hcju/proofwikiqa huggingface dataset:
=== Equality of Codomains === The codomains of $f$ and $f \circ I_S$ are both equal to $T$ from Codomain of Composite Relation. {{qed|lemma}} === Equality of Domains === From Domain of Composite Relation: :$\Dom {f \circ I_S} = \Dom {I_S}$ But from the definition of the identity mapping: :$\Dom {I_S} = \Img {I_S} = S$ {{qed|lemma}} === Equality of Mappings === The composite of $I_S$ and $f$ is defined as: :$f \circ I_S = \set {\tuple {x, z} \in S \times T: \exists y \in S: \tuple {x, y} \in I_S \land \tuple {y, z} \in f}$ But by definition of the identity mapping on $S$, we have that: :$\tuple {x, y} \in I_S \implies x = y$ Hence: :$f \circ I_S = \set {\tuple {y, z} \in S \times T: \exists y \in S: \tuple {y, y}\ \in I_S \land \tuple {y, z} \in f}$ But as $\forall y \in S: \tuple {y, y} \in I_S$, this means: :$f \circ I_S = \set {\tuple {y, z} \in S \times T: \tuple {y, z} \in f}$ That is: :$f \circ I_S = f$ {{qed|lemma}} Hence the result, by Equality of Mappings. {{qed}}
By definition, a mapping is also a relation. Also by definition, the identity mapping is the same as the diagonal relation. Thus Diagonal Relation is Right Identity can be applied directly. {{qed}}
We use the definition of relation equality, as follows: === Equality of Domains === The domains of $\RR$ and $\Delta_T \circ \RR$ are equal from Domain of Composite Relation: :$\Dom {\Delta_T \circ \RR} = \Dom \RR$ === Equality of Codomains === From Codomain of Composite Relation, the codomains of $\Delta_T \circ \RR$ and $\Delta_T$ are both equal to $T$. But from the definition of the diagonal relation, the codomain of $\Delta_T$ is $\Dom {\Delta_T} = T$. === Equality of Relations === The composite of $\RR$ and $\Delta_T$ is defined as: :$\Delta_T \circ \RR = \set {\tuple {x, z} \in S \times T: \exists y \in T: \tuple {x, y} \in \RR \land \tuple {y, z} \in \Delta_T}$ But by definition of the diagonal relation on $T$, we have that: :$\tuple {y, z} \in \Delta_T \implies y = z$ Hence: :$\Delta_T \circ \RR = \set {\tuple {x, y} \in S \times T: \exists y \in T: \tuple {x, y} \in \RR \land \tuple {y, y} \in \Delta_T}$ But as $\forall y \in T: \tuple {y, y} \in \Delta_T$, this means: :$\Delta_T \circ \RR = \set {\tuple {x, y} \in S \times T: \tuple {x, y} \in \RR}$ That is: :$\Delta_T \circ \RR = \RR$ Hence the result. {{qed}}
Let $\struct {S, \circ}$ be an algebraic structure in which $\circ$ is associative. Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism. As an isomorphism is surjective, it follows that: :$\forall u, v, w \in T: \exists x, y, z \in S: \map \phi x = u, \map \phi y = v, \map \phi z = w$ So: {{begin-eqn}} {{eqn | l = \paren {u * v} * w       | r = \paren {\map \phi x * \map \phi y} * \map \phi z       | c = as $\phi$ is a Surjection }} {{eqn | r = \map \phi {x \circ y} * \map \phi z       | c = {{Defof|Morphism Property}} }} {{eqn | r = \map \phi {\paren {x \circ y} \circ z}       | c = {{Defof|Morphism Property}} }} {{eqn | r = \map \phi {x \circ \paren {y \circ z} }       | c = Associativity of $\circ$ }} {{eqn | r = \map \phi x * \map \phi {y \circ z}       | c = {{Defof|Morphism Property}} }} {{eqn | r = \map \phi x * \paren {\map \phi y * \map \phi z}       | c = {{Defof|Morphism Property}} }} {{eqn | r = u * \paren {v * w}       | c = by definition as above }} {{end-eqn}} As $\phi$ is an isomorphism, it follows from Inverse of Algebraic Structure Isomorphism is Isomorphism that $\phi^{-1}$ is also an isomorphism. Thus the result for $\phi$ can be applied to $\phi^{-1}$. {{qed}}
