}} From Biconditional is Commutative: :$\forall x: \left({ x \in A \iff x \in B }\right) \implies \forall x: \left({ x \in B \iff x \in A }\right)$ Hence the result by definition of class equality.
{{NotZFC}} By Universal Generalisation and Biconditional is Transitive: :$\forall x: \left({ \left({ x \in A \iff x \in B }\right) \land \left({ x \in B \iff x \in C }\right) }\right) \implies \forall x: \left({ x \in A \iff x \in C }\right)$ by Universal Generalisation and Biconditional is Transitive {{explain|Further explanation needs to be added here to explain why this constitutes a proof.}} {{qed}}
{{NotZFC}} We have: :$x \in \left\{{x : x \in A}\right\} \iff x \in A$ by definition of class membership (applied to $\left\{{x : x \in A}\right\}$). By Universal Generalisation, it follows that: :$\forall x: \left({ x \in A \iff x \in \left\{{x : x \in A}\right\} }\right)$ Hence the result, by definition of class equality. {{qed}}
:$x = x$ Therefore by Existential Generalisation: :$\exists y: x = y$ {{qed}} {{explain|Add the appropriate words to link the above statements to the fact being proved, e.g. "Therefore by definition of ..." etc.}}
{{begin-eqn}} {{eqn|l = A \in \left\{ {x : P \left({x}\right)}\right\}      |o = \implies      |r = \exists x \in \left\{ {x : P \left({x}\right)}\right\}: A = x      |c = Definition of class membership }} {{eqn|l =       |o = \implies      |r = \exists x: \left({x = A \land P \left({x}\right)}\right)      |c = Definition of bounded existential quantifier }} {{eqn|l =       |o = \implies      |r = \exists x: \left({x = A \land P \left({A}\right)}\right)      |c = Substitutivity of Class Equality }} {{end-eqn}} {{qed}} {{questionable|Class Membership def can't be applied because, being in ZF, we can't quantify over classes. Use DefAbbrev on Definition:Class/Zermelo-Fraenkel instead}}
{{begin-eqn}} {{eqn|l = x =_1 y      |o = \iff      |r = \forall z: \left({ z \in x \iff z \in y }\right)      |c = Definition of set equality }} {{eqn|o = \iff      |r = x =_2 y      |c = Definition of class equality, Class Membership Extension of Set Membership }} {{end-eqn}} {{qed}} Category:Zermelo-Fraenkel Class Theory 0hzm5dujmkry222aczybvdxji4kjo2x
Let $a, b \in S$. It is required to show that: :$a *' b = a * b$ To achieve this, recall that $*'$ is defined on Group Induces $B$-Algebra to satisfy: :$a *' b = a \circ b^{-1}$ which, by the definition of $\circ$ on $B$-Algebra Induces Group comes down to: {{begin-eqn}} {{eqn|l = a *' b      |r = a * \left({0 * b^{-1} }\right) }} {{eqn|r = a * \left({0 * \left({0 * b}\right)}\right)      |c = $B$-Algebra Induces Group }} {{eqn|r = a * b      |c = Identity: $0 * \left({0 * x}\right) = x$ }} {{end-eqn}} Hence the result. {{qed}}
Let $a, b \in S$. It is required to show that: :$a \circ' b = a \circ b$ To achieve this, recall that $\circ'$ is defined on $B$-Algebra Induces Group to satisfy: :$a \circ' b = a * \left({e * b}\right)$ which, by the definition of $*$ on Group Induces $B$-Algebra comes down to: {{begin-eqn}} {{eqn|l = a \circ' b      |r = a \circ \left({e \circ b^{-1} }\right)^{-1} }} {{eqn|r = a \circ \left({b \circ e}\right)      |c = Inverse of Group Product }} {{eqn|r = a \circ b }} {{end-eqn}} Hence the result. {{qed}}
