hcju/proofwikiqa huggingface dataset:
As a mapping is by definition a left-total relation. Therefore Preimage of Image under Left-Total Relation is Superset applies: :$A \subseteq S \implies A \subseteq \paren {\mathcal R^{-1} \circ \mathcal R} \sqbrk A$ where $\mathcal R$ is a relation. Hence: :$A \subseteq S \implies A \subseteq \paren {f^{-1} \circ f} \sqbrk A$ {{qed}}
As $f$ is a mapping it follows by definition that $f$ is also a relation So we apply Image of Preimage under Relation is Subset directly: :$B \subseteq T \implies \paren {f \circ f^{-1} } \sqbrk B \subseteq B \implies f \sqbrk {f^{-1} \sqbrk B} \subseteq B$ But from Image of Subset under Relation is Subset of Image: Corollary 3 we also have: :$B \subseteq T \implies f^{-1} \sqbrk B \subseteq f^{-1} \sqbrk T$ But from Preimage of Mapping equals Domain $f$ we have that $f^{-1} \sqbrk T = S$ and so: :$B \subseteq T \implies f^{-1} \sqbrk B \subseteq S$ Applying Image of Subset under Relation is Subset of Image: Corollary 2 we have: :$f^{-1} \sqbrk B \subseteq S \implies f \sqbrk {f^{-1} \sqbrk B} \subseteq f \sqbrk S$ From Intersection is Largest Subset it follows that: :$B \subseteq T \implies \paren {f \circ f^{-1} } \sqbrk B \subseteq B \cap f \sqbrk S$ Now suppose $y \in B \cap f \sqbrk S$. Then: :$f^{-1} \sqbrk y \subseteq f^{-1} \sqbrk B$ and $f^{-1} \sqbrk y \subseteq f^{-1} \sqbrk {f \sqbrk S}$ and in particular: :$f^{-1} \sqbrk y \subseteq f^{-1} \sqbrk B$ Applying Image of Subset under Relation is Subset of Image: Corollary 2 again, we have: :$f \sqbrk {f^{-1} \sqbrk y} \subseteq f \sqbrk {f^{-1} \sqbrk B}$ But as $f$ is many-to-one, we have that: :$f \sqbrk {f^{-1} \sqbrk y} = y$ and so: :$y \in f \sqbrk {f^{-1} \sqbrk B} = \paren {f \circ f^{-1} } \sqbrk B$ So we have that: :$B \cap f \sqbrk S \subseteq \paren {f \circ f^{-1} } \sqbrk B$ Hence the result. {{qed}}
From Homomorphism of External Direct Products, we have that $\phi_1 \times \phi_2: \struct {S_1 \times S_2, \circ} \to \struct {T_1 \times T_2, *}$ is a homomorphism. From Cartesian Product of Bijections is Bijection, we have that $\phi_1 \times \phi_2$ is a bijection. Thus $\phi_1 \times \phi_2$ is a bijective homomorphism, and so an isomorphism. {{Qed}}
=== Definition 1 implies Definition 2 === Let $H_1, H_2, \ldots, H_n$ be independent by definition 1: :$\displaystyle \prod_{k \mathop = 1}^n h_k = e \iff \forall k \in \left[{1 \,.\,.\, n}\right]: h_k = e$ where $h_k \in H_k$ for all $k \in \left[{1 \,.\,.\, n}\right]$. Let $\displaystyle u \in \left({\prod_{j \mathop = 1}^{k-1} H_j}\right) \cap H_k$. Then: : $\displaystyle \exists x_1, x_2, \ldots, x_{k-1} \in H_1, \ldots, H_{k-1}: u = \prod_{j \mathop = 1}^{k-1} x_j$ We have that $u \in H_k$ and $H_k$ is a group. Therefore let $x_k = u^{-1}$, the inverse of $u$. For each $j \in \left\{{k+1, k+2, \ldots, n}\right\}$ let $x_j = e$. Then: {{begin-eqn}} {{eqn | l = \prod_{j \mathop = 1}^n x_j       | r = \left({\prod_{j \mathop = 1}^{k-1} x_j}\right) u^{-1} \left({\prod_{j \mathop = k+1}^{n} x_j}\right)       | c =  }} {{eqn | r = \left({\prod_{j \mathop = 1}^{k-1} x_j}\right) u^{-1}       | c =  }} {{eqn | r = e       | c =  }} {{end-eqn}} Thus by hypothesis $u^{-1} = e$ and hence $u = e$. Hence $H_1, H_2, \ldots, H_n$ are independent by definition 2: :$\displaystyle \forall k \in \left\{{2, 3, \ldots, n}\right\}: \left({\prod_{j \mathop = 