We have that an isomorphism is a homomorphism which is also a bijection. By definition, an epimorphism is a homomorphism which is also a surjection. That is, an isomorphism is an epimorphism which is also an injection. Thus Epimorphism Preserves Associativity can be applied. {{qed}}
Let $\struct {S, \circ}$ be an algebraic structure in which $\circ$ is commutative. Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism. As an isomorphism is surjective, it follows that: :$\forall u, v \in T: \exists x, y \in S: \map \phi x = u, \map \phi y = v$ So: {{begin-eqn}} {{eqn | l = u * v       | r = \map \phi x * \map \phi y       | c = as $\phi$ is a surjection }} {{eqn | r = \map \phi {x \circ y}       | c = Morphism Property }} {{eqn | r = \map \phi {y \circ x}       | c = Commutativity of $\circ$ }} {{eqn | r = \map \phi y * \map \phi x       | c = Morphism Property }} {{eqn | r = v * u       | c = by definition as above }} {{end-eqn}} As $\phi$ is an isomorphism, it follows from Inverse of Algebraic Structure Isomorphism is Isomorphism that $\phi^{-1}$ is also a isomorphism. Thus the result for $\phi$ can be applied to $\phi^{-1}$. {{qed}}
We have that an isomorphism is a homomorphism which is also a bijection. By definition, an epimorphism is a homomorphism which is also a surjection. That is, an isomorphism is an epimorphism which is also an injection. Thus Epimorphism Preserves Commutativity can be applied. {{qed}}
Let $\left({S, \circ}\right)$ be an algebraic structure in which $\circ$ has an identity $e_S$. From Epimorphism Preserves Identity, it follows that $\left({T, *}\right)$ also has an identity, which is $\phi \left({e_S}\right)$. Let $y$ be an inverse of $x$ in $\left({S, \circ}\right)$.  Then: {{begin-eqn}} {{eqn | l = \phi \left({x}\right) * \phi \left({y}\right)       | r = \phi \left({x \circ y}\right)       | c =  }} {{eqn | r = \phi \left({e_S}\right)       | c =  }} {{eqn | r = \phi \left({y \circ x}\right)       | c =  }} {{eqn | r = \phi \left({y}\right) * \phi \left({x}\right)       | c =  }} {{end-eqn}} So $\phi \left({y}\right)$ is an inverse of $\phi \left({x}\right)$ in $\left({T, *}\right)$. As $\phi$ is an isomorphism, it follows from Inverse of Algebraic Structure Isomorphism is Isomorphism that $\phi^{-1}$ is also a isomorphism. Thus the result for $\phi \left({x}\right)$ can be applied to $\phi^{-1} \left({\phi \left({x}\right)}\right)$. {{qed}}
We have that an isomorphism is a homomorphism which is also a bijection. By definition, an epimorphism is a homomorphism which is also a surjection. That is, an isomorphism is an epimorphism which is also an injection. Thus Epimorphism Preserves Inverses can be applied. {{qed}}
An group isomorphism is by definition a group epimorphism. The result follows from Epimorphism Preserves Inverses. {{Qed}}
Let $g \in G$. {{begin-eqn}} {{eqn | l = \phi \left({g}\right) * \phi \left({g^{-1} }\right)       | r = \phi \left({g \circ g^{-1} }\right)       | c = {{Defof|Group Isomorphism}} }} {{eqn | r = \phi \left({e_G}\right)       | c = {{Defof|Inverse Element}} }} {{eqn | r = e_H       | c = Group Isomorphism Preserves Identity }} {{end-eqn}} It follows from Inverse in Group is Unique that $\phi \left({g^{-1} }\right)$ is the unique inverse element of $\phi \left({g}\right)$ in $\left({H, *}\right)$. That is: : $\forall g \in G: \phi \left({g^{-1}}\right) = \left({\phi \left({g}\right)}\right)^{-1}$ {{qed}}
