avewright/proofwiki-math huggingface dataset:
Theorem: Equivalent Conditions for Entropic Structure/Pointwise Operation is Homomorphism

Let $\struct {T, \circledast}$ be an arbitrary algebraic structure.
Let $f$ and $g$ be mappings from $\struct {T, \circledast}$ to $\struct {S, \odot}$.
Let $f \odot g$ denote the pointwise operation on $S^T$ induced by $\odot$.
Then:
:If $f$ and $g$ are homomorphisms, then $f \odot g$ is also a homomorphism
{{iff}}:
:$\struct {S, \odot}$ is an entropic structure.

Proof:
=== Sufficient Condition ===
Let $\struct {S, \odot}$ be such that if $f$ and $g$ are homomorphisms, then $f \odot g$ is also a homomorphism.
So, let $f: T \to S$ and $g: T \to S$ be arbitrary homomorphisms.
Let $a, b \in T$ be arbitrary.
Because $T$ is arbitrary, and $f$ and $g$ are arbitrary, it follows that:
{{begin-eqn}}
{{eqn | q = \forall w, x, y, z \in S: \exists a, b \in T
      | l = \map f a
      | r = w
      | c = 
}}
{{eqn | lo= \land
      | l = \map f b
      | r = x
      | c = 
}}
{{eqn | l = \map g a
      | r = y
      | c = 
}}
{{eqn | lo= \land
      | l = \map g b
      | r = z
      | c = 
}}
{{end-eqn}}
Thus we have:
{{begin-eqn}}
{{eqn | l = \paren {w \odot x} \odot \paren {y \odot z}
      | r = \paren {\map f a \odot \map f b} \odot \paren {\map g a \odot \map g b}
      | c = {{apriori}}
}}
{{eqn | r = \paren {\map f {a \circledast b} }\odot \paren {\map g {a \circledast b} }
      | c = as $f$ and $g$ are both homomorphisms
}}
{{eqn | r = \map {\paren {f \odot g} } {a \circledast b}
      | c = {{Defof|Pointwise Operation}}
}}
{{eqn | r = \map {\paren {f \odot g} } a \odot \map {\paren {f \odot g} } b
      | c = as $f \odot g$ is a homomorphism
}}
{{eqn | r = \paren {\map f a \odot \map g a} \odot \paren {\map f b \odot \map g b}
      | c = {{Defof|Pointwise Operation}}
}}
{{eqn | r = \paren {w \odot y} \odot \paren {x \odot z}
      | c = {{Defof|Pointwise Operation}}
}}
{{end-eqn}}
As $w$, $x$, $y$ and $z$ are arbitrary, $\struct {S, \odot}$ is an entropic structure.
{{qed|lemma}}
Theorem: Composite with Constant Mapping is Constant Mapping

Let $f_c: S \to T$ be the constant mapping defined as:
:$\forall x \in S: \map {f_c} x = c$
where $c \in T$.
Then for all mappings $g: \Dom g \to S$:
:$f_c \circ g$ is a constant mapping
and for all mappings $h: T \to \Cdm h$:
:$h \circ f_c$ is a constant mapping
where:
:$\Dom g$ denotes the domain of $g$
:$\Cdm h$ denotes the codomain of $h$
:$\circ$ denotes composition of mappings.

Proof:
{{begin-eqn}}
{{eqn | q = \forall x \in \Dom g
      | l = \map {\paren {f_c \circ g} } x
      | r = \map {f_c} {\map g x}
      | c = {{Defof|Composition of Mappings}}
}}
{{eqn | r = c
      | c = {{Defof|Constant Mapping}}
}}
{{end-eqn}}
{{qed|lemma}}
{{begin-eqn}}
{{eqn | q = \forall x \in S
      | l = \map {\paren {h \circ f_c} } x
      | r = \map h {\map {f_c} x}
      | c = {{Defof|Composition of Mappings}}
}}
{{eqn | r = \map h c
      | c = {{Defof|Constant Mapping}}
}}
{{end-eqn}}
As $c$ is constant, so is $\map h c$.
{{qed}}
Category:Constant Mappings
Category:Composite Mappings
jsd8ppx12ybkqm9y9z2f7cpt2c94hp6
Theorem: Equivalent Conditions for Entropic Structure/Pointwise Operation of Homomorphisms from External Direct Product is Homomorphism

