hcju/proofwikiqa huggingface dataset:
{{BeginTableau|\vdash p \iff p}} {{TheoremIntro|1|p \implies p|Law of Identity: Formulation 2}} {{BiconditionalIntro|2||p \iff p|1|1}} {{EndTableau|qed}}
We apply the Method of Truth Tables. As can be seen by inspection, the truth values under the main connective match for both boolean interpretations. $\begin{array}{|ccc|} \hline p & \iff & p \\ \hline F & T & F \\ T & T & T \\ \hline \end{array}$ {{qed}}
Follows directly from Relation between Two Ordinals and Subset Relation is Ordering. {{qed}} Category:Ordinals 7j5gt4exlrw1dxv82z8s62oow1hw3bv
Suppose that $x \in S\,^+$. Then either $x \in S$ or $x = S$. If $x \in S$, it follows by the transitivity of $S$ that $x \subseteq S$. If $x = S$, then $x = S \subseteq S$ because a set is a subset of itself. Since $S \subseteq S\,^+$, it follows by the transitivity of the subset relation that $x \subseteq S\,^+$. {{qed}} Category:Set Theory fi6lvj26ufkd1bbuig50op5zdfncg0a
Since $S$ is transitive, it follows by Successor Set of Transitive Set is Transitive that $S^+$ is transitive. We now have to show that $S^+$ is strictly well-ordered by the epsilon restriction $\Epsilon \! \restriction_{S^+}$. So suppose that a subset $A \subseteq S^+$ is non-empty. Then: {{begin-eqn}} {{eqn | l = A       | r = A \cap \paren {S \cup \set S}       | c = Intersection with Subset is Subset and {{Defof|Successor Set}} }} {{eqn | r = \paren {A \cap S} \cup \paren {A \cap \set S}       | c = Intersection Distributes over Union }} {{end-eqn}} Let us refer to the above equation by the symbol $\paren 1$. We need to show that $A$ has a smallest element. We first consider the case where $A \cap S$ is empty. By equation $\paren 1$, it follows that $A \cap \set S$ is non-empty (because $A$ is non-empty). Therefore, $S \in A$. That is, $\set S \subseteq A$. By Union with Empty Set and Intersection with Subset is Subset, equation $\paren 1$ implies that $A \subseteq \set S$. Therefore, $A = \set S$ by the definition of set equality. So $S$ is the smallest element of $A$. We now consider the case where $A \cap S$ is non-empty. By Intersection is Subset, $A \cap S \subseteq S$; by the definition of a well-ordered set, there exists a smallest element $x$ of $A \cap S$. Let $y \in A$. If $y \in S$, then $y \in A \cap S$; therefore, by the definition of the smallest element, either $x \in y$ or $x = y$. Otherwise, $y = S$, and so $x \in S = y$. That is, $x$ is the smallest element of $A$. {{qed}}
Recall that the Ordinals are Totally Ordered. Let $A$ be a non-empty set of ordinals. Let $\alpha \in A$. Let $B = \alpha^+ \cap A$, where $\alpha^+$ denotes the successor set of $\alpha$. Recall that $\alpha^+$ is an ordinal. Note that $\alpha \in B$, so $B$ is non-empty. By Intersection is Subset, $B \subseteq \alpha^+$. It follows that there exists a smallest element $\kappa$ of $B$. We claim that $\kappa$ is the smallest element of $A$. So let $\beta \in A$. We need to show that $\kappa = \beta$ or $\kappa \in \beta$. By Ordinal Membership is Trichotomy, either $\beta \in \alpha^+$, $\alpha^+ = \beta$, or $\alpha^+ \in \beta$. If $\beta \in \alpha^+$, then $\beta \in B$; it follows by the definition of $\kappa$ that $\kappa = \beta$ or $\kappa \in \beta$. If $\alpha^+ = \beta$ or $\alpha^+ \in \beta$, then it follows by the transitivity of $\beta$ that $\alpha^+ \subseteq \beta$. Since $\kappa \in \alpha^+$, it follows that $\kappa \in \beta$. {{qed}}