{{begin-eqn}} {{eqn | o =        | r = z \in \bigcup \set {x, y} }} {{eqn | ll= \leadstoandfrom       | o =        | r = \exists w \in \set {x, y}: z \in w       | c = {{Defof|Union of Class}} }} {{eqn | ll= \leadstoandfrom       | o =        | r = \exists w: \paren {\paren {w = x \lor w = y} \land z \in w}       | c = {{Defof|Doubleton Class}} }} {{eqn | ll= \leadstoandfrom       | o =        | r = \paren {z \in x \lor z \in y}       | c = Equality implies Substitution }} {{eqn | ll= \leadstoandfrom       | o =        | r = z \in \paren {x \cup y}       | c = {{Defof|Class Union}} }} {{end-eqn}} {{qed}}
Let $\map {\mathscr M} A$ denote that $A$ is small. {{begin-eqn}} {{eqn | l = \map {\mathscr M} x \land \map {\mathscr M} y       | o = \leadsto       | r = \map {\mathscr M} {\set {x, y} }       | c = Axiom of Pairing }} {{eqn | o = \leadsto       | r = \map {\mathscr M} {\bigcup \set {x, y} }       | c = Axiom of Unions }} {{eqn | o = \leadsto       | r = \map {\mathscr M} {x \cup y}       | c = Union of Doubleton }} {{end-eqn}} {{qed}}
Mutatis mutandis the same as that of Category of Sets is Category. {{qed}} Category:Category of Finite Sets fa5ospwroqhp94rswnji0mqbk9nmn3d
Let us verify the axioms $(C1)$ up to $(C3)$ for a metacategory. For any two increasing mappings their composition (in the usual set theoretic sense) is again increasing by Composite of Increasing Mappings is Increasing. For any set $X$, we have the identity mapping $\operatorname{id}_X$. By Identity Mapping is Left Identity and Identity Mapping is Right Identity that this is the identity morphism for $X$. That it is increasing follows from Identity Mapping is Increasing. Finally by Composition of Mappings is Associative, the associative property is satisfied. Hence $\mathbf{OrdSet}$ is a metacategory. {{qed}}
{{NotZFC}} By Set Difference as Intersection with Relative Complement: :$\paren {x \setminus A} = \paren {x \cap \map \complement A}$ {{wtd|This rule must be proven for classes}} Next, by Axiom of Subsets Equivalents, $\paren {x \cap \map \complement A}$ is small. Therefore: :$\map \MM {x \setminus A}$ {{qed}}
{{NotZFC}} {{begin-eqn}} {{eqn | l = A \ne \O       | o = \leadstoandfrom       | r = \neg \forall x: \paren {x \in A \iff x \in \O}       | c = {{Defof|Class Equality}} }} {{eqn | o = \leadstoandfrom       | r = \neg \forall x: \neg x \in A       | c = {{Defof|Empty Set}} }} {{eqn | o = \leadstoandfrom       | r = \exists x: x \in A       | c = De Morgan's Laws (Predicate Logic) }} {{end-eqn}} {{qed}}
{{NotZFC}} $A \in A$ would create a relational loop for the class membership sign. By No Membership Loops, this is a contradiction. {{qed}}
Let us verify the axioms $(C1)$ up to $(C3)$ for a metacategory. We have already defined the composition of morphisms. For any set $X$, we have the diagonal relation $\operatorname{id}_X$. By Diagonal Relation is Left Identity and Diagonal Relation is Right Identity it follows that this is the identity morphism for $X$. Finally by Composition of Relations is Associative, the associative property is satisfied. Hence $\mathbf{Rel}$ is a metacategory. {{qed}}
{{NotZFC}} Let $I_\mathrm U$ be the identity mapping on $\mathrm U$. By Identity Mapping is Bijection it follows that $I_\mathrm U$ is a bijection. Therefore, by the Axiom of Limitation of Size, $\mathrm U$ is proper. {{qed}}
Assume that $\mathrm U$ is small. Note that $\operatorname{Ru} \subseteq \mathrm U$ where $\operatorname{Ru}$ denotes the Russell class. By Axiom of Subsets Equivalents, $\operatorname{Ru}$ is also small. This contradicts Russell's Paradox. {{qed}}
