hcju/proofwikiqa huggingface dataset:
Define a functor $H: \mathbf C \to \mathbf{Set}$ by: :$H C := \left\{{f \in \operatorname{mor} \mathbf C: \operatorname{cod} f = C}\right\}$ :$H f: H A \to H B, g \mapsto f \circ g$ for $f: A \to B$ a morphism of $\mathbf C$. It is immediate by the definition of identity morphism that: :$H \left({\operatorname{id}_A}\right) = \operatorname{id}_{H A}$ For $f: A \to B$ and $g: B \to C$, observe: {{begin-eqn}} {{eqn | l = H \left({g \circ f}\right) (h)       | r = \left({g \circ f}\right) \circ h }} {{eqn | r = g \circ \left({f \circ h}\right)       | c = Composition of Mappings is Associative }} {{eqn | r = g \circ \left({H f (h)}\right)       | c = Definition of $H f$ }} {{eqn | r = H g \left({H f (h)}\right) }} {{eqn | r = \left({H g \circ H f}\right) (h) }} {{end-eqn}} Thus by Equality of Mappings, $H \left({g \circ f}\right) = H g \circ H f$. It follows that $H$ is a functor. It is clear that $H$ is injective on objects. Suppose now that $H$ were not faithful. Then there would be morphisms $g, h: A \to B$ of $\mathbf C$ such that $H g = H h$. Since $\operatorname{id}_A \in H A$, this means in particular that: :$g \circ \operatorname{id}_A = h \circ \operatorname{id}_A$ by Equality of Mappings. But the definition of identity morphism then reduces this to $g = h$. Hence, $H$ is faithful. By Functor is Embedding iff Faithful and Injective on Objects, it follows that $H$ is an embedding. Thus $\mathbf C$ is isomorphic to a subcategory of $\mathbf{Set}$. {{qed}}
Let $A$ equal: :$\left\{{\left({\left({x, y}\right), x}\right): \left({x, y}\right) \in a}\right\}$ Then, $A$ maps $a$ to its domain. Thus, the domain of $a$ is the image of $a$ under $A$. By Image of Small Class under Mapping is Small, the domain of $a$ is small. {{explain|And exactly why is $A$ small, so that this result applies? Same qn for Range of Small Relation is Small}} {{qed}}
Let $A$ equal: :$\set {\tuple {\tuple {x, y}, y}: \tuple {x, y} \in a}$ Then, $A$ maps $a$ to its range. Thus, the range of $a$ is the image of $A$. By Image of Small Class under Mapping is Small, the range of $a$ is small. {{explain|Why is $A$ small, so that Image of Small Class under Mapping is Small of applies?}} {{qed}}
We have, for $g \in G$: {{begin-eqn}} {{eqn | l = g \in \map Z {G, \circ}       | o = \leadstoandfrom       | r = \forall h \in G: g \circ h = h \circ g       | c = {{Defof|Center of Group}} }} {{eqn | o = \leadstoandfrom       | r = \forall h \in G: h * g = g * h       | c = {{Defof|Opposite Group}} }} {{eqn | o = \leadstoandfrom       | r = g \in \map Z {G, *}       | c = {{Defof|Center of Group}} }} {{end-eqn}} Hence the result, by definition of set equality. {{qed}} Category:Opposite Groups Category:Centers of Groups plso6ifn0x3uxwi3s1d2porv2x0a12h
We have, for all $g, h \in G$: :$a \circ b = b * a = a \circ' b$ by definition of opposite group. Hence the result. {{qed}} Category:Opposite Groups fz0gqqmvg7kycy4vvm552knuvnbqbxb