By Ordinals are Totally Ordered, the ordinals are totally ordered by $\subseteq$: {{begin-eqn}} {{eqn | l = X       | o =\subsetneqq       | r = Y       | c =  }} {{eqn | lll=\leadstoandfrom       | ll= \exists a \in Y:       | l = X       | r = Y_a       | c = where $Y_a$ denotes the initial segment of $Y$ determined by $a$ }} {{eqn | lll=\leadstoandfrom       | l = X       | r = a       | c = since $Y_a = a$ }} {{eqn | lll=\leadstoandfrom       | l = X       | o = \in       | r = Y       | c =  }} {{end-eqn}} Thus: :the strict ordering $\subsetneqq$ on ordinals and :the strict ordering $\in$ on ordinals are the same. {{AimForCont}} the ordinals were not well-ordered by $\subsetneqq$. Then we could find a sequence $\sequence {X_n}_{n \mathop = 0}^\infty$ of ordinals such that: :$X_0 \supsetneqq X_1 \supsetneqq X_2 \cdots$ So for all $n > 0$: :$X_n \subsetneqq X_0$ so: :$X_n \in X_0$ Thus $\sequence {X_{n + 1} }_{n \mathop = 0}^\infty$ is a decreasing sequence under $\subsetneqq$ of elements of $\sequence {X_n}$. But since $X_0$ is an ordinal it is well-ordered by $\subsetneqq$. From Condition for Well-Foundedness, this demonstrates a contradiction. {{qed}}
Let $A \subseteq B$, and let $S$ be any set. From Set Intersection Preserves Subsets: :$A \subseteq B, \ S \subseteq T \implies A \cap S \subseteq B \cap T$ for arbitrary sets $S$ and $T$. Substituting $S$ for $T$: :$A \subseteq B, \ S \subseteq S \implies A \cap S \subseteq B \cap S$ From Set is Subset of Itself, $S \subseteq S$ for all sets $S$. Hence the result: :$A \subseteq B \implies A \cap S \subseteq B \cap S$ {{qed}}
For each $\alpha \in I$, define $A_\alpha := A$. Then by Intersection is Idempotent, it follows that: :$\displaystyle \bigcap_{\alpha \mathop \in I} A_\alpha = A$ Since $A \subseteq B_\alpha$ for all $\alpha \in I$, the premises of Set Intersection Preserves Subsets are satisfied. Applying this theorem gives: :$\displaystyle A = \bigcap_{\alpha \mathop \in I} A_\alpha \subseteq \bigcap_{\alpha \mathop \in I} B_\alpha$ which is precisely the desired result. {{qed}} Category:Set Intersection Preserves Subsets Category:Indexed Families ktv25tk9qpyi1uw94xao74a3m6bdqtx
Let $\mathbb T \subseteq \powerset T$. Suppose that $\forall X \in \mathbb T: S \subseteq X$. Consider any $x \in S$. By definition of subset, it follows that: :$\forall X \in \mathbb T: x \in X$ Thus it follows from definition of set intersection that: :$x \in \bigcap \mathbb T$ Thus by definition of subset, it follows that: :$S \subseteq \bigcap \mathbb T$ So: :$\paren {\forall X \in \mathbb T: S \subseteq X} \implies S \subseteq \bigcap \mathbb T$ {{qed|lemma}} Now suppose that $S \subseteq \bigcap \mathbb T$. From Intersection is Subset: General Result we have: :$\forall X \in \mathbb T: \bigcap \mathbb T \subseteq X$ So from Subset Relation is Transitive, it follows that: :$\forall X \in \mathbb T: S \subseteq \bigcap \mathbb T \subseteq X$ So it follows that $\forall X \in \mathbb T: S \subseteq X$. So: :$S \subseteq \bigcap \mathbb T \implies \paren {\forall X \in \mathbb T: S \subseteq X}$ {{qed|lemma}} Hence: :$\paren {\forall X \in \mathbb T: S \subseteq X} \iff S \subseteq \bigcap \mathbb T$ {{Qed}}
