hcju/proofwikiqa huggingface dataset:
By Identity Element is Idempotent, $e_S$ is idempotent. Let $x$ be a cancellable idempotent element of $\left({S, \circ}\right)$. {{begin-eqn}} {{eqn | l = x \circ x       | r = x       | c = $x$ is idempotent }} {{eqn | r = x \circ e_S       | c = Definition of Identity }} {{end-eqn}} So $x \circ x = x \circ e_S$. But because $x$ is also by hypothesis cancellable, it follows that $x = e_S$. {{Qed}}
By Set of All Self-Maps is Semigroup, $\struct {S^S, \circ}$ is a semigroup. By Identity Mapping is Left Identity and Identity Mapping is Right Identity the identity mapping on $S$ is the identity element of $\struct {S^S, \circ}$. Since $\struct {S^S, \circ}$ is a semigroup with an identity element, it is a monoid by definition. {{qed}}
From Subsemigroup Closure Test, $(1)$ and $(2)$ are sufficient to show that $\left({T, \circ}\right)$ is a subsemigroup of $\left({S, \circ}\right)$. Demonstrating the presence of an identity is then sufficient to show that it is a monoid. {{qed}} Category:Monoids 7f3509pyo2ofnhy6uphzcd05t64iw1c
By Identity of Monoid is Unique: : there is only one identity element of $\struct {S, \circ}$ and: : there is only one identity element of $\struct {T, \circ}$. Let $e_S$ be the identity of $\struct {S, \circ}$, and $e_T$ the identity of $\struct {T, \circ}$. From Identity is only Idempotent Cancellable Element, $e_S$ is the only cancellable element of $\struct {S, \circ}$ which is idempotent. But all elements of $\struct {S, \circ}$ are cancellable. Thus $e_S$ is the only idempotent element of $\struct {S, \circ}$. Again, all elements of $\struct {S, \circ}$ are cancellable. Thus from Cancellable Element is Cancellable in Subset, all elements of $\struct {T, \circ}$ are cancellable. Thus, $e_T$ is the only element of $\struct {T, \circ}$ which is idempotent. Thus, as $e_T \in S$, we have $e_S \circ e_T = e_T = e_T \circ e_T$ and thus $e_S = e_T$. {{qed}}
Let $C$ be the set of cancellable elements of $\struct {S, \circ}$. Obviously $C \subseteq S$. From Cancellable Elements of Semigroup form Subsemigroup, $\struct {C, \circ}$ is a subsemigroup of $S$. Let $e_S$ be the identity of $\struct {S, \circ}$. From Identity of Monoid is Cancellable, $e_S$ is cancellable, therefore $e_S \in C$. As $e_S$ is the identity of $S$, we have that: :$\forall x \in C: x \circ e_S = x = e_S \circ x$. Thus $e_S$ is the identity of $\struct {C, \circ}$, which is therefore a monoid. As $C \subseteq S$, the result follows. {{qed}} Category:Monoids kbtbh08ex31xlu4bbjua3p7ib14hisw
{{begin-eqn}} {{eqn | l = \paren {x \circ x_L} \circ \paren {x \circ x_L}       | r = x \circ \paren {x_L \circ x} \circ x_L       | c = as $\circ$ is associative }} {{eqn | r = x \circ e_L \circ x_L       | c = {{Defof|Left Inverse Element}} }} {{eqn | r = x \circ x_L       | c = {{Defof|Left Identity}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \paren {x_R \circ x} \circ \paren {x_R \circ x}       | r = x_R \circ \paren {x \circ x_R} \circ x       | c = as $\circ$ is associative }} {{eqn | r = x_R \circ e_R \circ x       | c = {{Defof|Right Inverse Element}} }} {{eqn | r = x_R \circ x       | c = {{Defof|Right Identity}} }} {{end-eqn}} {{qed}}