Let us verify the axioms $(C1)$ up to $(C3)$ for a metacategory. Suppose that $a: f \to g$ and $b: g \to h$ are morphisms in $\mathbf C / C$. To show that $b \circ a: f \to h$ is a morphism as well, compute: {{begin-eqn}} {{eqn|l = \left({b \circ a}\right) \circ f      |r = b \circ \left({a \circ f}\right)      |c = composition in $\mathbf C$ is associative }} {{eqn|r = b \circ g      |c = $a$ is a morphism }} {{eqn|r = h      |c = $b$ is a morphism }} {{end-eqn}} Hence $b \circ a: f \to h$ is a morphism. For $(C2)$, observe that for $a: f \to g$, with objects $f: X \to C$ and $g: Y \to C$, we have: {{begin-eqn}} {{eqn|l = a \circ \operatorname{id}_f      |r = a \circ \operatorname{id}_X      |c = Definition of $\operatorname{id}_f$ }} {{eqn|r = a      |c = Since $\mathbf C$ is a metacategory }} {{eqn|r = \operatorname{id}_Y \circ a }} {{eqn|r = \operatorname{id}_g \circ a }} {{end-eqn}} Hence $(C2)$ is shown to hold. Since the composition in $\mathbf C / C$ is inherited from $\mathbf C$, satisfaction of the associative property $(C3)$ is also inherited. Hence $\mathbf C / C$ is a metacategory. {{qed}} Category:Slice Categories bbizrlh6abyxjd6hfggwtb4uj9bekpi
Let us verify the axioms $(C1)$ up to $(C3)$ for a metacategory. Suppose that $a, b \in \mathbf S_1$ are morphisms. Since $\left({S, \circ}\right)$ is closed, $a \circ b$ is also a morphism. Axiom $(C2)$ follows directly from the axioms for the identity $e_S$ in a monoid. Axiom $(C3)$ holds because $\circ$ is associative. This is because $\left({S, \circ}\right)$ is a monoid. Hence $\mathbf S$ is a metacategory. As $\mathbf S$ has only one object, and $\mathbf S_1 = S$, it follows that $\mathbf S$ is also a category. {{qed}} Category:Monoid Categories dep2kxmmc62m5aot62our2164v8oi54
Let us verify the axioms $(C1)$ up to $(C3)$ for a metacategory. Let $f^{\text{op}}: C^{\text{op}} \to D^{\text{op}}$ and $g^{\text{op}}: D^{\text{op}} \to E^{\text{op}}$ be morphisms in $\mathbf C^{\text{op}}$. Then $f: D \to C$ and $g: E \to D$ are morphisms in $\mathbf C$, and so is $f \circ g: E \to C$. Therefore, also $g^{\text{op}} \circ f^{\text{op}}: C^{\text{op}} \to E^{\text{op}}$ is a morphism in $\mathbf C^{\text{op}}$, and $(C1)$ is shown to hold. For $(C2)$, observe that for $f^{\text{op}}: C^{\text{op}} \to D^{\text{op}}$, we have: {{begin-eqn}} {{eqn|l = f^{\text{op} } \circ \operatorname{id}_{C^{\text{op} } }      |r = f^{\text{op} } \circ \operatorname{id}_C^{\text{op} }      |c = Definition of $\operatorname{id}_{C^{\text{op} } }$ }} {{eqn|r = \left({\operatorname{id}_C \circ f}\right)^{\text{op} }      |c = Definition of composition }} {{eqn|r = f^{\text{op} }      |c = Since $\mathbf C$ is a metacategory }} {{end-eqn}} Similarly: {{begin-eqn}} {{eqn|l = \operatorname{id}_{D^{\text{op} } } \circ f^{\text{op} }      |r = \operatorname{id}_D^{\text{op} } \circ f^{\text{op} }      |c = Definition of $\operatorname{id}_{D^{\text{op} } }$ }} {{eqn|r = \left({f \circ \operatorname{id}_D}\right)^{\text{op} }      |c = Definition of composition }} {{eqn|r = f^{\text{op} }      |c = Since $\mathbf