1}^{k-1} H_j}\right) \cap H_k = \left\{{e}\right\}$ === Definition 2 implies Definition 1 === Let $H_1, H_2, \ldots, H_n$ be independent by definition 2: :$\displaystyle \forall k \in \left\{{2, 3, \ldots, n}\right\}: \left({\prod_{j \mathop = 1}^{k-1} H_j}\right) \cap H_k = \left\{{e}\right\}$ Aiming for a contradiction, suppose that: : $\displaystyle x_1, x_2, \ldots, x_n \in H_1, H_2, \ldots, H_n: \prod_{i \mathop = 1}^n x_i = e$ but that $x_j \ne e$ for some $j \in \left\{{1, 2, \ldots, n}\right\}$. There is bound to be more than one of them, because otherwise that would mean: : $\displaystyle \prod_{i \mathop = 1}^n x_i = x_j e = e \implies x_j = e$ So let $m$ be the largest of the $j$ such that $x_j \ne e$. Then $m > 1$, and: {{begin-eqn}} {{eqn | l = e       | r = x_1 \cdots x_n       | c =  }} {{eqn | r = x_1 \cdots x_m       | c =  }} {{eqn | r = \left({x_1 \cdots x_{m-1} }\right) x_m       | c =  }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = x_m^{-1}       | r = x_1 \cdots x_{m-1}       | c =  }} {{eqn | r = H_1 \cdots H_{m-1}       | c =  }} {{end-eqn}} So: : $x_m^{-1} \in H_m$ Therefore by hypothesis: : $x_m^{-1} = e$ and therefore: : $x_m = e$ But $m$ was defined to be the largest of the $j$ such that $x_j \ne e$. Thus by Proof by Contradiction it follows that there can be no such $j$ so that: : $\displaystyle x_1, x_2, \ldots, x_n \in H_1, H_2, \ldots, H_n: \prod_{i \mathop = 1}^n x_i = e$ but that $x_j \ne e$ for some $j \in \left\{{1, 2, \ldots, n}\right\}$. Hence $H_1, H_2, \ldots, H_n$ are independent by definition 1: :$\displaystyle \prod_{k \mathop = 1}^n h_k = e \iff \forall k \in \left[{1 \,.\,.\, n}\right]: h_k = e$ where $h_k \in H_k$ for all $k \in \left[{1 \,.\,.\, n}\right]$. {{qed}}
Let $\left({S, \preccurlyeq}\right)$ be a totally ordered set. From Dual Ordering is Ordering we have that $\succcurlyeq$ is an ordering. Let $x, y \in S$. Then $x \preccurlyeq y$ or $y \preccurlyeq x$ since $\preccurlyeq$ is total. By definition of dual ordering, also $y \succcurlyeq x$ or $x \succcurlyeq y$. Hence the result, by definition of total ordering. {{qed}}
Let $\left({S, \preccurlyeq}\right)$ be a lattice. It is to be shown that: :for all $x, y \in S$, the ordered set $\struct {\set {x, y}, \succcurlyeq}$ admits both a supremum and an infimum.  Let $x, y \in S$. Then $\struct {\set {x, y}, \preccurlyeq}$ admits both a supremum and an infimum. Let $c = \map \sup {\set {x, y}, \preccurlyeq}$. Then by definition of supremum: :$\forall s \in \set {x, y}: s \preccurlyeq c$ :$\forall d \in S: c \preccurlyeq d$ where $d$ is an upper bound of $\struct {\set {x, y}, \preccurlyeq} \subseteq S$. Hence by definition of dual ordering: :$\forall s \in \set {x, y}: c \succcurlyeq s$ :$\forall d \in S: d \succcurlyeq c$ where $d$ is an upper bound of $\struct {\set {x, y}, \preccurlyeq} \subseteq S$. By Upper Bound is Lower Bound for Inverse Ordering, $d$ is a lower bound of $\struct {\set {x, y}, \succcurlyeq} \subseteq S$. So by definition of infimum: :$c = \map \inf {\set {x, y}, \succcurlyeq}$ That is, $\struct {\set {x, y}, \succcurlyeq}$ admits an infimum. {{qed|lemma}} Let $c = \map \inf {\set {x, y}, \preccurlyeq}$. Then by definition of infimum: :$\forall s \in \set {x, y}: c \preccurlyeq s$ :$\forall d \in S: d \preccurlyeq c$ where $d$ is a lower bound of $\struct {\set {x, y}, \preccurlyeq} \subseteq S$. Hence