Let $\mathbb S \subseteq \powerset S$. By Union of Subsets is Subset: Subset of Power Set: :$\displaystyle \paren {\forall X \in \mathbb S: X \subseteq T} \implies \bigcup \mathbb S \subseteq T$ {{qed|lemma}} Now suppose that $\displaystyle \bigcup \mathbb S \subseteq T$. Consider any $X \in \mathbb S$ and take any $x \in X$. From Set is Subset of Union: General Result we have that: :$\displaystyle X \subseteq \bigcup \mathbb S$ Thus: :$\displaystyle x \in \bigcup \mathbb S$ But: :$\displaystyle \bigcup \mathbb S \subseteq T$ So it follows that: :$X \subseteq T$ So: :$\displaystyle \bigcup \mathbb S \subseteq T \implies \paren {\forall X \in \mathbb S: X \subseteq T}$ {{qed|lemma}} Hence: :$\displaystyle \paren {\forall X \in \mathbb S: X \subseteq T} \iff \bigcup \mathbb S \subseteq T$ {{Qed}}
By definition an involution on $S$ is a mapping such that: :$\forall x \in S: \map f {\map f x} = x$ === Necessary Condition === Let $f$ be an involution. By Involution is Permutation, $f$ is a permutation and therefore by definition a bijection. Then: {{begin-eqn}} {{eqn | l = \tuple {x, y}       | o = \in       | r = f       | c = considering $f$ as a relation: $f \subseteq S \times S$ }} {{eqn | ll= \leadsto       | l = \map f x       | r = y       | c = {{Defof|Mapping}} }} {{eqn | ll= \leadsto       | l = \map f {\map f x}       | r = \map f y       | c = {{Defof|Mapping}}: $a = b \implies \map f a = \map f b$ }} {{eqn | ll= \leadsto       | l = x       | r = \map f y       | c = {{Defof|Involution (Mapping)|Involution}} }} {{eqn | ll= \leadsto       | l = \tuple {y, x}       | o = \in       | r = f       | c = considering $f$ as a relation: $f \subseteq S \times S$ }} {{end-eqn}} Thus $f$, considered as a relation, is symmetric. Thus it has been shown that if $f$ is an involution, it is both a bijection and a symmetric relation. {{qed|lemma}} === Sufficient Condition === Let $f$ be a mapping which is both a bijection and a symmetric relation. Then: {{begin-eqn}} {{eqn | l = \map f x       | r = y       | c = for some unique $y \in S$ as $f$ is a bijection }} {{eqn | ll= \leadsto       | l = \tuple {x, y}       | o = \in       | r = f       | c = considering $f$ as a relation: $f \subseteq S \times S$ }} {{eqn | ll= \leadsto       | l = \tuple {y, x}       | o = \in       | r = f       | c = {{Defof|Symmetric Relation}} }} {{eqn | ll= \leadsto       | l = \map f y       | r = x       | c = {{Defof|Mapping}} }} {{eqn | ll= \leadsto       | l = \map f {\map f x}       | r = x       | c = as $y = \map f x$ }} {{end-eqn}} and so $f$ is shown to be an involution. {{qed}}
{{begin-eqn}} {{eqn | o =        | r = S \subseteq \paren {S \cup T}       | c = Set is Subset of Union }} {{eqn | o = \leadsto       | r = S \cap \paren {S \cup T} = S       | c = Intersection with Subset is Subset }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = x       | o = \in       | r = S \cap \paren {S \cup T} }} {{eqn | ll= \leadstoandfrom       | l = x       | o = \in       | r = S \land \paren {x \in S \lor x \in T}       | c = {{Defof|Set Intersection}} and {{Defof|Set Union}} }} {{eqn | ll= \leadstoandfrom       | l = x       | o = \in       | r = S       | c = Conjunction Absorbs Disjunction }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | o =        | r = \paren {S \cap T} \subseteq S       | c = Intersection is Subset }} {{eqn | o = \leadsto       | r = S \cup \paren {S \cap T} = S       | c = Union with Superset is Superset }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = x       | o = \in       | r = S \cup \paren {S \cap T} }} {{eqn | ll= \leadstoandfrom       | l = x       | o = \in       | r = S \lor \paren {x \in S \land x \in T}       | c = {{Defof|Set Intersection}} and {{Defof|Set Union}} }} {{eqn | ll= \leadstoandfrom       | l = x       | o = \in       | r = S       | c = Disjunction Absorbs Conjunction }} {{end-eqn}} {{qed}}