Let $y = x \circ x_L$. Then: {{begin-eqn}} {{eqn | l = e_L \circ y       | r = \paren {y_L \circ y} \circ y       | c = {{Defof|Left Inverse Element}} }} {{eqn | r = y_L \circ \paren {y \circ y}       | c = as $\circ$ is associative }} {{eqn | r = y_L \circ y       | c = Product of Semigroup Element with Left Inverse is Idempotent: $y = x \circ x_L$ }} {{eqn | r = e_L       | c = {{Defof|Left Inverse Element}} }} {{end-eqn}} So $x \circ x_L = e_L$, and $x_L$ behaves as a right inverse as well as a left inverse with respect to the left identity. {{qed}}
We note that as $\struct {S, \circ}$ is a monoid, $\circ$ is associative by definition. {{begin-eqn}} {{eqn | l = x_L       | r = x_L \circ e_S       | c = {{Defof|Identity Element}} }} {{eqn | r = x_L \circ \paren {x \circ x_R}       | c = {{Defof|Right Inverse Element}} }} {{eqn | r = \paren {x_L \circ x} \circ x_R       | c = {{Defof|Associative Operation}} }} {{eqn | r = e_S \circ x_R       | c = {{Defof|Left Inverse Element}} }} {{eqn | r = x_R       | c = {{Defof|Identity Element}} }} {{end-eqn}} Its uniqueness comes from Inverse in Monoid is Unique. {{qed}}
Let $z_L$ be the left inverse of $x \circ y$ and $z_R$ be the right inverse of $y \circ x$. Then: {{begin-eqn}} {{eqn | l = z_L \circ \paren {x \circ y}       | r = e_S       | c =  }} {{eqn | ll= \leadsto       | l = \paren {z_L \circ x} \circ y       | r = e_S       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \paren {y \circ x} \circ z_R       | r = e_S       | c =  }} {{eqn | ll= \leadsto       | l = y \circ \paren {x \circ z_R}       | r = e_S       | c =  }} {{end-eqn}} Thus $y$ has both a left inverse $z_L \circ x$ and a right inverse $x \circ z_R$. From Left Inverse and Right Inverse is Inverse: :$z_L \circ x = x \circ z_R$ and $y$ has an inverse, that is, is invertible. {{qed|lemma}} From the above, we have: {{begin-eqn}} {{eqn | l = \paren {z_L \circ x} \circ y       | r = e_S       | c =  }} {{eqn | ll= \leadsto       | l = \paren {x \circ z_R} \circ y       | r = e_S       | c =  }} {{eqn | ll= \leadsto       | l = x \circ \paren {z_R \circ y}       | r = e_S       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = y \circ \paren {x \circ z_R}       | r = e_S       | c =  }} {{eqn | ll= \leadsto       | l = y \circ \paren {z_L \circ x}       | r = e_S       | c =  }} {{eqn | ll= \leadsto       | l = \paren {y \circ z_L} \circ x       | r = e_S       | c =  }} {{end-eqn}} Thus $x$ has both a left inverse $y \circ z_L$ and a right inverse $z_R \circ y$. From Left Inverse and Right Inverse is Inverse: :$y \circ z_L = z_R \circ y$ and $x$ has an inverse, that is, is invertible. {{qed}}
First, note that: {{begin-eqn}} {{eqn | o =        | r = x \in X, y \in Y       | c =  }} {{eqn | o = \leadsto       | r = x^{-1} \in X^{-1}, y^{-1} \in Y^{-1}       | c = {{Defof|Inverse of Subset of Group}} }} {{eqn | o = \leadsto       | r = y^{-1} \circ x^{-1} \in Y^{-1} \circ X^{-1}       | c = {{Defof|Subset Product}} }} {{end-eqn}} Now: {{begin-eqn}} {{eqn | l = x \circ y       | o = \in       | r = X \circ Y       | c = {{Defof|Subset Product}} }} {{eqn | ll= \leadsto       | l = \paren {x \circ y}^{-1}       | o = \in       | r = \paren {X \circ Y}^{-1}       | c = {{Defof|Inverse of Subset of Group}} }} {{eqn | ll= \leadsto       | l = y^{-1} \circ x^{-1}       | o = \in       | r = \paren {X \circ Y}^{-1}       | c = Inverse of Group Product }} {{eqn | ll= \leadsto       | l = Y^{-1} \circ X^{-1}       | o = \subseteq       | r = \paren {X \circ Y}^{-1}       | c = {{Defof|Subset}} }} {{end-eqn}} By a similar argument we see that $\paren {X \circ Y}^{-1} \subseteq Y^{-1} \circ X^{-1}$. Hence the result. {{qed}}