Let $\struct {S, \circ}$ be an algebraic structure in which $\circ$ has an identity $e_S$. Then $\forall x \in S: x \circ e_S = x = e_S \circ x$. The result follows directly from the morphism property of $\circ$ under $\phi$: {{begin-eqn}} {{eqn | l = \map \phi {x \circ e_S}       | r = \map \phi x       | c = }} {{eqn | r = \map \phi {e_S \circ x}       | c = }} {{eqn | l = \map \phi x * \map \phi {e_S}       | r = \map \phi x       | c =  }} {{eqn | r = \map \phi {e_S} * \map \phi x       | c =  }} {{end-eqn}} As $\phi$ is an isomorphism, it follows from Inverse of Algebraic Structure Isomorphism is Isomorphism that $\phi^{-1}$ is also a isomorphism. Thus the result for $\map \phi {e_S}$ can be applied to $\map {\phi^{-1} } {\map \phi {e_S} }$. {{qed}}
We have that an isomorphism is a homomorphism which is also a bijection. By definition, an epimorphism is a homomorphism which is also a surjection. That is, an isomorphism is an epimorphism which is also an injection. Thus Epimorphism Preserves Identity can be applied. {{qed}}
This proof assumes the definition of $\exp$ as: :$\exp x = y \iff \ln y = x$ where: :$\ln y = \displaystyle \int_1^y \dfrac 1 t \rd t$ Let $X = \exp x$ and $Y = \exp y$. From Sum of Logarithms, we have: :$\ln X Y = \ln X + \ln Y = x + y$ From the Exponential of Natural Logarithm: :$\map \exp {\ln x} = x$ Thus: :$\map \exp {x + y} = \map \exp {\ln X Y} = X Y = \paren {\exp x} \paren {\exp y}$ {{qed}} Alternatively, this may be proved directly by investigating: :$\map D {\map \exp {x + y} / \exp x}$
This proof assumes the definition of $\exp$ as defined by a limit of a sequence: :$\exp x = \displaystyle \lim_{n \mathop \to +\infty} \paren {1 + \frac x n}^n$ From Powers of Group Elements we can presuppose the Exponent Combination Laws for natural number indices. First we introduce a lemma: By definition: {{begin-eqn}} {{eqn | l = \paren {\exp x} \paren {\exp y}       | r = \lim_{n \mathop \to +\infty} \paren {1 + \frac x n}^n \lim_{n \mathop \to +\infty} \paren {1 + \frac y n}^n }} {{eqn | r = \lim_{n \mathop \to +\infty} \paren {\paren {1 + \frac x n} \paren {1 + \frac y n} }^n       | c = Combination Theorem for Sequences }} {{eqn | r = \lim_{n \mathop \to +\infty} \paren {1 + \frac {x + y} n + \frac {x y} {n^2} }^n }} {{eqn | r = \lim_{n \mathop \to +\infty} \paren {\paren {1 + \frac {x + y} n} \paren {1 + \frac {\paren {\frac {x y} {n + x + y} } } n} }^n       | c = Lemma: {{WLOG}} let $n > - x - y$: therefore $n + x + y > 0$ }} {{eqn | r = \lim_{n \mathop \to +\infty} \paren {1 + \frac {x + y} n}^n \lim_{n \mathop \to +\infty} \paren {1 + \frac {\paren {\frac {x y} {n + x + y} } } n}^n       | c = Combination Theorem for Sequences }} {{eqn | r = \lim_{n \mathop \to +\infty} \paren {1 + \frac {x + y} n}^n       | c = Null Sequence in Exponential Sequence }} {{eqn | r = \map \exp {x + y} }} {{end-eqn}} {{qed}}