As $S_1, S_2, \ldots, S_n$ are pairwise disjoint, their intersections are all empty. The Cardinality of Set Union holds, but from Cardinality of Empty Set, all the terms apart from the first vanish. {{qed}}
By the definition of a correspondence it will be shown that $\mathcal R$ is both left-total and right-total. $\mathcal R$ is left-total iff: :$\forall x \in S: \exists y \in T: x \mathcal R y$ By the definition of the pre-image of $\mathcal R$: :$\Preimg {\mathcal R} = \set {x \in S: \exists y \in T: x \mathcal R y}$ Therefore $\mathcal R$ is left-total. $\mathcal R$ is right-total {{iff}}: :$\forall x \in T: \exists y \in S: x \mathcal R y$ By the definition of the image of $\mathcal R$: :$\Img {\mathcal R} = \set {x \in T: \exists y \in S: x \mathcal R y}$ Therefore $\mathcal R$ is right-total. Hence $\mathcal R$ is a correspondence. {{qed}} Category:Relation Theory 63cme2s5mqsyuouiic3tcstuaxx5ylw
{{begin-eqn}} {{eqn | l = \forall X \in \mathbb S: \bigcap \mathbb S       | o = \subseteq       | r = X       | c = Intersection is Subset: General Result }} {{eqn | ll= \leadsto       | l = \forall X \in \mathbb S: \RR \sqbrk {\bigcap \mathbb S}       | o = \subseteq       | r = \RR \sqbrk X       | c = Image of Subset is Subset of Image }} {{eqn | ll= \leadsto       | l = \RR \sqbrk {\bigcap \mathbb S}       | o = \subseteq       | r = \bigcap_{X \mathop \in \mathbb S} \RR \sqbrk X       | c = Intersection is Largest Subset: General Result }} {{end-eqn}} {{qed}} Category:Relation Theory Category:Set Intersection noktsv3g8aer4b4jghjbpksucmm06il
=== Sufficient Condition === Suppose: :$\ds \RR \sqbrk {\bigcap \mathbb S} = \bigcap_{X \mathop \in \mathbb S} \RR \sqbrk X$  where $\mathbb S$ is ''any'' subset of $\powerset S$. Then by definition of $\mathbb S$: :$\forall S_1, S_2 \in \mathbb S: \RR \sqbrk {S_1 \cap S_2} = \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$ and the sufficient condition applies for Image of Intersection under One-to-Many Relation. So $\RR$ is one-to-many. {{qed|lemma}} === Necessary Condition === Suppose $\RR$ is one-to-many. From Image of Intersection under Relation: General Result, we already have: :$\ds \RR \sqbrk {\bigcap \mathbb S} \subseteq \bigcap_{X \mathop \in \mathbb S} \RR \sqbrk X$ so we just need to show: :$\ds \forall \mathbb S \subseteq \powerset S: \bigcap_{X \mathop \in \mathbb S} \RR \sqbrk X \subseteq \RR \sqbrk {\bigcap \mathbb S}$ Let: :$\displaystyle t \in \bigcap_{X \mathop \in \mathbb S} \RR \sqbrk X$ Then: {{begin-eqn}} {{eqn | l = t       | o = \in       | r = \bigcap_{X \mathop \in \mathbb S} \RR \sqbrk X       | c =  }} {{eqn | ll= \leadsto       | l = \forall X \in \mathbb S: t       | o = \in       | r = \RR \sqbrk X       | c = {{Defof|Set Intersection}} }} {{eqn | ll= \leadsto       | l = \forall X \in \mathbb S: \exists x \in X: \tuple {x, t}       | o = \in       | r = \RR       | c = {{Defof|Relation}} }} {{eqn | ll= \leadsto       | l = x       | o = \in       | r = \bigcap \mathbb S       | c = {{Defof|One-to-Many Relation}} }} {{eqn | ll= \leadsto       | l = \RR \sqbrk x       | o = \in       | r = \RR \sqbrk {\bigcap \mathbb S}       | c = Image of Element is Subset }} {{eqn | ll= \leadsto       | l = \bigcap_{X \mathop \in \mathbb S} \RR \sqbrk X       | o = \subseteq       | r = \RR \sqbrk {\bigcap \mathbb S}       | c = {{Defof|Subset}} }} {{end-eqn}} So if $\RR$ is one-to-many, it follows that: :$\ds \forall \mathbb S \subseteq \powerset S: \RR \sqbrk {\bigcap \mathbb S} = \bigcap_{X \mathop \in \mathbb S} \RR \sqbrk X$  {{qed|lemma}} Putting the results together: $\RR$ is one-to-many {{iff}}: :$\ds \RR \sqbrk {\bigcap \mathbb S} = \bigcap_{X \mathop \in \mathbb S} \RR \sqbrk X$  where $\mathbb S$ is ''any'' subset of $\powerset S$. {{qed}} Category:Relation Theory Category:Set Intersection p47boeielt1dixkgaw26hmlwp75lz0m
{{begin-eqn}} {{eqn | l = t       | o = \in       | r = \RR \sqbrk {\bigcup \mathbb S} }} {{eqn | ll= \leadstoandfrom       | l = \exists s \in \bigcup \mathbb S: t       | o = \in       | r = \map \RR s       | c = Image of Subset under Relation equals Union of Images of Elements }} {{eqn | ll= \leadstoandfrom       | l = \exists X \in \mathbb S: \exists s \in X: t       | o = \in       | r = \map \RR s       | c = {{Defof|Union of Set of Sets}} }} {{eqn | ll= \leadstoandfrom       | l = \exists X \in \mathbb S: t       | o = \in       | r = \RR \sqbrk X       | c = {{Defof|Image of Subset under Relation}} }} {{eqn | ll= \leadstoandfrom       | l = t       | o = \in       | r = \bigcup_{X \mathop \in \mathbb S} \RR \sqbrk X       | c = {{Defof|Union of Set of Sets}} }} {{end-eqn}} {{Qed}} Category:Image of Union under Relation mnf0d9qopv4id9aid2wimeqyfqmjdvk
As $f$, being a mapping, is also a relation, we can apply Image of Union under Relation: General Result: :$\displaystyle \RR \sqbrk {\bigcup \mathbb S} = \bigcup_{X \mathop \in \mathbb S} \RR \sqbrk X$ {{qed}}
As $f$, being a mapping, is also a relation, we can apply Preimage of Union under Relation: General Result: :$\displaystyle \RR^{-1} \sqbrk {\bigcup \mathbb T} = \bigcup_{X \mathop \in \mathbb T} \RR^{-1} \sqbrk X$ {{qed}}
We have that $\RR^{-1}$ is a relation. The result follows from Image of Union under Relation: General Result. {{qed}} Category:Preimage of Union under Relation gv13kk17pp72fpfp0cbp6khnlic96xr
As $f$, being a mapping, is also a relation, we can apply Image of Intersection under Relation: General Result: :$\displaystyle \RR \sqbrk {\bigcap \mathbb S} \subseteq \bigcap_{X \mathop \in \mathbb S} \RR \sqbrk X$ {{qed}}
$f$ is a mapping. Therefore it is by definition a many-to-one relation. It follows from Inverse of Many-to-One Relation is One-to-Many that its inverse $f^{-1}$ is a one-to-many relation. Thus Image of Intersection under One-to-Many Relation: General Result applies: :$\displaystyle \mathcal R \left[{\bigcap \mathbb T}\right] = \bigcap_{X \mathop \in \mathbb T} \mathcal R \left[{X}\right]$ where here $\mathcal R = f^{-1}$. {{Qed}}
Suppose $\mathcal R$ is a bijection. Then by definition: :$(1): \quad \mathcal R$ is a surjection and therefore right-total. :$(2): \quad \mathcal R$ is a mapping and therefore left-total. :$(3): \quad \mathcal R$ is a one-to-one relation and therefore also both a many-to-one relation and a one-to-many relation. By Inverses of Right-Total and Left-Total Relations, we have that $\mathcal R^{-1}$ is also both right-total and left-total. By Inverse of Many-to-One Relation is One-to-Many we have that $\mathcal R^{-1}$ is also both a many-to-one relation and a one-to-many relation. From Condition for Composite Relation with Inverse to be Identity, it follows that: :$\mathcal R^{-1} \circ \mathcal R = I_S$ and :$\mathcal R \circ \mathcal R^{-1} = I_T$ Hence the result. {{qed}}