=== Superset === We will show that $\forall z \in Y^{-1} \circ X^{-1}: z \in \left({X \circ Y}\right)^{-1}$, from which: :$Y^{-1} \circ X^{-1} \subseteq \left({X \circ Y}\right)^{-1}$ Let $z \in Y^{-1} \circ X^{-1}$. By the definition of subset product: :$\exists x' \in X^{-1}, y' \in Y^{-1}: z = y' \circ x'$ Then by Inverse of Group Product: :$(2)\quad z^{-1} = x'^{-1} \circ y'^{-1}$ By the definition of inverse of subset: :$x'^{-1} \in X$ and $y'^{-1} \in Y$ By the definition of subset product: :$x'^{-1} \circ y'^{-1} \in X \circ Y$ Thus by $(2)$: :$z^{-1} \in X \circ Y$ By the definition of inverse of subset: :$z \in \left({X \circ Y}\right)^{-1}$ === Subset === We will show that $\forall z \in \left({X \circ Y}\right)^{-1}: z \in Y^{-1} \circ X^{-1}$, from which: :$\left({X \circ Y}\right)^{-1} \subseteq Y^{-1} \circ X^{-1}$ Let $z \in \left({X \circ Y}\right)^{-1}$. By the definition of inverse of subset: :$z^{-1} \in X \circ Y$ From Inverse of Inverse of Subset of Group: :$z^{-1} \in \left({ X^{-1} }\right)^{-1} \circ \left({ Y^{-1} }\right)^{-1}$ Thus by the superset proof above: :$z^{-1} \in \left({ Y^{-1} \circ X^{-1} }\right)^{-1}$ From the definition of inverse of subset: :$z \in Y^{-1} \circ X^{-1}$ {{qed}}
From Identity Element is Idempotent: :$e \circ e = e$ Hence the result by definition of identity element. {{qed}}
Let $a \in S$ be invertible. Suppose $a \circ x = a \circ y$. Then: {{begin-eqn}} {{eqn | l = x       | r = e_S \circ x       | c = Behaviour of Identity }} {{eqn | r = \left({a^{-1} \circ a}\right) \circ x       | c = Behaviour of Inverse }} {{eqn | r = a^{-1} \circ \left({a \circ x}\right)       | c = Associativity of $\circ$ }} {{eqn | r = a^{-1} \circ \left({a \circ y}\right)       | c = By Hypothesis }} {{eqn | r = \left({a^{-1} \circ a}\right) \circ y       | c = [Definition:Associative Operation|Associativity]] of $\circ$ }} {{eqn | r = e_S \circ y       | c = Behaviour of Inverse }} {{eqn | r = y       | c = Behaviour of Identity }} {{end-eqn}} A similar argument shows that $x \circ a = y \circ a \implies x = y$. {{Qed}}
Suppose $a \in \struct {S, \circ}$ is invertible. A permutations is a bijection from a set to itself. As $\lambda_a: S \to S$ and $\rho_a: S \to S$ are defined from $S$ to $S$, all we need to do is show that they are bijections. To do that we can show that they are both injective and surjective. === Injectivity === From Invertible Element of Monoid is Cancellable, as $a$ is invertible, it is also cancellable. From Cancellable iff Regular Representations Injective, it follows that both $\lambda_a$ and $\rho_a$ are injective. {{qed|lemma}} === Surjectivity === Let $y \in S$. Then: {{begin-eqn}} {{eqn | n = 1       | l = y       | r = \paren {a \circ a^{-1} } \circ y       | c = (as $a$ is invertible) }} {{eqn | r = a \circ \paren {a^{-1} \circ y}       | c = Associativity of $\circ$ }} {{eqn | r = \lambda_a \paren {a^{-1} \circ y}       | c = Definition of $\lambda_a$ }} {{end-eqn}} {{begin-eqn}} {{eqn | n = 2       | l = y       | r = y \circ \paren {a^{-1} \circ a} \circ y       | c = (as $a$ is invertible) }} {{eqn | r = \paren {y \circ a^{-1} } \circ a       | c = Associativity of $\circ$ }} {{eqn | r = \rho_a \paren {y \circ a^{-1} }       | c = Definition of $\rho_a$ }} {{end-eqn}} Thus both $\lambda_a$ and $\rho_a$ are surjective. {{qed|lemma}} So $\lambda_a$ and $\rho_a$ are injective and surjective, and therefore bijections, and thus permutationss of $S$. {{qed}}
Let $\left({S, \circ}\right)$ be a group. Then for all $a \in S$: {{begin-eqn}} {{eqn | l=e       | r=a \circ a^{-1}       | c=Every $a \in G$ is Invertible }} {{eqn | ll=\implies       | l=e \circ a       | r=a \circ a^{-1} \circ a       | c= }} {{eqn | ll=\implies       | l=a       | r=a \circ a^{-1} \circ a       | c=Definition of Identity }} {{end-eqn}} and {{begin-eqn}} {{eqn | l=e       | r=a \circ a^{-1}       | c=Every $a \in G$ is Invertible }} {{eqn | ll=\implies       | l=a^{-1} \circ e       | r=a^{-1} \circ a \circ a^{-1}       | c= }} {{eqn | ll=\implies       | l=a^{-1}       | r=a^{-1} \circ a \circ a^{-1}       | c=Definition of Identity }} {{end-eqn}} Thus the criteria of an inverse semigroup are fulfilled. {{qed}} Category:Group Theory gkhnvfry9jz2o0x0xbb4iy7aq2uz4nt
From Cancellable Elements of Monoid form Submonoid, $\struct {C, \circ}$ is a submonoid of $\struct {S, \circ}$. Let its identity be $e_C$ (which may or may not be the same as $e_S$). Let $T$ be the set of all invertible elements of $S$. From Invertible Element of Monoid is Cancellable, all the invertible elements of $S$ are also all cancellable, so $T \subseteq C$. Let $x, y \in T$. Clearly $x^{-1}, y^{-1} \in T$, as if $x, y$ are invertible, then so are their inverses. Taking the group axioms in turn: === G0: Closure === By Inverse of Product, $x^{-1} \circ y^{-1} \in T$. Thus $\struct {T, \circ}$ is closed. {{qed|lemma}} === G1: Associativity === This is inherited from $S$, by Subsemigroup Closure Test. Thus $\struct {T, \circ}$ is associative. {{qed|lemma}} === G2: Identity === All the elements of $\struct {C, \circ}$ are by definition cancellable So, by Identity of Cancellable Monoid is Identity of Submonoid: : $e_C \in T$ Thus $\struct {T, \circ}$ has an identity element. {{qed|lemma}} === G3: Inverses === By Inverse of Inverse in Monoid: :$\struct {x^{-1} \circ y^{-1} }^{-1} \in T$ Thus every element of $\struct {T, \circ}$ has an inverse. {{qed|lemma}} All the group axioms are thus seen to be fulfilled, and so $\struct {T, \circ}$ is a group. {{qed}}
Let $f, g, h \in T^S$. Let $\struct {T, \circ}$ be an associative algebraic structure. Then: {{begin-eqn}} {{eqn | l = \map {\paren {\paren {f \oplus g} \oplus h} } x       | r = \map {\paren {f \oplus g} } x \circ \map h x       | c = {{Defof|Pointwise Operation}} }} {{eqn | r = \paren {\map f x \circ \map g x} \circ \map h x       | c = {{Defof|Pointwise Operation}} }} {{eqn | r = \map f x \circ \paren {\map g x \circ \map h x}       | c = $\circ$ is associative }} {{eqn | r = \map f x \circ \map {\paren {g \oplus h} } x       | c = {{Defof|Pointwise Operation}} }} {{eqn | r = \map {\paren {f \oplus \paren {g \oplus h} } } x       | c = {{Defof|Pointwise Operation}} }} {{end-eqn}} {{qed}}