by definition of dual ordering: :$\forall s \in \set{x, y}: s \succcurlyeq c$ :$\forall d \in S: c \succcurlyeq d$ where $d$ is a lower bound of $\struct {\set {x, y}, \preccurlyeq} \subseteq S$. By Lower Bound is Upper Bound for Inverse Ordering, $d$ is an upper bound of $\struct {\set {x, y}, \succcurlyeq} \subseteq S$. So by definition of supremum: :$c = \map \sup {\set {x, y}, \succcurlyeq}$ That is, $\struct {\set {x, y}, \succcurlyeq}$ admits a supremum. {{qed|lemma}} Hence $\struct {\set {x, y}, \succcurlyeq}$ admits both a supremum and an infimum. That is, $\succcurlyeq$ is a lattice ordering. {{qed}}
=== Necessary Condition === Let $\left({S, \preccurlyeq_1}\right)$ be a totally ordered set Then by definition $\preccurlyeq_1$ is a total ordering. Let $x, y \in S$. Then either $x \preccurlyeq_1 y$ or $y \preccurlyeq_1 x$. From the definition of order isomorphism, either: :$\phi \left({x}\right) \preccurlyeq_2 \phi \left({y}\right)$ or: :$\phi \left({y}\right) \preccurlyeq_2 \phi \left({x}\right)$ and so by definition $\preccurlyeq_2$ is also a total ordering. So by definition $\left({T, \preccurlyeq_2}\right)$ is also a totally ordered set. {{qed|lemma}} === Sufficient Condition === By Inverse of Order Isomorphism is Order Isomorphism, if $\phi$ is an order isomorphism then so is $\phi^{-1}$. Applying the other implication (proven above) to $\phi^{-1}$, it follows that if $\left({T, \preccurlyeq_2}\right)$ is a totally ordered set then so is $\left({S, \preccurlyeq_1}\right)$. {{qed}}
Let $\struct {S_1, \preccurlyeq_1}$ be a well-ordered set. Then by definition $\preccurlyeq_1$ is a well-ordering. By Well-Ordering is Total Ordering, $\preccurlyeq_1$ is a total ordering. From Order Isomorphism on Totally Ordered Set preserves Total Ordering, it follows that $\struct {S_2, \preccurlyeq_2}$ is a totally ordered set. All that remains is to show that $\preccurlyeq$ is well-founded. Let $T_1 \subseteq S_1$. As $\struct {S_1, \preccurlyeq_1}$ is a well-ordered set: :$\exists m_1 \in T_1: \forall t \in T_1: t = m_1 \lor \neg \paren {t \preccurlyeq_1 m_1}$ Now consider the image $m_2$ of $m_1$ in $S_2$: :$m_2 = \map \phi {m_1}$ Suppose $\struct {S_2, \preccurlyeq_2}$ is not a well-ordered set. Let there exist $T_2 \subseteq S_2$ such that: :$\exists x \in T_2: x \ne m_2, x \preccurlyeq_2 m_2$ By Inverse of Order Isomorphism is Order Isomorphism, if $\phi$ is an order isomorphism then so is $\phi^{-1}$. Hence: :$\map {\phi^{-1} } x \ne \map {\phi^{-1} } {m_2}, \map {\phi^{-1} } x \preccurlyeq_2 \map {\phi^{-1} } {m_2}$ But this contradicts the fact that $\struct {S_1, \preccurlyeq_1}$ is well-ordered. Hence $\struct {S_2, \preccurlyeq_2}$ must be a well-ordered set. The same technique is used to show that if $\struct {S_2, \preccurlyeq_2}$ is a well-ordered set then so is $\struct {S_1, \preccurlyeq_1}$. {{qed}}
Let $\left({S, \preccurlyeq_1}\right)$ be a lattice Then by definition $\preccurlyeq_1$ is a lattice ordering. We need to show that for all $x, y \in S$, the ordered set $\left({\left\{{\phi \left({x}\right), \phi \left({y}\right)}\right\}, \preccurlyeq_2}\right)$ admits both a supremum and an infimum.  