Let $\struct {S \times S, \otimes}$ denote the external direct product of $\struct {S, \odot}$ with itself:
:$\forall \tuple {x_1, y_1}, \tuple {x_2, y_2} \in S \times S: \tuple {x_1, y_1} \otimes \tuple {x_2, y_2} = \tuple {x_1 \odot x_2, y_1 \odot y_2}$
Let $f$ and $g$ be mappings from $\struct {S \times S, \otimes}$ to $\struct {S, \odot}$.
Let $f \odot g$ denote the pointwise operation on $S^{S \times S}$ induced by $\odot$.
Then:
:If $f$ and $g$ are homomorphisms, then $f \odot g$ is also a homomorphism
{{iff}}:
:$\struct {S, \odot}$ is an entropic structure.

Proof:
=== Sufficient Condition ===
Let $\struct {S, \odot}$ be such that if $f$ and $g$ are homomorphisms, then $f \odot g$ is also a homomorphism.
So, let $f: S \times S \to S$ and $g: S \times S \to S$ be arbitrary homomorphisms.
Let $w$, $x$, $y$ and $d$ in $S$ be arbitrary.
Then as $f$ and $g$ are arbitrary:
{{begin-eqn}}
{{eqn | q = \exists \tuple {x_1, y_1}, \tuple {x_2, y_2} \in S \times S
      | l = \map f {x_1, y_1}
      | r = w
      | c = 
}}
{{eqn | lo= \land
      | l = \map f {x_2, y_2}
      | r = x
      | c = 
}}
{{eqn | l = \map g {x_1, y_1}
      | r = y
      | c = 
}}
{{eqn | lo= \land
      | l = \map g {x_2, y_2}
      | r = z
      | c = 
}}
{{end-eqn}}
Then we have:
{{begin-eqn}}
{{eqn | o = 
      | r = \paren {w \odot x} \odot \paren {y \odot z}
      | c = 
}}
{{eqn | r = \paren {\map f {x_1, y_1} \odot \map f {x_2, y_2} } \odot \paren {\map g {x_1, y_1} \odot \map g {x_2, y_2} }
      | c = {{apriori}}
}}
{{eqn | r = \paren {\map f {\tuple {x_1, y_1} \otimes \tuple {x_2, y_2} } } \odot \paren {\map g {\tuple {x_1, y_1} \otimes \tuple {x_2, y_2} } }
      | c = as $f$ and $g$ are both homomorphisms
}}
{{eqn | r = \map {\paren {f \odot g} } {\tuple {x_1, y_1} \otimes \tuple {x_2, y_2} }
      | c = {{Defof|Pointwise Operation}}
}}
{{eqn | r = \map {\paren {f \odot g} } {x_1, y_1} \odot \map {\paren {f \odot g} } {x_2, y_2}
      | c = as $f \odot g$ is a homomorphism
}}
{{eqn | r = \paren {w \odot y} \odot \paren {x \odot z}
      | c = {{Defof|Pointwise Operation}}
}}
{{end-eqn}}
As $w$, $x$, $y$ and $z$ are arbitrary, $\struct {S, \odot}$ is an entropic structure.
{{qed|lemma}}
Theorem: Equivalent Conditions for Entropic Structure/Mapping from External Direct Product is Homomorphism

Let $\struct {S \times S, \otimes}$ denote the external direct product of $\struct {S, \odot}$ with itself:
:$\forall \tuple {x_1, y_1}, \tuple {x_2, y_2} \in S \times S: \tuple {x_1, y_1} \otimes \tuple {x_2, y_2} = \tuple {x_1 \odot x_2, y_1 \odot y_2}$
Consider the operation $\odot$ as a mapping from $S \times S$ to $S$.
That is:
:$\forall a, b \in S: \map \odot {a, b} = a \odot b$
Then:
:$\odot: S \times S \to S$ is a homomorphism from $\struct {S \times S, \otimes}$ to $\struct {S, \odot}$
{{iff}}:
:$\struct {S, \odot}$ is an entropic structure.