Let $\struct {T, \circ}$ be a commutative algebraic structure. Let $f, g \in T^S$. Then: {{begin-eqn}} {{eqn | l = \map {\paren {f \oplus g} } x       | r = \map f x \circ \map g x       | c = {{Defof|Pointwise Operation}} }} {{eqn | r = \map g x \circ \map f x       | c = $\circ$ is commutative }} {{eqn | r = \map {\paren {g \oplus f} } x       | c = {{Defof|Pointwise Operation}} }} {{end-eqn}} {{qed}}
Let $\struct {T, \circ}$ be an algebraic structure with an identity $e$. Let $f \in T^S$. Then: {{begin-eqn}} {{eqn | l = \map {\paren {f \oplus f_e} } x       | r = \map f x \circ \map {f_e} x       | c = {{Defof|Pointwise Operation}} }} {{eqn | r = \map f x \circ e       | c = {{Defof|Constant Mapping}} $f_e$ }} {{eqn | r = \map f x       | c = $e$ is the identity of $\struct {T, \circ}$ }} {{eqn | r = e \circ \map f x       | c = $e$ is the identity of $\struct {T, \circ}$ }} {{eqn | r = \map {f_e} x \circ \map f x       | c = {{Defof|Constant Mapping}} $f_e$ }} {{eqn | r = \map {\paren {f \oplus f_e} } x       | c = {{Defof|Pointwise Operation}} }} {{end-eqn}} {{qed}}
From Structure Induced by Group Operation is Group, $\struct {G^S, \oplus}$ is a group. From Structure Induced by Commutative Operation is Commutative, so is the pointwise operation it induces on $G^S$. Hence the result. {{qed}} Category:Abelian Groups Category:Pointwise Operations gffz66j74cdmaqb4seldfypi6jcr7vr
From Power Set is Closed under Union: :$\forall A, B \in \powerset S: A \cup B \in \powerset S$ From Set System Closed under Union is Commutative Semigroup, $\struct {\powerset S, \cup}$ is a commutative semigroup. From Identity of Power Set with Union, $\O$ acts as the identity of $\struct {\powerset S, \cup}$. It remains to be shown that only $\O$ has an inverse: For $T \subseteq S$ to have an inverse under $\cup$, we require $T^{-1} \cup T = \O$. From this it follows that $T = \O = T^{-1}$. The result follows by definition of commutative monoid. {{qed}}
From Power Set is Closed under Intersection: :$\forall A, B \in \powerset S: A \cap B \in \powerset S$ From Set System Closed under Intersection is Commutative Semigroup, $\struct {\powerset S, \cap}$ is a commutative semigroup. From Identity of Power Set with Intersection, we have that $S$ acts as the identity. It remains to be shown that only $S$ has an inverse: For $T \subseteq S$ to have an inverse under $\cap$, we require $T^{-1} \cap T = S$. From this it follows that $T = S = T^{-1}$. The result follows by definition of commutative monoid. {{qed}}
{{questionable|It needs to be pointed out that the definition of universally compatible relation may not be correct as such. I will address it when I work up to that chapter in my book.}} Let $\struct {S, \circ}$ be any algebraic structure. {{begin-eqn}} {{eqn | o =        | r = x_1 \Delta_S x_2 \land y_1 \Delta_S y_2       | c =  }} {{eqn | o = \leadsto       | r = x_1 = x_2 \land y_1 = y_2       | c = {{Defof|Diagonal Relation}} }} {{eqn | o = \leadsto       | r = x_1 \circ y_1 = x_2 \circ y_2       | c = (consequence of equality) }} {{eqn | o = \leadsto       | r = \paren {x_1 \circ y_1} \mathrel {\Delta_S} \paren {x_2 \circ y_2}       | c = {{Defof|Diagonal Relation}} }} {{end-eqn}} {{qed}} Category:Compatible Relations pug8w8n3as389p9zzsr5bp1d4o64vbt