C$ is a metacategory }} {{end-eqn}} Hence $(C2)$ is shown to hold. To show $(C3)$, reason as follows: {{begin-eqn}} {{eqn|l = \left({f^{\text{op} } \circ g^{\text{op} } }\right) \circ h^{\text{op} }      |r = \left({g \circ f}\right)^{\text{op} } \circ h^{\text{op} }      |c = Definition of composition }} {{eqn|r = \left({h \circ \left({g \circ f}\right)}\right)^{\text{op} }      |c = Definition of composition }} {{eqn|r = \left({\left({h \circ g}\right) \circ f}\right)^{\text{op} }      |c = Since $\mathbf C$ is a metacategory }} {{eqn|r = f^{\text{op} } \circ \left({h \circ g}\right)^{\text{op} } }} {{eqn|r = f^{\text{op} } \circ \left({g^{\text{op} } \circ h^{\text{op} } }\right) }} {{end-eqn}} Hence $\mathbf C^{\text{op}}$ is a metacategory. {{qed}} Category:Dual Categories 9i5b7ic5zh4k5h7ylvkmimpf1r80359
Let $n \in \Z: n \ge 0$. Then by definition of the signum function: :$\map {\sgn_\Z} n = \begin {cases} 0 & : n = 0 \\ 1 & : n > 0 \end {cases}$ So by definition of the signum function on the natural numbers: :$\forall n \in \N: \map {\sgn_\Z} n = \map {\sgn_\N} n$ Hence the result, by definition of extension. {{qed}} Category:Number Theory 1fee7396mz20grb6ycct0hlcaubyj8u
{{begin-eqn}} {{eqn | l = A \times B       | r = \left\{ { \left({ x, y }\right) : x \in A \land y \in B }\right\}       | c = Definition of Cartesian Product }} {{eqn | r = \left\{ { \left({ y, x }\right) : x \in A \land y \in B }\right\}^{-1}       | c = Definition of Inverse Relation }} {{eqn | r = \left({ B \times A }\right)^{-1}       | c = Definition of Cartesian Product }} {{end-eqn}} Let $B \times A$ be a small class. Then, by Inverse of Small Relation is Small, $A \times B$ is also small. Similarly, let $A \times B$ be small. Then, by Inverse of Small Relation is Small, $B \times A$ is also small. {{qed}}
{{NotZFC}} Suppose, to the contrary, that $\paren {A \times B}$ is small. By Domain of Small Relation is Small, the domain of $\paren {A \times B}$ is small. Since $B \ne \O$, Nonempty Class has Members shows that $\exists y: y \in B$. The domain of $\paren {A \times B}$ is the collection of all $x \in A$ such that $\exists y: y \in B$. The domain of $\paren {A \times B}$ is $A$. Therefore, $A$ is small, contradicting the fact that it is a proper class. {{qed}}
{{begin-eqn}} {{eqn | n = 1       | l = z \in \mathcal R \left({ x }\right)       | o = \implies       | r = \exists y: \left({z \in y \land \left({x, y}\right) \in \mathcal R }\right) \land \exists ! y: \left({x, y}\right) \in \mathcal R       | c = Definition of Image of Element under Relation }} {{eqn | o = \implies       | r = \forall w: \left({\left({x, w}\right) \in \mathcal R \iff w = y}\right)       | c = Definition of Unique and by hypothesis that $\left({x, y}\right) \in \mathcal R$ }} {{eqn | o = \implies       | r = \left({z \in w \land \left({x, w}\right) \in \mathcal R}\right)       | c = Existential Instantiation from $(1)$ }} {{eqn | o = \implies       | r = z \in y       | c = Substitutivity of Equality from $z \in w \land w = y$ }} {{end-eqn}} Conversely: {{begin-eqn}} {{eqn | l = z \in y       | o = \implies       | r = \left({z \in y \land \left({x, y}\right) \in \mathcal R}\right)       | c = by hypothesis that $\left({x, y}\right) \in \mathcal R$ }} {{eqn | o =        | r = \exists ! y: \left({x, y}\right) \in \mathcal R       | c = by hypothesis }} {{eqn | o = \implies       | r = z \in \mathcal R \left({x}\right)       | c = Definition of Image of Element under relation }} {{end-eqn}} Generalizing: :$\forall z: \left({z \in y \iff z \in \mathcal R \left({x}\right) }\right)$ Therefore: :$y = \mathcal R \left({x}\right)$ by the definition of class equality. {{qed|lemma}} Suppose that $\neg \exists ! y: \left({x, y}\right) \in \mathcal R$. Then: {{begin-eqn}} {{eqn | l = \neg \exists ! y: \left({x, y}\right) \in \mathcal R       | o = \implies       | r = z \notin \mathcal R \left({x}\right)       | c = Definition of Image of Element under relation }} {{end-eqn}} Thus: :$\forall z: z \notin \mathcal R \left({ x }\right)$ Therefore: :$\mathcal R \left({ x }\right) = \varnothing$ {{qed}}
Let $\RR$ be an arbitrary relation. Let $s$ be any set. The value of a relation is either equal to some set $y$ or $\O$ by Uniqueness Condition for Relation Value. If it is equal to some set $y$, then the value of $s$ under $\RR$ is a small class by the definition of small class. If it is equal to $\O$, then the result follows from Empty Set is Small.
{{NotZFC}} Let $A$ denote the domain of $F$. Let $B$ denote the image of $F$. Since $F$ is a mapping, $F$ is also a relation.  Therefore: :$F \subseteq A \times B$ where $A \times B$ denotes the Cartesian product of $A$ and $B$. $B$ is the image of $A$ under $F$ and is therefore a small class by Image of Small Class under Mapping is Small. Since $A$ and $B$ are both small, their Cartesian Product is Small. By Axiom of Subsets Equivalents, it follows that $F$ is small.
The domain of $F {\restriction_A}$ is a subset of $A$. By Axiom of Subsets Equivalents, the domain is a small class. By Mapping whose Domain is Small Class is Small, it follows that $F {\restriction_A}$ is a small class. {{qed}}
{{begin-eqn}} {{eqn | l = \map {\RR^{-1} } t       | r = \set {s: \paren {s, t} \in \RR}       | c = {{Defof|Preimage of Element under Relation}} }} {{eqn | r = \set {s: \exists y: \paren {y = t \land \tuple {s, y} \in \RR} }       | c = Equality implies Substitution }} {{eqn | r = \set {s: \exists y \in \set t: \tuple {s, y} \in \RR}       | c = {{Defof|Singleton}} }} {{eqn | r = \RR^{-1} \sqbrk {\set t}       | c = {{Defof|Preimage of Subset under Relation}} }} {{end-eqn}} {{qed}}
Suppose $x$ is a minimal element of $B$.  That is: :$\neg \exists y \in B: x \prec_1 y$ This is equivalent to the statement: :$\neg \exists z \in \phi\left({B}\right): \phi\left({x}\right) \prec_2 z$ since $z$ is of the form $\phi\left({y}\right)$ by the definition of order isomorphism. {{handwaving|Too much corners cut}} Thus, this statement is true iff $\phi\left({x}\right)$ is a minimal element of $B$.