Let $\mathcal R \subseteq S \times T$ be such that: : $\mathcal R^{-1} \circ \mathcal R = I_S$ and : $\mathcal R \circ \mathcal R^{-1} = I_T$. From Condition for Composite Relation with Inverse to be Identity, we have that: :$\mathcal R$ is many-to-one :$\mathcal R$ is right-total :$\mathcal R^{-1}$ is many-to-one :$\mathcal R^{-1}$ is right-total. From Inverse of Many-to-One Relation is One-to-Many, it follows that both $\mathcal R$ and $\mathcal R^{-1}$ are by definition one-to-one. From Inverses of Right-Total and Left-Total Relations, it also follows that both $\mathcal R$ and $\mathcal R^{-1}$ are left-total. By definition, an injection is a relation which is: :One-to-one :left-total. Also by definition, a surjection is a relation which is: :Left-total :Many-to-one :Right-total. It follows that $\mathcal R$ is both an injection and a surjection, and so by definition a bijection. By the same coin, the same applies to $\mathcal R^{-1}$. {{qed}}
Let $f$ be a bijection. Then it is both an injection and a surjection, thus both the described $g_1$ and $g_2$ must exist from Injection iff Left Inverse and Surjection iff Right Inverse. The fact that $g_1 = g_2 = f^{-1}$ follows from Left and Right Inverses of Mapping are Inverse Mapping. {{qed}}
Suppose $f$ is a bijection. From Bijection iff Inverse is Bijection and Composite of Bijection with Inverse is Identity Mapping, it is shown that the inverse mapping $f^{-1}$ such that: : $f^{-1} \circ f = I_S$ : $f \circ f^{-1} = I_T$ is a bijection. {{qed}}
Let $f$ be a bijection. By definition, $f$ is a mapping, and hence also by definition a relation. Hence the result Bijective Relation has Left and Right Inverse applies directly and so: : $f^{-1} \circ f = I_S$ and : $f \circ f^{-1} = I_T$ {{qed}}
From the definition of generalized momentum and the Euler-Lagrange Equations: {{begin-eqn}} {{eqn | l = 0       | r = \frac \d {\d t} \frac {\partial \mathcal L} {\partial \dot x} - \frac {\partial \mathcal L} {\partial x}       | c =  }} {{eqn | r = \dot p_i - \frac {\partial \mathcal L} {\partial x}       | c =  }} {{eqn | ll= \leadsto       | l = \dot p_i       | r = \frac {\partial \mathcal L} {\partial x}       | c =   }} {{end-eqn}} Therefore, via the definition of action, Definite Integral of Partial Derivative and the Fundamental Theorem of Calculus: {{begin-eqn}} {{eqn | l = \frac {\d S_{12} } {\d x}       | r = \frac \d {\d x} \int_{t_1}^{t_2} \mathcal L \rd t       | c =  }} {{eqn | r = \int_{t_1}^{t_2} \frac {\partial \mathcal L} {\partial x} \rd t       | c =  }} {{eqn | r = \int_{t_1}^{t_2} \dot p_i \rd t       | c =  }} {{eqn | r = p_2 - p_1       | c =  }} {{end-eqn}} {{qed}} Category:Laws of Conservation kge2yazzdb8mmu7bf20py9ll0offsid