Let $\left({S, \circ}\right)$ be any algebraic structure which is closed for $\circ$. By definition of trivial relation, $x \in S \land y \in S \implies x \mathcal R y$. So: {{begin-eqn}} {{eqn | l=x_1, x_2, y_1, y_2       | o=\in       | r=S       | c= }} {{eqn | ll=\implies       | l=x_1 \circ y_1, x_2 \circ y_2       | o=\in       | r=S       | c=Definition of closed algebraic structure }} {{eqn | ll=\implies       | l=\left({x_1 \circ y_1}\right)       | o=\mathcal R       | r=\left({x_2 \circ y_2}\right)       | c=Definition of Trivial Relation }} {{end-eqn}} {{qed}}
Let $c \in S$. By the definition of the constant operation: :$\forall x, y \in S: x \sqbrk c y = c$ Let $\RR$ be an equivalence relation on $S$. Every equivalence relation is reflexive, so: :$c \mathrel \RR c$ So: {{begin-eqn}} {{eqn | l = x_1 \mathrel \RR x_2       | o = \land       | r = y_1 \mathrel \RR y_2       | c =  }} {{eqn | ll= \leadsto       | l = \paren {x_1 \sqbrk c y_1}       | o = \RR       | r = \paren {x_2 \sqbrk c y_2}       | c = True Statement is implied by Every Statement }} {{end-eqn}} Hence the result. {{Qed}}
By definition of subset product: :$X \circ_\mathcal P Y = \left\{{x \circ y: x \in X, y \in Y}\right\}$ Thus: :$X \circ_\mathcal R Y = \left\{{x \circ y: x \in X, y \in Y}\right\} \cup \left\{{x \circ y: x \in \left[\!\left[{X}\right]\!\right]_\mathcal R, y \in \left[\!\left[{Y}\right]\!\right]_\mathcal R}\right\}$ The result follows from Subset of Union. {{qed}}
Let $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$ be closed. Let $\tuple {s_1, t_1} \in S \times T$ and $\tuple {s_2, t_2} \in S \times T$. Then: {{begin-eqn}} {{eqn | l = \tuple {s_1, t_1} \circ \tuple {s_2, t_2}       | r = \tuple {s_1 \circ_1 s_2, t_1 \circ_1 t_2}       | c = {{Defof|External Direct Product}} }} {{eqn | o = \in       | r = S \times T       | c = as $S$ and $T$ are closed: $s_1 \circ_1 s_2 \in S, t_1 \circ_1 t_2 \in T$ }} {{end-eqn}} demonstrating that $\struct {S \times T, \circ}$ is closed. {{qed}} Category:Direct Products Category:External Direct Products snavudfreayq56xw6zpqcy7nt5h6dok
{{begin-eqn}} {{eqn | l = \paren {\tuple {s_1, t_1} \circ \tuple {s_2, t_2} } \circ \tuple {s_3, t_3}       | r = \tuple {\paren {s_1 \circ_1 s_2} \circ_1 s_3, \paren {t_1 \circ_2 t_2} \circ_2 t_3}       | c =  }} {{eqn | r = \tuple {s_1 \circ_1 s_2 \circ_1 s_3, t_1 \circ_2 t_2 \circ_2 t_3}       | c = {{Defof|Associative Operation}} }} {{eqn | r = \tuple {s_1 \circ_1 \paren {s_2 \circ_1 s_3}, t_1 \circ_2 \paren {t_2 \circ_2 t_3} }       | c =  }} {{eqn | r = \tuple {s_1, t_1} \circ \paren {\tuple {s_2, t_2} \circ \tuple {s_3, t_3} }       | c =  }} {{end-eqn}} Thus $\circ$ is associative. {{qed}}
* From External Direct Product Closure, the external direct product of two closed algebraic structures is closed. * From External Direct Product Associativity, the external direct product of two associative algebraic structures is associative. The result follows. {{qed}} Category:Direct Products Category:Semigroups hfejmyry60s1o3wdbvteyfdeqxcglrt