Let $f, g$ be morphisms of $\mathbf C$ such that $g \circ f$ is defined. Then: {{begin-eqn}} {{eqn | l = \operatorname{id}_{\mathbf C} \left({g \circ f}\right)       | r = g \circ f       | c = {{Defof|Identity Functor}} }} {{eqn | r = \operatorname{id}_{\mathbf C} g \circ 1_{\mathbf C} f       | c = {{Defof|Identity Functor}} }} {{end-eqn}} Also, for any object $C$ of $\mathbf C$: {{begin-eqn}} {{eqn | l = \operatorname{id}_{\mathbf C} \operatorname{id}_C       | r = \operatorname{id}_C       | c = {{Defof|Identity Functor}} }} {{eqn | r = \operatorname{id}_{\operatorname{id}_{\mathbf C} C}       | c = {{Defof|Identity Functor}} }} {{end-eqn}} Hence $\operatorname{id}_{\mathbf C}$ is shown to be a functor. {{qed}} Category:Category Theory bwu2tng0tenw2r9n2jm6jj9voefd9uf
Let us verify the axioms $(C1)$ up to $(C3)$ for a metacategory. For any two functors their composition is again a functor by Composite Functor is Functor. For any small category $\mathbf C$, we have the identity functor $\operatorname{id}_{\mathbf C}$. By Identity Functor is Left Identity and Identity Functor is Right Identity this is the identity morphism for $\mathbf C$. Finally by Composition of Functors is Associative, the associative property is satisfied. Hence $\mathbf{Cat}$ is a metacategory. {{qed}} Category:Category of Categories cel40mbnt7v5yjn7tcqnhb8l0bj314v
We have, for all objects $C$ of $\mathbf C$: :$\operatorname{id}_{\mathbf D}F C = \operatorname{id}_{\mathbf D} \left({F C}\right) = F C$ by definition of composition of functors and of identity functor. Similarly, we have, for a morphism $f$ of $\mathbf C$: :$\operatorname{id}_{\mathbf D}F f = \operatorname{id}_{\mathbf D} \left({F f}\right) = F f$ Hence the result. {{qed}} Category:Category Theory howawdznxulifwdyotqykit2dip4rro
We have, for all objects $C$ of $\mathbf C$: :$F \operatorname{id}_{\mathbf C} C = F \left({\operatorname{id}_{\mathbf C} C}\right) = F C$ by definition of composition of functors and of identity functor. Similarly, we have, for a morphism $f$ of $\mathbf C$: :$F \operatorname{id}_{\mathbf C} f = F \left({\operatorname{id}_{\mathbf C} f}\right) = F f$ Hence the result. {{qed}} Category:Category Theory qgn4dq1bsly6qx0fpmu5tpipyabxbf3
Let $A$ be an object of $\mathbf A$. Then, solely by the definition of composite functor: {{begin-eqn}} {{eqn | l = H \paren {G F} A       | r = H \paren {G F A} }} {{eqn | r = H \paren {G \paren {F A} } }} {{eqn | r = H G \paren {F A} }} {{eqn | r = \paren {H G} F A }} {{end-eqn}} Mutatis mutandis, the same proof works for a morphism $f$ of $\mathbf A$ as well. Hence the result. {{qed}} Category:Category Theory d5jcufeskokfnp1g2orbto1u6wvkb4l
{{begin-eqn}} {{eqn | l = \left({a \times b}\right)       | r = \left\{ {\left({x, y}\right) : x \in a \land y \in b}\right\}       | c = Definition of Cartesian Product }} {{eqn | r = \left\{ {\left({x, y}\right) : \left\{ {x}\right\} \subseteq a \land \left\{ {y}\right\} \subseteq b} \right\}       | c = Singleton of Element is Subset }} {{eqn | o = \subseteq       | r = \left\{ {\left({x, y}\right) : \left\{ {x, y}\right\} \subseteq \left({a \cup b}\right)}\right\}       | c = Set Union Preserves Subsets }} {{eqn | o = \subseteq       | r = \left\{ {\left({x, y}\right) : \left({x, y}\right) \subseteq \mathcal P \left({a \cup b}\right)}\right\}       | c = Power Set of Subset and Subset Relation is Transitive }} {{end-eqn}} {{explain|It's not obvious how $\left({x, y}\right) \subseteq \mathcal P \left({a \cup b}\right)$}} So by the definition of power set: : $\left({a \times b}\right) \subseteq \mathcal P \left({\mathcal P \left({a \cup b}\right)}\right)$ By Union of Small Classes is Small, $\left({a \cup b}\right)$ is small. By the Axiom of Powers, $\mathcal P \left({\mathcal P \left({a \cup b}\right)}\right)$ is small. By Axiom of Subsets Equivalents, $\left({a \times b}\right)$ is small. {{qed}}
Let us verify the axioms $(C1)$ up to $(C3)$ for a metacategory. Note that for objects $a, b$ of $\mathbf S$, there is at most ''one'' morphism $a \to b$, by definition of $\mathbf S$. Suppose that $a \to b$ and $b \to c$ are morphisms of $\mathbf S$. Then we have $a \precsim b$ and $b \precsim c$, and as $\precsim$ is a preordering: :$a \precsim c$ which is to say that there is a unique morphism $a \to c$. By uniqueness, this morphism is the composite of $a \to b$ and $b \to c$. Also, for any $a \in S$, there is a morphism $a \to a$ by virtue of: :$a \precsim a$ since $\precsim$ is a preordering. The uniqueness of the morphisms mentioned above implies directly that $a \to a$ classifies as the identity morphism $\operatorname{id}_a$ for $a$. The uniqueness of morphisms also immediately implies associativity. Hence $\mathbf S$ is a metacategory. Since both $S$ and $\precsim$ are sets, $\mathbf S$ is also a category. {{qed}} Category:Preorder Categories 6pf8wlzx1ehy1xy0j1t5okdfkqjjeeb
It suffices to establish $\precsim$ is reflexive and transitive. === Reflexivity === For any $a \in S$, we have the identity morphism $\operatorname{id}_a : a \to a$. Hence $a \precsim a$. {{qed|lemma}} === Transitivity === For $a, b, c \in S$, let $a \precsim b$ and $b \precsim c$. Then we have: :$f: a \to b$ and $g: b \to c$ Therefore, we have the composite morphism $g \circ f: a \to c$ as well. Hence $a \precsim c$. {{qed|lemma}} It follows that $\precsim$ is a preorder. {{qed}} Category:Category Theory Category:Preorder Theory k9mv82nlbz986dbu81b6ousdel2v23y
Suppose that for some $a, b \in S$, we have: :$a \precsim' b$ By Category Induces Preorder, this happens iff there exists an $\mathbf S$-morphism $f: a \to b$. By definition of $\mathbf S$, this $f: a \to b$ exists iff: :$a \precsim b$ Hence the result. {{qed}} Category:Preorder Categories i1lwvmh0h3n1rag5svb7tkuanyelyrm
Suppose that for some $a, b \in S$, we have: :$a \preceq b$ Then there is a morphism $a \to b$ in $\mathbf S$. As $F$ is a functor, it follows that there is a morphism: :$F a \to F b$ in $\mathbf T$ as well, that is: :$F a \preceq' F b$ Hence the result. {{qed}}
We have, for any morphism $a \to b$ in $\mathbf{Dis} \left({S}\right)$ that $a = b$. Thus we see that $\mathbf{Dis} \left({S}\right)$ will be an order category iff: :$\forall a, b \in S: a \preceq b \iff a = b$ holds for some ordering $\preceq$ on $S$. The trivial ordering on $S$ accomplishes this. Hence the result. {{qed}}