Proof:
=== Sufficient Condition ===
Let $\struct {S, \odot}$ be such that $\odot: S \times S \to S$ is a homomorphism.
Let $\tuple {x_1, y_1}, \tuple {x_2, y_2} \in S \times S$ be arbitrary.
We have:
{{begin-eqn}}
{{eqn | o = 
      | r = \paren {x_1 \odot x_2} \odot \paren {y_1 \odot y_2}
      | c = 
}}
{{eqn | r = \paren {\map \odot {x_1, x_2} } \odot \paren {\map \odot {y_1, y_2} }
      | c = Definition of $\odot$
}}
{{eqn | r = \map \odot {\tuple {x_1, x_2} \otimes \tuple {y_1, y_2} }
      | c = {{hypothesis}} $\odot$ is a homomorphism
}}
{{eqn | r = \map \odot {\tuple {x_1 \odot y_1, x_2 \odot y_2} }
      | c = {{Defof|External Direct Product}}
}}
{{eqn | r = \paren {x_1 \odot y_1} \odot \paren {x_2 \odot y_2}
      | c = Definition of $\odot$
}}
{{end-eqn}}
and it is seen $\struct {S, \odot}$ is an entropic structure by definition.
{{qed|lemma}}
Theorem: Condition for Algebra Loop to be Abelian Group

Let $\struct {S, \odot}$ be an algebra loop.
Let $\struct {S \times S, \otimes}$ denote the external direct product of $\struct {S, \odot}$ with itself:
:$\forall \tuple {x_1, y_1}, \tuple {x_2, y_2} \in S \times S: \tuple {x_1, y_1} \otimes \tuple {x_2, y_2} = \tuple {x_1 \odot x_2, y_1 \odot y_2}$
Let $f$ and $g$ be mappings from $\struct {S \times S, \otimes}$ to $\struct {S, \odot}$.
Let $f \odot g$ denote the pointwise operation on $S^{S \times S}$ induced by $\odot$.
Let $\struct {S, \odot}$ satisfy the condition that:
If $f$ and $g$ are homomorphisms, then $f \odot g$ is also a homomorphism.
Then $\struct {S, \odot}$ is an abelian group.

Proof:
Recall that an '''algebra loop''' $\struct {S, \circ}$ is a quasigroup with an identity element:
:$\exists e \in S: \forall x \in S: x \circ e = x = e \circ x$
From Equivalent Conditions for Entropic Structure: Pointwise Operation of Homomorphisms from External Direct Product is Homomorphism:
:$\struct {S, \odot}$ is an entropic structure.
From Entropic Structure with Identity is Commutative Monoid, $\struct {S, \odot}$ is a commutative monoid.
Hence {{afortiori}} $\struct {S, \odot}$ is a commutative semigroup.
From Structure is Group iff Semigroup and Quasigroup, it follows that $\struct {S, \odot}$ is a group.
As $\odot$ is a commutative operation, it follows that $\struct {S, \odot}$ is an abelian group.
{{qed}}
Theorem: Cartesian Product of Equivalence Relations

Let $A$ and $B$ be sets.
Let $\RR$ and $\SS$ be equivalence relations on $A$ and $B$ respectively.
Let $\TT$ be the relation on $A \times B$ defined as:
:$\forall \tuple {u, v}, \tuple {x, y} \in A \times B: \tuple {u, v} \mathrel \TT \tuple {x, y} \iff u \mathrel \RR x \land v \mathrel \SS y$
Then $\TT$ is an equivalence relation on $A \times B$.
The equivalence class $\eqclass {\tuple {a, b} } \TT$ of an element $\tuple {a, b}$ of $\TT$ is:
:$\eqclass {\tuple {a, b} } \TT = \set {\tuple {x, y}: x \in \eqclass a \RR, y \in \eqclass b \SS}$

Proof:
We have that $\RR$ and $\SS$ are equivalence relations on $A$ and $B$ respectively.
Thus they are both reflexive, symmetric and transitive.
Checking in turn each of the criteria for equivalence:
Theorem: Condition for Mapping between Structures to be Homomorphism

Let $\struct {A, \odot}$ and $\struct {B, \circledast}$ be magmas.
Let $\struct {A \times B, \otimes}$ be the external direct product of $\struct {A, \odot}$ and $\struct {B, \circledast}$.
Let $\phi: A \to B$ be a mapping.
Let $\phi$ be considered as a subset of the Cartesian product $A \times B$.
Then:
:$\phi$ is a homomorphism
{{iff}}:
:the algebraic structure $\struct {\phi, \otimes_\phi}$ is a submagma of $\struct {A \times B, \otimes}$.