Let $x, y \in S$. Then $\left({\left\{{x, y}\right\}, \preccurlyeq_1}\right)$ admits both a supremum and an infimum. Let $c = \sup \left({\left\{{x, y}\right\}, \preccurlyeq_1}\right)$. Then by definition of supremum: : $\forall s \in \left\{{x, y}\right\}: s \preccurlyeq_1 c$ : $\forall d \in S: c \preccurlyeq_1 d$ ::where $d$ is an upper bound of $\left({\left\{{x, y}\right\}, \preccurlyeq_1}\right) \subseteq S$. Now consider the image of $\left\{{x, y}\right\}$ under $\phi$. By definition of order isomorphism: : $\forall \phi \left({s}\right) \in \left\{{\phi \left({x}\right), \phi \left({y}\right)}\right\}: \phi \left({c}\right) \preccurlyeq_2 \phi \left({s}\right)$ : $\forall \phi \left({d}\right) \in S_2: \phi \left({d}\right) \succcurlyeq \phi \left({c}\right)$ ::where $\phi \left({d}\right)$ is an upper bound of $\left({\left\{{\phi \left({x}\right), \phi \left({y}\right)}\right\}, \preccurlyeq_2}\right) \subseteq T$. So by definition of supremum: :$\phi \left({c}\right) = \sup \left({\left\{{\phi \left({x}\right), \phi \left({y}\right)}\right\}, \preccurlyeq_2}\right)$ That is, $\left({\left\{{\phi \left({x}\right), \phi \left({y}\right)}\right\}, \preccurlyeq_2}\right)$ admits a supremum. Using a similar technique it can be shown that: :If $c = \inf \left({\left\{{x, y}\right\}, \preccurlyeq}\right)$, then: :$\phi \left({c}\right) = \inf \left({\left\{{\phi \left({x}\right), \phi \left({y}\right)}\right\}, \preccurlyeq_2}\right)$ Hence $\left({\left\{{\phi \left({x}\right), \phi \left({y}\right)}\right\}, \preccurlyeq_2}\right)$ admits both a supremum and an infimum. That is, $\preccurlyeq_2$ is a lattice ordering and so $\left({T, \preccurlyeq_2}\right)$ is a lattice. By Inverse of Order Isomorphism is Order Isomorphism, if $\phi$ is an order isomorphism then so is $\phi^{-1}$. So the same technique is used to show that if $\left({T, \preccurlyeq_2}\right)$ is a lattice then so is $\left({S, \preccurlyeq_1}\right)$. {{qed}}
Let $\O \subset \mathbb H \subseteq \mathbb G$. By Intersection of Subgroups: General Result, $\bigcap \mathbb H$ is the largest subgroup of $G$ contained in each of the elements of $\mathbb H$. Thus, not only is $\bigcap \mathbb H$ a lower bound of $\mathbb H$, but also the largest, and therefore an infimum. The supremum of $\mathbb H$ is the smallest subgroup of $G$ containing $\bigcup \mathbb H$. Therefore $\struct {\mathbb G, \subseteq}$ is a complete lattice. {{qed}}
Because $\left({S, \circ}\right)$ is a monoid, it is a fortiori also a semigroup. From Index Laws for Semigroup: Sum of Indices: :$\forall m, n \in \N_{>0}: \circ^{n + m} a = \left({\circ^n a}\right) \circ \left({\circ^m a}\right)$ That is: :$\forall m, n \in \N_{>0}: a^{n + m} = a^n \circ a^m$ It remains to be shown that the result holds for the cases where $m = 0$ and $n = 0$. Let $n \in \N$: {{begin-eqn}} {{eqn | l = a^{n + 0}       | r = a^n       | c = Definition of Zero }} {{eqn | r = a^n \circ e       | c = Definition of Identity Element }} {{eqn | r = a^n \circ a^0       | c = Definition of $a^0$ }} {{end-eqn}} Similarly, let $m \in \N$: {{begin-eqn}} {{eqn | l = a^{0 + m}       | r = a^m       | c = Definition of Zero }} {{eqn | r = e \circ a_m       | c = Definition of Identity Element }} {{eqn | r = a^0 \circ a^m       | c = Definition of $a^0$ }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = a^{0 + 0}       | r = a^0       | c = Definition of Zero }} {{eqn | r = e       | c = Definition of $a^0$ }} {{eqn | r = e \circ e       | c = Definition of Identity Element }} {{eqn | r = a^0 \circ a^0       | c = Definition of $a^0$ }} {{end-eqn}} Thus: :$a^{n + m} = a^n \circ a^m$ holds for $n = 0$ and $m = 0$. Thus: :$\forall m, n \in \N: a^{n + m} = a^n \circ a^m$ {{qed}}