This proof assumes the definition of $\exp$ as defined by a limit of a sequence: :$\exp x = \displaystyle \lim_{n \mathop \to +\infty} \paren {1 + \frac x n}^n$ From Powers of Group Elements we can presuppose the Exponent Combination Laws for natural number indices. By definition: {{begin-eqn}} {{eqn | l = \paren {\exp x} \paren {\exp y}       | r = \lim_{n \mathop \to +\infty} \paren {1 + \frac x n}^n \lim_{n \mathop \to +\infty} \paren {1 + \frac y n}^n }} {{eqn | r = \lim_{n \mathop \to +\infty} \paren {\paren {1 + \frac x n} \paren {1 + \frac y n} }^n       | c = Combination Theorem for Limits of Functions }} {{eqn | r = \lim_{n \mathop \to +\infty} \paren {1 + \frac{x + y} n + \frac{x y} {n^2} }^n }} {{end-eqn}} Intuitively, the $\paren {1 + \dfrac {x + y} n}$ term is the most influential of the terms involved in the limit, and: :$\paren {1 + \dfrac {x + y} n + \dfrac {x y} {n^2} }^n \to \paren {1 + \dfrac {x + y} n}^n$ as $n \to +\infty$ To formalize this claim: :$\map \exp {x + y} = \exp x \cdot \exp y \iff \dfrac {\exp x \cdot \exp y} {\map \exp {x + y} } = 1$ {{begin-eqn}} {{eqn | l = \frac {\paren {1 + \frac {x + y} n + \frac {x y} {n^2} }^n} {\paren {1 + \frac{x + y} n}^n}       | r = \paren {1 + \frac{x y} {n^2 + n x + n y} }^n       | c = Lemma }} {{eqn | r = \sum_{k \mathop = 0}^n \binom n k \paren {\frac {x y} {n^2 + n x + n y} }^k       | c = Binomial Theorem }} {{eqn | r = 1 + \sum_{k \mathop = 1}^n \binom n k n^{-k} \paren {\frac {x y} {n + x + y} }^k }} {{end-eqn}} Now, as $n \to +\infty$, we use the Combination Theorem for Limits of Functions to investigate the behavior of this sequence, term by term. As $1$ trivially converges to $1$, consider now the other terms of the sequence. We invoke the Squeeze Theorem for Absolutely Convergent Series. Hence it will suffice to investigate the limit behaviour of: :$\displaystyle \sum_{k \mathop = 1}^n \, \size {\binom n k n^{-k} \paren {\frac {x y} {n + x + y} }^k}$ From $\dbinom n k$ is not greater than $n^k$: {{begin-eqn}} {{eqn | l = \binom n k       | o = \le       | r = n^k       | c = for all $n, k$ here considered }} {{eqn | ll= \leadsto       | l = \binom n k n^{-k}        | o = \le       | r = 1       | c = divide both sides by $n^k$ }} {{end-eqn}} Therefore, we may conclude, using Absolute Value is Bounded Below by Zero: :$\displaystyle 0 \le \sum_{k \mathop = 1}^n \size { {n \choose k} n^{-k} \paren {\frac {x y} {n + x + y} }^k} \le \sum_{k \mathop = 1}^n \size {\frac {x y} {n + x + y} }^k$ From Sum of Infinite Geometric Sequence, the right hand term converges to: {{begin-eqn}} {{eqn | l = \frac {\dfrac {x y} {n + x + y} } {1 - \dfrac {x y} {n + x + y} }       | r = \frac {x y} {n + x + y - x y} }} {{eqn | o = \to       | r = 0       | c = as $n \to +\infty$ }} {{end-eqn}} :$0 \to 0$ as $n \to +\infty$, trivially. This means: :$\dfrac {\paren {1 + \dfrac {x + y} n + \dfrac {x y} {n^2} }^n} {\paren {1 + \dfrac {x + y} n}^n} \to 1$ as $n \to +\infty$ which is equivalent to our hypothesis: :$\paren {1 + \dfrac {x + y} n + \dfrac {x y} {n^2} }^n \to \paren {1 + \dfrac {x + y} n}^n$ as $n \to +\infty$ {{qed}}
=== Closure === From Natural Numbers have No Proper Zero Divisors: :$\forall m, n \in \N: m \times n = 0 \iff m = 0 \lor n = 0$ It follows that: :$\forall m, n \in \N_{>0}: m \times n \ne 0$ and so: :$\forall m, n \in \N_{>0}: m \times n \in \N_{>0}$ That is, $\struct {\N_{>0}, \times}$ is closed. {{qed|lemma}} === Associativity === Natural Number Multiplication is Associative. {{qed|lemma}} === Commutativity === 	 Natural Number Multiplication is Commutative. {{qed|lemma}} === Infinite === We have that the Natural Numbers are Infinite. Then we have that Infinite if Injection from Natural Numbers. The mapping $s: \N \to \N_{>0}: \map s n = n + 1$ is such an injection. Hence $\N_{>0}$ is infinite. {{qed}}