Proof:
Let $\phi$ be a homomorphism
Let $\tuple {a, b}, \tuple {c, d} \in A \times B$ such that:
{{begin-eqn}}
{{eqn | l = \tuple {a, b}
      | o = \in
      | r = \phi
}}
{{eqn | l = \tuple {c, d}
      | o = \in
      | r = \phi
      | c = 
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \map \phi a
      | r = b
      | c = {{Defof|Mapping}}
}}
{{eqn | lo= \land
      | l = \map \phi c
      | r = d
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = \map \phi {a \odot c}
      | r = b \circledast d
      | c = as $\phi$ is a homomorphism
}}
{{eqn | ll= \leadstoandfrom
      | l = \tuple {a \odot c, b \circledast d}
      | o = \in
      | r = \phi
      | c = {{Defof|Mapping}}
}}
{{eqn | ll= \leadstoandfrom
      | l = \tuple {a, b} \otimes_\phi \tuple {c, d}
      | o = \in
      | r = \phi
      | c = {{Defof|External Direct Product}}
}}
{{end-eqn}}
That is:
:$\struct {\phi, \otimes_\phi}$ is a closed subset of $\struct {A \times B, \otimes}$
which means the same thing as:
:$\struct {\phi, \otimes_\phi}$ is a submagma of $\struct {A \times B, \otimes}$.
{{qed|lemma}}
The argument reverses, so:
:if $\struct {\phi, \otimes_\phi}$ is a submagma of $\struct {A \times B, \otimes}$
then:
:$\phi$ is a homomorphism.
{{qed}}
Theorem: Condition for Semigroup to be Internal Direct Product of Subgroup and Subsemigroup with Right Operation

Let $\struct {S, \odot}$ be a semigroup.
Then:
:$\struct {S, \odot}$ is the internal direct product of a subgroup $\struct {G, \odot_G}$ and subsemigroup $\struct {H, \odot_H}$ such that $\odot_H$ is the right operation
{{iff}}:
:for all $x \in S$ there exists a left identity $a$ such that $x \odot a = x$, and element $y$ such that $y \odot x = a$.

Proof:
=== Sufficient Condition ===
Let $\struct {S, \odot}$ be the internal direct product of a subgroup $\struct {G, \odot_G}$ and subsemigroup $\struct {H, \odot_H}$ such that $\odot_H$ is the right operation.
Hence, by definition, the mapping $\phi: G \times H \to S$ defined as:
:$\forall g \in H, h \in H: \map \phi {g, h} = g \odot h$
is an isomorphism from the '''(external) direct product''' $\struct {G, \odot_G} \times \struct {H, \odot_H}$ onto $\struct {S, \odot}$.
Thus {{afortiori}} $\phi$ is a bijection.
From Condition for Mapping between Structure and Cartesian Product of Substructures to be Bijection:
:for all $s \in S$: there exists a unique $\tuple {x, y} \in A \times B$ such that $x \odot y = s$.
So, let $e \in G$ be the identity element of $G$.
As $e \in G \implies e \in S$, the above condition therefore applies to $e$.
We have that:
:$\forall g \in G: g \odot e = g = e \odot g$
This also applies to $e$:
:$e \odot e = e$
But we have:
:$\forall z \in G: e = z \odot z^{-1}$
where $z^{-1} \in G$.
So, as $e$ can be represented uniquely in the form $e = x \odot y$, it follows that that unique representation is:
:$e = e \odot e$
where $e \in G \cap H$.
We have that:
:$\forall g \in G: g \odot e = g$
is the unique representation of $g$ such that $\tuple {g, e} \in G \odot H$.
Then we have by definition of right operation:
:$\forall h \in H: e \odot h = h$
which is the unique representation of $h$ such that $\tuple {e, h} \in G \odot H$.
We have by definition of the identity element of $G$ that:
:$\forall g \in G: e \odot g = g$
Now let $z \in G \odot H$ such that $z \notin G$ and $z \notin H$.
Then we have that:
:$z = g \odot h$
where $g \in G$ and $h \in H$ uniquely.
Thus:
{{begin-eqn}}
{{eqn | l = e \odot z
      | r = e \odot \paren {g \odot h}
      | c = Definition of $z$
}}
{{eqn | r = \paren {e \odot g} \odot h
      | c = {{Semigroup-axiom|1}}
}}
{{eqn | r = g \odot h
      | c = as $e$ is the identity element of $G$
}}
{{eqn | r = z
      | c = Definition of $z$
}}
{{end-eqn}}
So we have shown that:
:$\forall s \in S: e \odot s = s$
Hence, by definition, $e$ is a left identity in $S$.
Hence there exists at least one left identity in $S$.
{{qed|lemma}}
Let $x \in S$ be arbitrary.
It is to be shown that:
:there exists a left identity $a$ in $S$ such that $x \odot a = x$
:$\exists y \in S: y \odot x = a$
By {{Group-axiom|2}}, for all $g \in G$:
:$\exists e \in G: g \odot e = g$
By {{Group-axiom|3}}, for all $g \in G$:
:$\exists g^{-1} \in G: g^{-1} \odot g = e$
and so for elements $g$ of $G$:
:$e$ fulfils the condition of $a$
:$g^{-1}$ fulfils the condition of $y$.
{{ProofWanted}}
Theorem: Pointwise Addition on Continuous Real Functions on Closed Unit Interval forms Group