The proof proceeds by the Principle of Mathematical Induction: Let $\map P n$ be the proposition: :$\map {\circ^n} {a \circ b} = \paren {\circ^n a} \circ \paren {\circ^n b}$ === Basis of the Induction === {{begin-eqn}} {{eqn | l = \circ^1 \paren {a \circ b}       | r = a \circ b       | c = Definition of $\circ^1$ }} {{eqn | r = \paren {\circ^1 a} \circ \paren {\circ^1 b}       | c = Definition of $\circ^1$ }} {{end-eqn}} So $\map P 1$ holds. This is the basis for the induction. === Induction Hypothesis === Suppose that $\map P k$ holds: :$\map {\circ^k} {a \circ b} = \paren {\circ^k a} \circ \paren {\circ^k b}$ This is the induction hypothesis. It remains to be shown that: :$\map P k \implies \map P {k + 1}$ That is, that: :$\map {\circ^{k + 1} } {a \circ b} = \paren {\circ^{k + 1} a} \circ \paren {\circ^{k + 1} b}$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \map {\circ^{k + 1} } {a \circ b}       | r = \paren {\map {\circ^k} {a \circ b} } \circ \paren {a \circ b}       | c = Definition of $\circ^{k + 1}$ }} {{eqn | r = \paren {\paren {\circ^k a} \circ \paren {\circ^k b} } \circ \paren {a \circ b}       | c = Induction Hypothesis }} {{eqn | r = \paren {\circ^k a} \circ \paren {\paren {\paren {\circ^k b} \circ a} \circ b}       | c = $\circ$ is associative }} {{eqn | r = \paren {\circ^k a} \circ \paren {\paren {a \circ \paren {\circ^k b} } \circ b}       | c = Powers of Commuting Elements of Semigroup Commute }} {{eqn | r = \paren {\paren {\circ^k a} \circ a} \circ \paren {\paren {\circ^k b} \circ b}       | c = $\circ$ is associative }} {{eqn | r = \paren {\circ^{k + 1} a} \circ \paren {\circ^{k + 1} b}       | c = Definition of $\circ^{k + 1}$ }} {{end-eqn}} So $\map P {k + 1}$ holds. Thus by the Principle of Mathematical Induction, the result holds for all $n \in \N_{>0}$: :$\forall n \in \N_{>0}: \map {\circ^n} {a \circ b} = \paren {\circ^n a} \circ \paren {\circ^n b}$ {{qed}}
Consider the following array consisting of $n$ ones and $n-1$ blanks: :$\begin{bmatrix} 1 \ \_ \ 1 \ \_ \ \cdots \ \_ \ 1 \ \_ \ 1 \end{bmatrix}$ In each blank we can either put a comma or a plus sign. Each way of choosing $,$ or $+$ will give a composition of $n$ with the commas separating the individual $c_i$'s. It follows easily that there are $2^{n-1}$ ways of doing this, since there are two choices for each of $n-1$ blanks. The result follows from the Product Rule for Counting. Similarly if we want specifically $k$ different $c_i$'s then we are left with choosing $k-1$ out of $n-1$ blanks to place the $k-1$ commas. The number of ways of doing so is $\displaystyle \binom{n-1}{k-1}$ from the Binomial Theorem. {{qed}} Category:Combinatorics 4nm3jvlnlo0x3lru49v1qpfbieyzqxp
Proof by induction: For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition: :$\displaystyle \sum_{i \mathop = 1}^n i^5 + \sum_{i \mathop = 1}^n i^7 = 2 \paren {\sum_{i \mathop = 1}^n i}^4$ === Basis for the Induction === {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 1}^1 i^5 + \sum_{i \mathop = 1}^1 i^7       | r = 1^5 + 1^7       | c =  }} {{eqn | r = 2 \times 1^4       | c =  }} {{eqn | r = 2 \paren {\sum_{i \mathop = 1}^1 i}^4       | c =  }} {{end-eqn}} So $\map P 1$ has been demonstrated to hold. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\displaystyle \sum_{i \mathop = 1}^k i^5 + \sum_{i \mathop = 1}^k i^7 = 2 \paren {\sum_{i \mathop = 1}^k i}^4$ Then we need to show: :$\displaystyle \sum_{i \mathop = 1}^{k + 1} i^5 + \sum_{i \mathop = 1}^{k + 1} i^7 = 2 \paren {\sum_{i \mathop = 1}^{k + 1} i}^4$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = 2 \paren {\sum_{i \mathop = 1}^{k + 1} i}^4       | r = 2 \paren {\sum_{i \mathop = 1}^{k + 1} i^3}^2       | c = Sum of Sequence of Cubes }} {{eqn | r = 2 \paren {\paren {k + 1}^3 + \sum_{i \mathop = 1}^k i^3}^2       | c = {{Defof|Summation}} }} {{eqn | r = 2 \paren {k + 1}^6 + 4 \paren {\paren {k + 1}^3 \sum_{i \mathop = 1}^k i^3} + 2 \paren {\sum_{i \mathop = 1}^k i^3}^2       | c = Square of Sum }} {{eqn | r = 2 \paren {k + 1}^6 + 4 \paren {\paren {k + 1}^3 \frac {k^2 \paren {k + 1}^2} 4} + 2 \paren {\sum_{i \mathop = 1}^k i}^4       | c = Sum of Sequence of Cubes }} {{eqn | r = 2 \paren {k + 1}^6 + \paren {k^2 \paren {k + 1}^5} + \sum_{i \mathop = 1}^k i^7 + \sum_{i \mathop = 1}^k i^5       | c = Induction Hypothesis }} {{eqn | r = \paren {k + 1}^5 \paren {2 \paren {k + 1} + k^2} + \sum_{i \mathop = 1}^k i^7 + \sum_{i \mathop = 1}^k i^5       | c = simplifying first $2$ terms }} {{eqn | r = \paren {k + 1}^5 \paren {\paren {k^2 + 2 k + 1} + 1} + \sum_{i \mathop = 1}^k i^7 + \sum_{i \mathop = 1}^k i^5       | c = simplifying }} {{eqn | r = \paren {k + 1}^5 \paren {\paren {k + 1}^2 + 1} + \sum_{i \mathop = 1}^k i^7 + \sum_{i \mathop = 1}^k i^5       | c = Square of Sum }} {{eqn | r = \paren {k + 1}^7 + \paren {k + 1}^5 + \sum_{i \mathop = 1}^k i^7 + \sum_{i \mathop = 1}^k i^5       | c = simplifying }} {{eqn | r = \sum_{i \mathop = 1}^{k + 1} i^7 + \sum_{i \mathop = 1}^{k + 1} i^5       | c = {{Defof|Summation}} }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \N_{>0}: \sum_{i \mathop = 1}^n i^5 + \sum_{i \mathop = 1}^n i^7 = 2 \paren {\sum_{i \mathop = 1}^n i}^4$ {{qed}}
From: :Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing and: :Strictly Monotone Mapping with Totally Ordered Domain is Injective a strictly increasing mapping $\phi$ from $S$ to $T$ is an order isomorphism from $S$ to $\map \phi S$. Let $\mathbb F$ be the set of all strictly increasing mappings from $S$ to $T$. Let $\mathbb G$ be the set of all subsets of $T$ with $m$ elements. By Unique Isomorphism between Finite Totally Ordered Sets, the mapping $\Phi: \mathbb F \to \mathbb G$ defined as: :$\forall \phi \in \mathbb F: \Phi: \phi \to \map \phi S$ is a bijection. The result follows from Cardinality of Set of Subsets. {{qed}}
By Cancellable Elements of Semigroup form Subsemigroup, $\left({C, \circ {\restriction_C}}\right)$ is a subsemigroup of $\left({S, \circ}\right)$, where $\circ {\restriction_C}$ is the restriction of $\circ$ to $C$. By Restriction of Commutative Operation is Commutative, as $\left({C, \circ {\restriction_C}}\right)$ is a substructure of a commutative structure, it is also commutative. From: : the external direct product preserves the nature of semigroups : the external direct product preserves commutativity we see that $\left({S \times C, \oplus}\right)$ is a commutative semigroup. {{Qed}}
From Semigroup is Subsemigroup of Itself, $\left({S, \circ}\right)$ is a subsemigroup of $\left({S, \circ}\right)$. Also from Semigroup is Subsemigroup of Itself, $\left({C, \circ {\restriction_C}}\right)$ is a subsemigroup of $\left({C, \circ {\restriction_C}}\right)$. The result follows from Cross-Relation is Equivalence Relation. {{qed}}