$\phi$ maps the $\prec_1$-initial segment of $x$ to: {{begin-eqn}} {{eqn|l = \phi \left[{\left\{y \in A: y \prec_1 x \right\} }\right]      |r = \phi \left[{\left\{y \in A: \phi \left({y}\right) \prec_2 \phi \left({x}\right) \right\} }\right]      |c = by the definition of initial segment and order isomorphism }} {{eqn|l =       |r = \left\{ \phi\left({y}\right) \in \phi \left[{A}\right]: \phi \left({y}\right) \prec_2 \phi \left({x}\right) \right\}      |c = by the definition of order isomorphism and image }} {{end-eqn}} This is the $\prec_2$-initial segment of $\phi \left({x}\right)$. {{qed}}
Take any nonempty subset $B \subseteq A$. Then, $x$ is a minimal element of $B$ iff $\phi \left({ x }\right)$ is a minimal element of $\phi \left({ B }\right)$ by the fact that an Order Isomorphism Preserves Minimal Elements. By the definition of foundational relation, $\left({ A_1 , \prec_1 }\right)$ is foundational if and only if $\left({ A_2 , \prec_2 }\right)$ is foundational. {{proofread}} {{qed}}
We have that $\map \sgn {S_n}$ is onto $C_2$. Thus from the First Isomorphism Theorem, $A_n$ consists of the set of even permutations of $S_n$. {{Qed}}
Take any $x,y \in A_1$ such that $x \preceq_1 y$. Since $x, y \in A_1$, it follows by the definition of a mapping that: : $\phi\left({x}\right), \phi\left({y}\right) \in A_2$ So $x \in A_1$ and $y \in A_1$ and $x \preceq_1 y$. It follows from the definition of $\preceq_2$ that: : $\phi\left({x}\right) \preceq_2 \phi\left({y}\right)$ Conversely, suppose that: : $\phi\left({x}\right) \preceq_2 \phi\left({y}\right)$ By the definition of $\preceq_2$, it follows that: : $x \preceq_1 y$ Therefore, the biconditional holds: :$x \preceq_1 y \iff \phi\left({x}\right) \preceq_2 \phi\left({y}\right)$ By definition, it follows that: : $\phi: \left({A_1, \preceq_1}\right) \to \left({A_2, \preceq_2 }\right)$ is an order isomorphism. {{qed}}
By Ordinal Multiplication is Left Distributive, it follows that: :$m \times \left({ x + n }\right) = \left({ m \times x }\right) + \left({ m \times n }\right)$ It remains to prove that $x = \left({ m \times x }\right)$. Since $x$ is a limit ordinal, it follows that: {{begin-eqn}} {{eqn | lo= \exists y \in \operatorname{On}:       | l = x       | r = \left({\omega \times y}\right)       | c = Factorization of Limit Ordinals }} {{eqn | l = m \times x       | r = m \times \left({\omega \times y}\right)       | c = Substitutivity of Class Equality }} {{eqn | l = m \times x       | r = \left({m \times \omega}\right) \times y       | c = Ordinal Multiplication is Associative }} {{eqn | r = \omega \times y       | c = by lemma }} {{end-eqn}} {{qed}}
By the Division Theorem for Ordinals, :$x = ( \omega \times y ) + z$ for some unique $y$ and $z \in \omega$ Aiming for a contradiction, suppose $z \ne 0$. Since $z \in \omega$, $z$ is not a limit ordinal. Therefore, by the definition of limit ordinal, $z = w^+$ for some $w \in \omega$. But this means that: {{begin-eqn}} {{eqn|l = x       |r = ( \omega \times y ) + w^+      |c = by Division Theorem for Ordinals }} {{eqn|l =       |r = ( ( \omega \times y ) + w )^+      |c = definition of ordinal addition }} {{end-eqn}} This means that $x$ is the successor of some ordinal and cannot be a limit ordinal. But this contradicts the fact that $x$ is a limit ordinal, so $z = 0$. Therefore: {{begin-eqn}} {{eqn|l = x      |r = ( \omega \times y ) + z      |c = by Division Theorem for Ordinals }} {{eqn|l =       |r = \omega \times y      |c = by Ordinal Addition by Zero }} {{end-eqn}} {{qed}} Category:Ordinal Arithmetic 3v4xwywyiryqgwvui5qjabjj3megzd8