Let $J \subseteq \R$ denote the closed unit interval $\closedint 0 1$.
Let $\map {\mathscr C} J$ denote the set of all continuous real functions from $J$ to $\R$.
Let $\R^J$ denote the set of all mappings from $J$ to $\R$.
Let $\struct {\R^J, +}$ denote the algebraic structure on $\R^J$ induced by addition:
:$\forall f, g \in \R^J: \map {\paren {f + g} } x = \map f x + \map g x$
Then $\struct {\map {\mathscr C} J, +}$ is a subgroup of $\struct {\R^J, +}$.

Proof:
Taking the group axioms in turn:
Theorem: Pointwise Addition on Differentiable Real Functions on Closed Unit Interval forms Group

Let $J \subseteq \R$ denote the closed unit interval $\closedint 0 1$.
Let $\map {\mathscr D} J$ denote the set of all differentiable real functions from $J$ to $\R$.
Let $\R^J$ denote the set of all mappings from $J$ to $\R$.
Let $\struct {\R^J, +}$ denote the algebraic structure on $\R^J$ induced by addition:
:$\forall f, g \in \R^J: \map {\paren {f + g} } x = \map f x + \map g x$
Then $\struct {\map {\mathscr D} J, +}$ is a subgroup of $\struct {\R^J, +}$.

Proof:
Taking the group axioms in turn:
Theorem: Integral of Compactly Supported Function

Let $f : \R \to \R$ be a continuous real function.
Let $K \subset \R$ be a compact subset, say, $\closedint a b$.
Let $K$ be the support of $f$:
:$\map \supp f = K$.
Then:
:$\ds \int_{- \infty}^\infty \map f x \rd x = \int_a^b \map f x \rd x$

Proof:
{{begin-eqn}}
{{eqn | l = \int_{-\infty}^\infty \map f x \rd x
      | r = \int_{\overline \R} \map f x \rd x
      | c = {{Defof|Extended Real Number Line}}
}}
{{eqn | r = \int_{K \cup \paren { {\overline \R} \setminus K} } \map f x \rd x
      | c = {{Defof|Set Difference}}, {{Defof|Set Complement}}
}}
{{eqn | r = \int_K \map f x \rd x + \int_{\R \setminus K} \map f x \rd x
      | c = Riemann Integral Operator is Linear Mapping
}}
{{eqn | r = \int_K \map f x \rd x
      | c = {{Defof|Support of Continuous Mapping}}
}}
{{eqn | r = \int_a^b \map f x \rd x
}}
{{end-eqn}}
{{qed}}
Category:Calculus
kb6lbwqqmojo1o53ai7wbf4haxkzoar
Theorem: Value of Compactly Supported Function outside its Support