From Isomorphism Preserves Semigroups, if $\left({S, \circ}\right)$ is a semigroup then so is $\left({T, *}\right)$. From Isomorphism Preserves Identity, if $\left({S, \circ}\right)$ has an identity $e_S$, then $\phi \left({e_S}\right)$ is the identity for $*$. From Isomorphism Preserves Inverses, if $x^{-1}$ is an inverse of $x$ for $\circ$, then $\phi \left({x^{-1}}\right)$ is an inverse of $\phi \left({x}\right)$ for $*$. The result follows from the definition of group. {{qed}}
An isomorphism is an epimorphism. The result follows as a direct corollary of Epimorphism Preserves Groups. {{qed}}
If $\left({S, \circ}\right)$ is a semigroup, then by definition it is closed. From Morphism Property Preserves Closure, $\left({T, *}\right)$ is therefore also closed. If $\left({S, \circ}\right)$ is a semigroup, then by definition $\circ$ is associative. From Isomorphism Preserves Associativity, $*$ is therefore also associative. So $\left({T, *}\right)$ is closed, and $*$ is associative, and therefore by definition, $\left({T, *}\right)$ is a semigroup. {{qed}}
An isomorphism is an epimorphism. The result follows as a direct corollary of Epimorphism Preserves Semigroups. {{qed}}
Suppose $a \in S$ is left cancellable. Then: :$a \circ x = a \circ y \implies x = y$ From the definition of the left regular representation: :$\map {\lambda_a} x = a \circ x$ Thus: :$\map {\lambda_a} x = \map {\lambda_a} y \implies x = y$ and so the left regular representation is injective. {{qed|lemma}} Suppose $\map {\lambda_a} x$ is injective. Then: :$\map {\lambda_a} x = \map {\lambda_a} y \implies x = y$ From the definition of the left regular representation: :$\map {\lambda_a} x = a \circ x$ Thus: : $a \circ x = a \circ y \implies x = y$ and so $a$ is left cancellable. {{Qed}}
Suppose $a \in S$ is right cancellable. Then: :$x \circ a = y \circ a \implies x = y$ From the definition of the right regular representation: :$\map {\rho_a} x = x \circ a$ Thus: :$\map {\rho_a} x = \map {\rho_a} y \implies x = y$ and so the right regular representation is injective. {{qed|lemma}} Suppose $\map {\rho_a} x$ is injective. Then: :$\map {\rho_a} x = \map {\rho_a} y \implies x = y$ From the definition of the right regular representation: :$\map {\rho_a} x = x \circ a$ Thus: : $x \circ a = y \circ a \implies x = y$ and so $a$ is right cancellable. {{Qed}}
Let $\left({S, \circ}\right)$ be an algebraic structure in which $a$ is cancellable. From Isomorphism Preserves Left Cancellability and Isomorphism Preserves Right Cancellability: : $a \in S$ is left cancellable in $\left({S, \circ}\right)$ iff $\phi \left({a}\right) \in T$ is left cancellable in $\left({T, *}\right)$ and : $a \in S$ is right cancellable in $\left({S, \circ}\right)$ iff $\phi \left({a}\right) \in T$ is right cancellable in $\left({T, *}\right)$. It follows that: : $a \in S$ is cancellable in $\left({S, \circ}\right)$ iff : $\phi \left({a}\right) \in T$ is cancellable in $\left({T, *}\right)$. {{qed}}
Let $a$ be left cancellable in $\struct {S, \circ}$. Let $x \in S$ and $y \in S$ be arbitrary. Then: {{begin-eqn}} {{eqn | l = \map \phi a * \map \phi x       | r = \map \phi a * \map \phi y       | c =  }} {{eqn | ll= \leadsto       | l = \map \phi {a \circ x}       | r = \map \phi {a \circ y}       | c = Morphism Property }} {{eqn | ll= \leadsto       | l = \map \phi x       | r = \map \phi y       | c = as $a$ is left cancellable }} {{end-eqn}} That is, $\map \phi a$ is left cancellable in $\struct {T, *}$. As $\phi$ is an isomorphism, then so is $\phi^{-1}$. So the same proof works in reverse in exactly the same way. {{qed}}