Let $x = 0$ and $y = 0$. Then by the definition of ordinal exponentiation: :$x^y = 1$ Let $x = 0$ and $y \ne 0$. Then by the definition of ordinal exponentiation: :$x^y = 0$ In either case, $x^y$ is an ordinal. Now suppose that $x \ne 0$. The proof shall proceed by Transfinite Induction on $y$. === Basis for the Induction === By definition of ordinal exponentiation: :$x^0 = 1$ which is an ordinal. This proves the basis for the induction. === Induction Step === The inductive hypothesis states that $x^y \in \operatorname{On}$. Suppose the inductive hypothesis holds. Then: {{begin-eqn}} {{eqn | l = x^{y^+}       | r = x^y \times x       | c = Definition of Ordinal Exponentiation }} {{eqn | o = \in       | r = \operatorname{On}       | c = Ordinal Multiplication is Closed }} {{end-eqn}} This proves the induction step. === Limit Case === The inductive hypothesis states that $x^z \in \operatorname{On}$ for all $z \in y$. {{begin-eqn}} {{eqn | l = x^y       | r = \bigcup_{z \mathop \in y} x^z       | c = Definition of Ordinal Exponentiation }} {{eqn | o = \in       | r = \operatorname{On}       | c = Union of Ordinals is Ordinal }} {{end-eqn}} This proves the limit case. {{qed}} Category:Ordinal Arithmetic jy5dxitvmkzghrwitas1wjz7tnx7n4u
The proof shall proceed by Transfinite Induction on $x$. === Basis for the Induction === {{begin-eqn}} {{eqn|l = 1^0      |r = 1      |c = definition of ordinal exponentiation }} {{end-eqn}} This proves the basis for the induction. === Induction Step === The inductive hypothesis supposes that $1^x = 1$ for some $x$. {{begin-eqn}} {{eqn|l = 1^{x^+}      |r = 1^x \times 1      |c = definition of ordinal exponentiation }} {{eqn|l =       |r = 1 \times 1      |c = Inductive Hypothesis }} {{eqn|l =       |r = 1      |c = Ordinal Multiplication by One }} {{end-eqn}} This proves the induction step. === Limit Case === The inductive hypothesis supposes that $\forall y \in x: 1^y = 1$. It follows that: {{begin-eqn}} {{eqn|ll = \forall y \in x:      |l = 1^y      |o = =      |r = 1      |c =  }} {{eqn|ll = \implies      |l = \bigcup_{y \in x} 1^y      |r = \bigcup_{y \in x} 1      |c = by Indexed Union Equality }} {{eqn|ll = \implies      |l = 1^x      |r = 1      |c = definition of ordinal exponentiation }} {{end-eqn}} This proves the limit case. {{qed}}
The proof shall proceed by Transfinite Induction on $y$. === Basis for the Induction === :$x^0 = 1$ by the definition of ordinal exponentiation. Therefore, $x^0 \ne 0$. This proves the basis for the induction. === Induction Step === The inductive hypothesis supposes that $x^y \ne 0$. {{begin-eqn}} {{eqn|l = x^{y^+}      |r = x^y \times x      |c = definition of Definition:Ordinal Exponentiation }} {{eqn|l = x^y      |o = \ne      |r = 0      |c = Inductive Hypothesis }} {{eqn|l = x      |o = \ne      |r = 0      |c = Hypothesis }} {{eqn|l = x^y \times x      |o = \ne      |r = 0      |c = by Ordinals have No Zero Divisors }} {{end-eqn}} This proves the induction step. === Limit Case === The inductive hypothesis says that: :$\forall z \in y: x^z \ne 0$ {{begin-eqn}} {{eqn|ll = \forall z \in y:      |l = x^z      |o = \ne      |r = 0      |c = Inductive Hypothesis }} {{eqn|ll = \implies      |l = 0      |o = \in      |r = x^z      |c = by Ordinal Membership is Trichotomy }} {{eqn|ll = \implies      |l = 0      |o = \in      |r = \bigcup_{z \in y} x^z      |c = by the fact that limit ordinals are nonempty }} {{eqn|ll = \implies      |l = 0      |o = \in      |r = x^y      |c = definition of ordinal exponentiation }} {{eqn|ll = \implies      |l = x^y      |o = \ne      |r = 0      |c = definition of empty set }} {{end-eqn}} This proves the limit case. {{qed}}