{{NotZFC}} Since $A$ is a mapping: :$\forall y: \exists x: \forall z: \left({ y A z \implies z = x }\right)$ This satisfies the antecedent of the axiom of replacement.  Therefore: :$\forall w: \exists x: \forall y: \left({ y \in w \implies \forall z: \left({ y A z \implies z \in x }\right) }\right)$ Universal Instantiation yields: :$\exists x: \forall y: \left({ y \in a \implies \forall z: \left({ y A z \implies z \in x }\right) }\right)$ By applying the definition for the restricted universal quantifier and rearranging quantifiers: :$\exists x: \forall z: \left({ \exists y \in a: y A z \implies z \in x }\right)$ Applying the definition for image: :$\exists x: \operatorname{Im} \left({a}\right) \subseteq x$ By Axiom of Subsets Equivalents, the image of $a$ under $A$ must be small. {{qed}}
Let $A$ equal: :$\set {\tuple {\tuple {x, y}, \tuple {y, x} } : \tuple {x, y} \in a}$ Then $A$ maps $a$ to its inverse. {{link wanted|$A$ in general has a name but I can't find it atm}} Thus, the inverse of $a$ is the image of $a$ under $A$. By Image of Small Class under Mapping is Small, the inverse of $a$ is small. {{qed}}
By definition of monoid category, we have: :$\mathbf S_1 = S, \mathbf T_1 = T$ which in particular are sets. Hence the morphism functor $F_1: \mathbf S_1 \to \mathbf T_1$ is a mapping $F_1: S \to T$. For $F_1$ to be a monoid homomorphism, it suffices to show that the following hold: :$F_1 \left({e_S}\right) = e_T$ :$\forall s_1, s_2 \in S: F_1 \left({s_1 \circ s_2}\right) = F_1 \left({s_1}\right) * F_1 \left({s_2}\right)$ Since all $s_1, s_2 \in S$ are composable morphisms of $\mathbf S$, these conditions are exactly equivalent to $F$ being a functor. Hence the result. {{qed}}
Let us verify the axioms $(C1)$ up to $(C3)$ for a metacategory. We have Composite of Homomorphisms on Algebraic Structure is Homomorphism, verifying $(C1)$. We have Identity Mapping is Automorphism providing $\operatorname{id}_S$ for every monoid $\left({S, \circ}\right)$. Now, $(C2)$ follows from Identity Mapping is Left Identity and Identity Mapping is Right Identity. Finally, $(C3)$ follows from Composition of Mappings is Associative. Hence $\mathbf{Mon}$ is a metacategory. {{qed}} Category:Category of Monoids sppgcr1pruam0jkf2euhyvt6y3jpnt0
Let $G$ be a group and let $a \in G$. Consider the left regular representation $\lambda_a: G \to G$ defined as: :$\map {\lambda_a} x = a \cdot x$ From Regular Representations in Group are Permutations we have that $\lambda_a$ is a permutation. Now let $b \in G$ and consider $\lambda_b: G \to G$ defined as: :$\map {\lambda_b} x = b \cdot x$ From the Cancellation Laws it follows that $\lambda_a \ne \lambda_b \iff a \ne b$. Let $H = \set {\lambda_x: x \in G}$. Consider the mapping $\Phi: G \to H$ defined as: :$\forall a \in G: \map \Phi a = \lambda_a$ From the above we have that $\Phi$ is a bijection. Let $a, b \in G$. From Composition of Regular Representations we have that: :$\lambda_a \circ \lambda_b = \lambda_{a \cdot b}$ where $\circ$ denotes composition of mappings. That is, $\Phi$ has the morphism property. Thus $\Phi$ is seen to be a group isomorphism. We also have that: :$\paren {\lambda_a}^{-1} = \lambda_\paren {a^{-1} }$ because: :$\lambda_a \circ \paren {\lambda_a}^{-1} = \lambda_\paren {a \cdot a^{-1} }$ Hence the set of left regular representations $\set {\lambda_x: x \in G}$ is a group which is isomorphic to $G$. {{qed}}
Apply Permutation of Cosets to $H$ and let $N = \map \ker \theta$. Then: :$N \lhd G$ and $N \lhd H$ so from the Correspondence Theorem: :$H / N \le G / N$ such that: :$\index {G / N} {H / N} = n$ Thus: :$n \divides \index G N$ Also by Permutation of Cosets: :$\exists K \in S_n: G / N \cong K$ Thus: :$\index G N \divides n!$ as required. {{Qed}}