Let $a$ be right cancellable in $\struct {S, \circ}$. Let $x \in S$ and $y \in S$ be arbitrary. Then: {{begin-eqn}} {{eqn | l = \map \phi x * \map \phi a       | r = \map \phi y * \map \phi a       | c =  }} {{eqn | ll= \implies       | l = \map \phi {x \circ a}       | r = \map \phi {y \circ a}       | c = Morphism property }} {{eqn | ll= \implies       | l = \map \phi x       | r = \map \phi y       | c = as $a$ is right cancellable }} {{end-eqn}} That is, $\map \phi a$ is right cancellable in $T$. As $\phi$ is an isomorphism, then so is $\phi^{-1}$. So the same proof works in reverse in exactly the same way. {{qed}}
First we can make the substitution $z = x - \xi$ and convert the expression into: :$\displaystyle \frac {\mathrm d^n} {\mathrm d x^n} \sum_{m \mathop \ge 0} a_m z^m$ We then use Nth Derivative of Mth Power: :$\displaystyle \frac {\mathrm d^n} {\mathrm d z^n} z^m = \begin{cases} m^{\underline n} z^{m - n} & : n \le m \\ 0 & : n > m \end{cases}$ By hypothesis $x$ is within the interval of convergence. It follows from Abel's Theorem that: :$\displaystyle \frac {\mathrm d^n}{\mathrm dz^n} \sum_{m \mathop \ge 0} a_m z^m = \sum_{m \mathop \ge n} a_m m^{\underline n} z^{m - n}$ Then from Derivative of Identity Function and others, we have: : $\displaystyle \frac {\mathrm d} {\mathrm d x} \left({x - \xi}\right) = 1$ The result follows from the Chain Rule for Derivatives. {{qed}} {{Handwaving|An explanation needed for how Abel's Theorem allows term-by-term differentiation of the infinite power series.}} Category:Differential Calculus Category:Power Series 8xgvqg2hwbch8uw5h77lw0hmg688v5d
From Equivalence of Definitions of Self-Inverse, $x \circ x = e \implies x = x^{-1}$. From Inverse in Monoid is Unique, it follows that every element of $\left({S, \circ}\right)$ has a unique inverse. So by definition, $\left({S, \circ}\right)$ is a group. From All Elements Self-Inverse then Abelian, it follows that $\left({S, \circ}\right)$ is an abelian group. {{qed}}
Let $x, y \in S$ be arbitrary. Then: {{begin-eqn}} {{eqn | l = a \circ x       | r = a \circ y       | c =  }} {{eqn | ll= \implies       | l = b \circ \paren {a \circ x}       | r = b \circ \paren {a \circ y}       | c = }} {{eqn | ll= \implies       | l = \paren {b \circ a} \circ x       | r = \paren {b \circ a} \circ y       | c = as $\struct {S, \circ}$ is a semigroup, $\circ$ is associative }} {{eqn | ll= \implies       | l = e_L \circ x       | r = e_L \circ y       | c = by hypothesis }} {{eqn | ll= \implies       | l = x       | r = y       | c = Definition of Left Identity }} {{end-eqn}} The result follows by definition of left cancellable. {{qed}}
Let $x, y \in S$ be arbitrary. Then: {{begin-eqn}} {{eqn | l = x \circ a       | r = y \circ a       | c =  }} {{eqn | ll= \implies       | l = \left({x \circ a}\right) \circ b       | r = \left({y \circ a}\right) \circ b       | c =  }} {{eqn | ll= \implies       | l = \left({x \circ a}\right) \circ b       | r = \left({y \circ a}\right) \circ b       | c = as $\left({S, \circ}\right)$ is a semigroup, $\circ$ is associative }} {{eqn | ll= \implies       | l = x \circ e_R       | r = y \circ e_R       | c = By hypothesis }} {{eqn | ll= \implies       | l = x       | r = y       | c = Definition of Right Identity }} {{end-eqn}} The result follows by definition of right cancellable. {{qed}}