Let $f : \R \to \R$ be a continuous real function.
Let $K \subseteq \R$ be a compact subset.
Let $K$ be the support of $f$:
:$\map \supp f = K$.
Then:
:$\forall x \notin K : \map f x = 0$

Proof:
We have that:
:$\R = K \cup \paren {\R \setminus K}$.
By definition of the support:
:$x \in \map \supp f \iff \map f x \ne 0$
By Biconditional Equivalent to Biconditional of Negations:
:$\neg \paren {x \in \map \supp f} \iff \neg \paren {\map f x \ne 0}$
That is:
:$x \notin K \iff  \map f x = 0$
or
:$x \in \R \setminus K \iff  \map f x = 0$
Hence:
:$\forall x \notin K : \map f x = 0$
{{qed}}
Category:Real Analysis
abg8jppmi6qe6ut34p9vro6a445kzcc
Theorem: Dirichlet's Principle for Harmonic Functions/Riemannian Manifold

Let $\struct {M, g}$ be a compact connected $n$-dimensional Riemannian manifold with nonempty boundary. 
Let $\map {C^\infty} M$ be the smooth function space.
Let $u \in \map {C^\infty} M$ be a smooth real function.
Let $\rd V_g$ be the Riemannian volume form.
Let $\grad$ be the gradient operator.
Let $\size {\, \cdot \,}$ be the Riemannian inner product norm.
Then $u$ is harmonic {{iff}} $u$ minimizes
:$\ds \int_M \size {\grad u}^2 \rd V_g$
among all smooth real function with the same boundary values.

Proof:
{{ProofWanted}}
{{Namedfor|Johann Peter Gustav Lejeune Dirichlet|cat = Dirichlet}}
Theorem: Hasse Diagram of Dual Ordering

Let $D$ be the Hasse diagram of an ordering $\RR$.
Then the Hasse diagram $D'$ of the dual ordering $\RR^{-1}$ of $\RR$ is obtained by turning $D$ upside down.

Proof:
Let $\tuple {x, y}$ be an ordered pair in $\RR$ which is represented by a line in $D$ joining $x$ and $y$ such that $x$ is below $y$.
Then $x \mathrel \RR y$ such that:
:$\not \exists z: x \mathrel \RR z \land \mathrel \RR y$
As $\RR^{-1}$ is the dual of $\RR$, we have that $y \mathrel {\RR^{-1} } x$ such that:
:$\not \exists z: y \mathrel {\RR^{-1} } z \land \mathrel {\RR^{-1} }$
and so there will be a line in $D'$ joining $x$ and $y$ such that $x$ is above $y$.
The result follows.
{{qed}}
Theorem: Supremum of Elements of Sublattice not necessarily Same as for Lattice

Let $\struct {S, \preceq}$ be a lattice.
Let $\struct {T, \preceq_T}$ be a sublattice of $S$.
Let $a, b \in T$.
Then it is not necessarily the case that:
:$\sup_S \set {a, b}$
is the same as:
:$\sup_T \set {a, b}$

Proof:
Proof by Counterexample:
Let $\struct {G, \circ}$ be a group.
Let $\mathbb G$ be the set of all subgroups of $G$.
Let $\powerset G$ denote the power set of $G$.
Let $\struct {\powerset G, \subseteq}$ be the complete lattice formed by $\powerset G$ and $\subseteq$.
From Power Set is Complete Lattice, $\struct {\powerset G, \subseteq}$ indeed forms a complete lattice.
Let $\struct {\mathbb G, \subseteq}$ be the complete lattice formed by $\mathbb G$ and $\subseteq$.
From Set of Subgroups forms Complete Lattice, $\struct {\mathbb G, \subseteq}$ indeed forms a complete lattice.
By definition, $\struct {\mathbb G, \subseteq}$ is a sublattice of $\struct {\powerset G, \subseteq}$.
Let $H, K \in \mathbb G$.
We have that:
:$H \cup K$ is the supremum of $H$ and $K$ in $\struct {\powerset G, \subseteq}$.
But from Union of Subgroups, $H \cup K$ is not a subgroup of $G$ element of $\mathbb G$.
However, from Supremum of Subgroups in Lattice, we do have that the subset product $H K$ in $G$ is a subgroup of $G$ such that:
:$\sup_{\mathbb G} \set {H, K} = H K$
{{qed}}
Theorem: Generated Subsemigroup is not necessarily Same as Generated Group