Let $\circ_1$ and $\circ_2$ be commutative operations. {{begin-eqn}} {{eqn | l = \left({s_1, t_1}\right) \circ \left({s_2, t_2}\right)       | r = \left({s_1 \circ_1 s_2, t_1 \circ_2 t_2}\right)       | c =  }} {{eqn | r = \left({s_2 \circ_1 s_1, t_2 \circ_2 t_1}\right)       | c =  }} {{eqn | r = \left({s_2, t_2}\right) \circ \left({s_1, t_1}\right)       | c =  }} {{end-eqn}} Thus $\circ$ is commutative. {{qed}}
{{begin-eqn}} {{eqn | l = \tuple {s, t} \circ \tuple {e_S, e_T}       | r = \tuple {s \circ_1 e_S, t \circ_2 e_T}       | c =  }} {{eqn | r = \tuple {s, t}       | c =  }} {{eqn | l = \tuple {e_S, e_T} \circ \tuple {s, t}       | r = \tuple {e_S \circ_1 s, e_T \circ_2 t} = \left({s, t}\right)       | c =  }} {{eqn | r = \tuple {s, t}       | c =  }} {{end-eqn}} Thus $\tuple {e_S, e_T}$ is the identity of $\struct {S \times T, \circ}$. {{qed}}
Let: :$e_S$ be the identity for $\struct {S, \circ_1}$ and: :$e_T$ be the identity for $\struct {T, \circ_2}$. Also let: :$s^{-1}$ be the inverse of $s \in \struct {S, \circ_1}$ and :$t^{-1}$ be the inverse of $t \in \struct {T, \circ_2}$. Then: {{begin-eqn}} {{eqn | l = \tuple {s, t} \circ \tuple {s^{-1}, t^{-1} }       | r = \tuple {s \circ_1 s^{-1}, t \circ_2 t^{-1} }       | c =  }} {{eqn | r = \tuple {e_S, e_T}       | c =  }} {{eqn | r = \tuple {s^{-1} \circ_1 s, t^{-1} \circ_2 t}       | c =  }} {{eqn | r = \tuple {s^{-1}, t^{-1} } \circ \tuple {s, t}       | c =  }} {{end-eqn}} Thus the inverse of $\tuple {s, t}$ is $\tuple {s^{-1}, t^{-1} }$. {{qed}}
Suppose that $\circ_k$ has the morphism property under $\phi$. Suppose that $S' \subseteq S$ is closed under $\circ_k$. Thus: :$s_1, s_2 \in S' \implies s_1 \circ_k s_2 \in S'$ Similarly, suppose that  $T' \subseteq T$ is closed under $*_k$. Thus: :$t_1, t_2 \in T' \implies t_1 *_k t_2 \in T'$ First we prove that $\phi \left[{S'}\right]$ is closed under $*_k$: {{begin-eqn}} {{eqn | l = t_1, t_2       | o = \in       | r = \phi \left[{S'}\right] }} {{eqn | ll= \implies       | l = \exists s_1 \in S': t_1       | r = \phi \left({s_1}\right)       | c = Definition of Image of Subset }} {{eqn | ll= \land       | l = \exists s_2 \in S': t_2       | r = \phi \left({s_2}\right) }} {{eqn | ll= \implies       | l = t_1 *_k t_2       | r = \phi \left({s_1}\right) *_k \phi \left({s_2}\right)       | c = }} {{eqn | r = \phi \left({s_1 \circ_k s_2}\right)       | c = Definition of Morphism Property }} {{eqn | ll= \implies       | l = t_1 *_k t_2       | o = \in       | r = \phi \left[{S'}\right]       | c = $S'$ is closed under $\circ$ }} {{end-eqn}} Then we prove that $\phi^{-1} \left({T'}\right)$ is closed under $\circ_k$: {{begin-eqn}} {{eqn | l = s_1, s_2       | o = \in       | r = \phi^{-1} \left[{T'}\right] }} {{eqn | ll= \implies       | l = \phi \left({s_1}\right), \phi \left({s_2}\right)       | o = \in       | r = T'       | c = Definition of Inverse Mapping }} {{eqn | l = \phi \left({s_1}\right) *_k \phi \left({s_2}\right)       | o = \in       | r = T'       | c = }} {{eqn | ll= \implies       | l = \phi \left({s_1 \circ_k s_2}\right)       | o = \in       | r = T'       | c = Definition of Morphism Property }} {{eqn | ll= \implies       | l = s_1 \circ_k s_2       | o = \in       | r = \phi^{-1} \left[{T'}\right]       | c = Definition of Inverse Mapping }} {{end-eqn}} {{qed}}