Let $x, y \in S$. From Integer Multiples Closed under Addition, $x + y \in n \Z$. As $x, y > p$ we have that: :$\exists r \in \Z_{>0}: x = p + r$ :$\exists s \in \Z_{>0}: y = p + s$ Thus it follows that: {{begin-eqn}} {{eqn | l = x + y       | r = \paren {p + r} + \paren {p + s}       | c =  }} {{eqn | r = 2 p + r + s       | c =  }} {{eqn | o = >       | r = p       | c = as $r, s > 0$ }} {{end-eqn}} So $x + y > p$ and $x + y \in n \Z$. Hence by definition $x + y \in S$, and so $S$ is closed under addition. {{qed}}
Let $x, y \in S$. From Integer Multiples Closed under Multiplication, $x y \in n \Z$. As $x, y > p$ we have that: :$\exists r \in \Z_{>0}: x = p + r$ :$\exists s \in \Z_{>0}: y = p + s$ Thus it follows that: {{begin-eqn}} {{eqn | l = x + y       | r = \paren {p + r} \paren {p + s}       | c =  }} {{eqn | r = p^2 + p \paren {r + s} + r s       | c =  }} {{eqn | o = >       | r = p       | c = as $r, s > 0$ }} {{end-eqn}} So $x y > p$ and $x y \in n \Z$. Hence by definition $x y \in S$, and so $S$ is closed under multiplication. {{qed}}
Recall that Rational Numbers form Field under the operations of addition and multiplication. By definition of a field, the algebraic structure $\struct {\Q_{\ne 0}, \times}$ is a group. Thus, by definition, $\times$ is closed in $\struct {\Q_{\ne 0}, \times}$. {{qed}}
Recall that Real Numbers form Field under the operations of addition and multiplication. By definition of a field, the algebraic structure $\left({\R_{\ne 0}, \times}\right)$ is a group. Thus, by definition, $\times$ is closed in $\left({\R_{\ne 0}, \times}\right)$. {{qed}}
Let $x \times y = 0$. {{WLOG}}, suppose that $x \ne 0$. Then: {{begin-eqn}} {{eqn | l = x \times y       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \frac 1 x \times \paren {x \times y}       | r = \frac 1 x \times 0       | c = as $x \ne 0$ }} {{eqn | ll= \leadsto       | l = \paren {\frac 1 x \times x} \times y       | r = \frac 1 x \times 0       | c = Real Number Axioms: $\R M 1$: Associativity }} {{eqn | ll= \leadsto       | l = 1 \times y       | r = \frac 1 x \times 0       | c = Real Number Axioms: $\R M 4$: Inverse }} {{eqn | ll= \leadsto       | l = y       | r = \frac 1 x \times 0       | c = Real Number Axioms: $\R M 3$: Identity }} {{eqn | ll= \leadsto       | l = y       | r = 0       | c = Real Zero is Zero Element }} {{end-eqn}} Thus: :$x \times y = 0, x \ne 0 \implies y = 0$ Mutatis mutandis :$x \times y = 0, y \ne 0 \implies x = 0$ and so: :$x \times y = 0 \implies y = 0 \lor x = 0$ So: {{begin-eqn}} {{eqn | o =        | r = x \times y = 0       | c =  }} {{eqn | o = \leadsto       | r = \paren {x = 0} \lor \paren {y = 0}       | c =  }} {{eqn | o = \leadsto       | r = \neg \paren {x \ne 0 \land y \ne 0}       | c = De Morgan's Laws: Disjunction }} {{end-eqn}} The result follows by the Rule of Transposition. {{qed}}
Recall that Complex Numbers form Field under the operations of addition and multiplication. By definition of a field, the algebraic structure $\left({\C_{\ne 0}, \times}\right)$ is a group. Thus, by definition, $\times$ is closed in $\left({\C_{\ne 0}, \times}\right)$. {{qed}}