Let $\struct {G, \circ}$ be a group.
Let $A \subseteq G$ be a subset of $G$.
Let $\struct {S, \circ}$ be the subsemigroup of $\struct {G, \circ}$ generated by $S$.
Let $\struct {H, \circ}$ be the subgroup of $\struct {G, \circ}$ generated by $S$.
Then it is not necessarily the case that $\struct {S, \circ}$ is the same as $\struct {H, \circ}$.

Proof:
Proof by Counterexample:
Let $\struct {\Z, +}$ be the additive group of integers.
Let $A$ be the set of positive odd integers.
From Generator of Subsemigroup: Positive Odd Numbers:
:the subsemigroup of $\struct {\Z, +}$ generated by $A$ is the semigroup of strictly positive integers under addition
while from Generator of Subgroup: Positive Odd Numbers
:the subgroup of $\struct {\Z, +}$ generated by $A$ is $\struct {\Z, +}$ itself.
{{qed}}
Theorem: Cardinality of Set of Relations

Let $S$ and $T$ be finite sets.
Let $\card S = n$ and $\card T = m$, where $\card {\, \cdot \,}$ denotes cardinality (that is, the number of elements).
Let $\RR$ denote the set of all relations from $S$ to $T$.
Then the cardinality of $\RR$ is given by:
:$\card \RR = 2^{n m}$

Proof:
By definition, a relation from $S$ to $T$ is a subset of the cartesian product $S \times T$ of $S$ and $T$.
From Cardinality of Cartesian Product of Finite Sets, we have:
:$\card {S \times T} = n m$
The number of subsets of $S \times T$ is the cardinality of the power set $\powerset {S \times T}$ of $S \times T$.
From Cardinality of Power Set of Finite Set:
:$\card {\powerset {S \times T} } = 2^{\card {S \times T} } = 2^{n m}$
Hence the result.
{{qed}}
Category:Combinatorics
Category:Cardinality
Category:Relations
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Theorem: Cardinality of Set of Endorelations

Let $S$ be a finite set.
Let $\card S = n$, where $\card {\, \cdot \,}$ denotes cardinality (that is, the number of elements).
Let $\RR$ denote the set of all endorelations on $S$.
Then the cardinality of $\RR$ is given by:
:$\card \RR = 2^{\paren {n^2} }$

Proof:
By definition, an endorelation on $S$ is a relation from $S$ to itself.
From Cardinality of Set of Relations, the number of relations from $S$ to $T$ is equal to $2^{\card S \times \card T}$.
In this case $S = T$ and the result follows.
{{qed}}
Category:Combinatorics
Category:Cardinality
Category:Endorelations
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Theorem: Number of Total Orderings on Finite Set

Let $S$ be a finite set with $n$ elements.
Then there are $n!$ different total orderings that can be applied to $S$.

Proof:
A total ordering on $S$ is by definition a permutation on $S$ in the sense of an ordered selection.
The result follows from Number of Permutations.
{{qed}}
Theorem: Isomorphism Class of Total Orderings

Let $S$ be a finite set with $n$ elements.
There is exactly one isomorphism class containing the total orderings on $S$.
That is, every total ordering on $S$ is (order) isomorphic to every other total ordering.

Proof:
{{ProofWanted|Intuitively obvious, probably needs to be hammered out in a proof by induction. I want to move on to other stuff today.}}
Theorem: Reflexive Reduction of Transitive Relation is Transitive

Let $S$ be a set.
Let $\RR$ be a transitive relation on $S$.
Let $\RR^\ne$ be the reflexive reduction of $\RR$.
Then $\RR^\ne$ is transitive.

Proof:
Let $a, b, c \in S$.
Let $a \mathrel {\RR^\ne} b$ and $b \mathrel {\RR^\ne} c$.
By the definition of reflexive reduction:
:$a \ne b$ and $a \mathrel \RR b$
:$b \ne c$ and $b \mathrel \RR c$
Since $\mathrel \RR$ is transitive:
:$a \mathrel \RR c$
{{qed}}