=== Sufficient Condition === The proof shall proceed by Transfinite Induction on $y$. === Basis for the Induction === It is a contradiction that $x < 0$ by the definition of empty set. This proves the basis for the induction. === Induction Step === The inductive step states that $x < y \implies z^x < z^y$. Then, if $x < y^+$, then $x = y \lor x < y$. ==== Lemma ==== $z^y < z^y \times y$ by Membership is Left Compatible with Ordinal Multiplication. So $z^y < z^{y^+}$ by the definition of ordinal exponentiation. {{qed|lemma}} Suppose $x = y$.  Then $z^x < z^{y^+}$. Suppose $x < y$.  Then: {{begin-eqn}} {{eqn | l = z^x       | o = <       | r = z^y       | c = Inductive Hypothesis }} {{eqn | o = <       | r = z^{y^+}       | c = Lemma above }} {{end-eqn}} In either case, the inductive step holds. This proves the induction step. === Limit Case === The inductive hypothesis states that: :$\forall w \in y: z^x < z^w$ for $x < w$ Take any $x < y$. By Union of Limit Ordinal, it follows that $x < w$ for some $w \in y$. By the inductive hypothesis, $z^x < z^w$. Since $w \in y$, it follows that: {{begin-eqn}} {{eqn | l = z^x       | o = <       | r = z^w       | c = above }} {{eqn | l =        | o = \le       | r = \bigcup_{w \in y} z^w       | c =  }} {{eqn | l =        | o = =       | r = z^y       | c = {{Defof|Ordinal Exponentiation}} }} {{end-eqn}} This proves the limit case. {{qed|lemma}} === Necessary Condition === Note that the last part proves that: :$x \in y \implies z^x \in z^y$ Moreover: :$x = y \implies z^x = z^y$ Therefore, by the fact that $\in$ is a strong ordering: :$x \notin y \implies z^x \notin z^y$ By the Rule of Transposition: :$z^x < z^y \implies x < y$ {{qed}}
Clearly GA-1 is fulfilled as $e * H = H$. GA-2 is shown to be fulfilled thus: {{begin-eqn}} {{eqn | l = \paren {g_1 \circ g_2} * H       | r = \paren {g_1 \circ g_2} \circ H \circ \paren {g_1 \circ g_2}^{-1}       | c =  }} {{eqn | r = g_1 \circ g_2 \circ H \circ g_2^{-1} \circ g_1^{-1}       | c =  }} {{eqn | r = g_1 * \paren {g_2 \circ H \circ g_2^{-1} }       | c =  }} {{eqn | r = g_1 * \paren {g_2 * H}       | c =  }} {{end-eqn}} {{Qed}}
The objects of $\mathbf P \mathbin / p$ are morphisms $q \to p$ of $\mathbf P$. The morphisms are $q \to r$ fitting into a commutative diagram: ::$\begin{xy}\xymatrix@C=1em{  q   \ar[rr]   \ar[rd] & &  r   \ar[ld] \\ &  p }\end{xy}$ Define a functor $U: \mathbf P \mathbin / p \to \mathbf P$ by: :$U \left({q \to p}\right) := q$ :$U \left({q \to r}\right) := q \to r$ Because there is at most one morphism $q \to p$ for each $q$, $U$ is injective on objects. That $U$ is faithful is trivial. Hence by Functor is Embedding iff Faithful and Injective on Objects, $U$ is an embedding. That is, $\mathbf P \mathbin / p$ is isomorphic to the image of $U$. Now the objects of the image of $U$ in $\mathbf P$ are those $q$ such that $q \preceq p$. That is, the image of $U$ has as objects $p^\preceq$. By Commutative Diagram in Preorder Category, every morphism $q \to r$ in $p^\preceq$ fits into a commutative triangle as above. Thus all morphisms of $p^\preceq$ are in the image of $U$, so that $p^\preceq$ is precisely the image of $U$. Hence the result. {{qed}}