From the defined equivalence relation, we have that: :$\left({x_1, y_1}\right) \boxtimes \left({x_2, y_2}\right) \iff x_1 \circ y_2 = x_2 \circ y_1$ is a congruence relation on $\left({S \times C, \oplus}\right)$. From the definition of the members of the equivalence classes: : $(1) \quad \forall x, y \in S, a, b \in C: \left({x \circ a, a}\right) \boxtimes \left({y \circ b, b}\right) \iff x = y$ : $(2) \quad \forall x, y \in S, a, b \in C: \left[\!\left[{\left({x \circ a, y \circ a}\right)}\right]\!\right]_\boxtimes = \left[\!\left[{\left({x, y}\right)}\right]\!\right]_\boxtimes$ From the definition of the equivalence class of equal elements: : $(3) \quad \forall c, d \in C: \left({c, c}\right) \boxtimes \left({d, d}\right)$ where $\left[\!\left[{\left({x, y}\right)}\right]\!\right]_\boxtimes$ is the equivalence class of $\left({x, y}\right)$ under $\boxtimes$. Hence we are justified in asserting the existence of the quotient structure: : $\displaystyle \left({T', \oplus'}\right) = \left({\frac {S \times C} \boxtimes, \oplus_\boxtimes}\right)$ {{qed}}
We have that: : From Integers under Addition form Abelian Group, the algebraic structure $\struct {\Z, +}$ is an abelian group. : From Integers under Multiplication form Countably Infinite Commutative Monoid, the algebraic structure $\struct {\Z, \times}$ is a commutative monoid and therefore a commutative semigroup. : Integer Multiplication Distributes over Addition. Thus all the ring axioms are fulfilled, and $\struct {\Z, +, \times}$ is a commutative ring. By Integer Multiplication has Zero, the zero is $0$. {{Qed}}
Taking the group axioms in turn: === G0: Closure === By Composite of Permutations is Permutation, $S$ is itself a permutation on $S$. Thus $\struct {\map \Gamma S, \circ}$ is closed. {{qed|lemma}} === G1: Associativity === From Set of all Self-Maps is Monoid, we have that $\struct {\map \Gamma S, \circ}$ is associative. {{qed|lemma}} === G2: Identity === From Set of all Self-Maps is Monoid, we have that $\struct {\map \Gamma S, \circ}$ has an identity, that is, the identity mapping. {{qed|lemma}} === G3: Inverses === By Inverse of Permutation is Permutation, if $f$ is a permutation of $S$, then so is its inverse $f^{-1}$. {{qed|lemma}} Thus all the group axioms have been fulfilled, and the result follows. {{qed}}
A direct application of Set of Invertible Mappings forms Symmetric Group. {{qed}}
From Ring is Ideal of Itself, it is clear we can form the quotient ring $\struct {R / R, +, \circ}$. But $R / R = 0_R$ and so is the null ring. {{qed}}
Let $\P \subset \mathbb S \subseteq \mathbb K$. By Intersection of Subrings is Largest Subring Contained in all Subrings: :$\bigcap \mathbb S$ is the largest subring of $K$ contained in each of the elements of $\mathbb S$. By Intersection of Subrings Containing Subset is Smallest: :The intersection of the set of all subrings of $K$ containing $\bigcup \mathbb S$ is the smallest subring of $K$ containing $\bigcup \mathbb S$. Thus: :Not only is $\bigcap \mathbb S$ a lower bound of $\mathbb S$, but also the largest, and therefore an infimum. :The supremum of $\mathbb S$ is the intersection of the set of all subrings of $K$. Therefore $\struct {\mathbb K, \subseteq}$ is a complete lattice. {{qed}}