An injection is a type of one-to-one relation, and therefore also a one-to-many relation. Therefore Image of Intersection under One-to-Many Relation applies: :$\displaystyle \forall \mathbb S \subseteq \mathcal P \left({S}\right): \mathcal R \left[{\bigcap \mathbb S}\right] = \bigcap_{X \mathop \in \mathbb S} \mathcal R \left[{\mathbb S}\right]$ {{iff}} $\mathcal R$ is a one-to-many relation. We have that $f$ is a mapping and therefore a many-to-one relation. So $f$ is a one-to-many relation {{iff}} $f$ is also an injection. It follows that: :$\displaystyle \forall \mathbb S \subseteq \mathcal P \left({S}\right): f \left[{\bigcap \mathbb S}\right] = \bigcap_{X \mathop \in \mathbb S} f \left[{X}\right]$  {{iff}} $f$ is an injection. {{qed}} Category:Image of Intersection under Injection m7cd4l2j5vrgqdxts2t332l9w5q26um
From the definition: :For all sets $X$ and all mappings $f_1: X \to S$ and $f_2: X \to T$ there exists a unique mapping $h: X \to P$ such that: ::$\phi_1 \circ h = f_1$ ::$\phi_2 \circ h = f_2$ Let $X = S$ and let $f_1 = I_S$ where $I_S$ is the identity mapping on $S$. Then we have: :$\phi_1 \circ h = I_S$ We have from Identity Mapping is Surjection that $I_S$ is a surjection. From Surjection if Composite is Surjection it follows that $\phi_1$ is a surjection. Similarly, let $X = T$ and let $f_2 = I_T$ where $I_T$ is the identity mapping on $T$. Then we have: :$\phi_2 \circ h = I_T$ We have from Identity Mapping is Surjection that $I_T$ is a surjection. From Surjection if Composite is Surjection it follows that $\phi_2$ is a surjection. {{qed}}
Consider any set $X$ and mappings $f_1: X \to S$ and $f_2: X \to T$. Define $h: X \to S \times T$ by: :$\forall x \in X: \map h x = \tuple {\map {f_1} x, \map {f_2} x}$ Then for all $x \in X$ we have: :$\map {\paren {\pr_1 \mathop \circ h} } x = \pr_1 \tuple {\map {f_1} x, \map {f_2} x} = \map {f_1} x)$ and :$\map {\paren {\pr_2 \mathop \circ h} } x = \pr_2 \tuple {\map {f_1} x, \map {f_2} x} = \map {f_2} x$ So: :$\pr_1 \mathop \circ h = f_1$ :$\pr_2 \mathop \circ h = f_2$ Thus $h$ is shown to exist. Suppose there exists a mapping $k: X \to S \times T$ such that: :$\pr_1 \mathop \circ k = f_1$ :$\pr_2 \mathop \circ k = f_2$ Let $x \in X$ and let $\map k x = \tuple {s, t}$. Then: :$\map {f_1} x = \map {\paren {\pr_1 \mathop \circ k} } x = \map {\pr_1} {s, t} = s$ :$\map {f_2} x = \map {\paren {\pr_2 \mathop \circ k} } x = \map {\pr_2} {s, t} = t$ and so: {{begin-eqn}} {{eqn | l = \map k x       | r = \tuple {s, t}       | c =  }} {{eqn | r = \tuple {\map {f_1} x, \map {f_2} x}       | c =  }} {{eqn | r = \map h x       | c =  }} {{end-eqn}} and so $k = h$. So $h$ is unique. Hence the result. {{qed}}
We have that $\struct {P, \phi_1, \phi_2}$ is a set product. From the definition of set product $\chi: Q \to P$ is the unique mapping such that: :$\phi_1 \circ \chi = \psi_1$ :$\phi_2 \circ \chi = \psi_2$ Similarly, we have that $\struct {Q, \psi_1, \psi_2}$ is a set product. So from the definition of set product there exists a unique mapping $\xi: P \to Q$ such that: :$\psi_1 \circ \xi = \phi_1$ :$\psi_2 \circ \xi = \phi_2$ It can be seen that $\chi \circ \xi: P \to P$ is the unique mapping such that: :$\phi_1 \circ \chi \circ \xi = \phi_1$ :$\phi_2 \circ \chi \circ \xi = \phi_2$ and similarly that $\xi \circ \chi: Q \to Q$ is the unique mapping such that: :$\psi_1 \circ \xi \circ \chi = \psi_1$ :$\psi_2 \circ \xi \circ \chi = \psi_2$ So it appears that: :$\chi \circ \xi = I_P$ :$\xi \circ \chi = I_Q$ where $I_P$ and $I_Q$ are the identity mappings on $P$ and $Q$ respectively. From the corollary to Bijection iff Left and Right Inverse it follows that $\chi$ and $\xi$ are bijections such that $\chi = \xi^{-1}$. {{qed}}