hcju/proofwikips huggingface dataset:
Characteristic Function Measurable iff Set Measurable
Let $\left({X, \Sigma}\right)$ be a measurable space. Let $E \subseteq X$. Then the following are equivalent: :$(1): \quad E \in \Sigma$; i.e., $E$ is a $\Sigma$-measurable set :$(2): \quad \chi_E: X \to \left\{{0, 1}\right\}$, the characteristic function of $E$, is $\Sigma$-measurable
Simple Function is Measurable
Let $\left({X, \Sigma}\right)$ be a measurable space. Let $f: X \to \R$ be a simple function. Then $f$ is $\Sigma$-measurable.
Identity Mapping is Relation Isomorphism
Let $\struct {S, \RR}$ be a relational structure. Then the identity mapping $I_S: S \to S$ is a relation isomorphism from $\struct {S, \RR}$ to itself.
Inverse of Relation Isomorphism is Relation Isomorphism
Let $\struct {S, \RR_1}$ and $\struct {T, \RR_2}$ be relational structures. Let $\phi: \struct {S, \RR_1} \to \struct {T, \RR_2}$ be a bijection. Then: :$\phi: \struct {S, \RR_1} \to \struct {T, \RR_2}$ is a relation isomorphism : :$\phi^{-1}: \struct {T, \RR_2} \to \struct {S, \RR_1}$ is also a relation isomorphism.
Measurable Function is Simple Function iff Finite Image Set
Let $\struct {X, \Sigma}$ be a measurable space. Let $f: X \to \R$ be a measurable function. Then $f$ is a simple function  its image is finite: :$\card {\Img f} < \infty$
Composite of Relation Isomorphisms is Relation Isomorphism
Let $\struct {S_1, \RR_1}$, $\struct {S_2, \RR_2}$ and $\struct {S_3, \RR_3}$ be relational structures. Let: :$\phi: \struct {S_1, \RR_1} \to \struct {S_2, \RR_2}$ and: :$\psi: \struct {S_2, \RR_2} \to \struct {S_3, \RR_3}$ be relation isomorphisms. Then $\psi \circ \phi: \struct {S_1, \RR_1} \to \struct {S_3, \RR_3}$ is also a relation isomorphism.
Pointwise Sum of Simple Functions is Simple Function
Let $\struct {X, \Sigma}$ be a measurable space. Let $f, g : X \to \R$ be simple functions. Then $f + g: X \to \R, \map {\paren {f + g} } x := \map f x + \map g x$ is also a simple function.
Pointwise Product of Simple Functions is Simple Function
Let $\struct {X, \Sigma}$ be a measurable space. Let $f, g : X \to \R$ be simple functions. Then $f \cdot g: X \to \R, \map {\paren {f \cdot g} } x := \map f x \cdot \map g x$ is also a simple function.
Positive Part of Simple Function is Simple Function
Let $\left({X, \Sigma}\right)$ be a measurable space. Let $f: X \to \R$ be a simple function. Then $f^+: X \to \R$, the positive part of $f$, is also a simple function.
Negative Part of Simple Function is Simple Function
Let $\left({X, \Sigma}\right)$ be a measurable space. Let $f: X \to \R$ be a simple function. Then $f^-: X \to \R$, the negative part of $f$ is also a simple function.
Difference of Positive and Negative Parts
Let $X$ be a set, and let $f: X \to \overline{\R}$ be an extended real-valued function. Let $f^+$, $f^-: X \to \overline{\R}$ be the positive and negative parts of $f$, respectively. Then $f = f^+ - f^-$.
Sum of Positive and Negative Parts
Let $X$ be a set, and let $f: X \to \overline{\R}$ be an extended real-valued function. Let $f^+, f^-: X \to \overline{\R}$ be the positive and negative parts of $f$, respectively. Then $\left\vert{f}\right\vert = f^+ + f^-$, where $\left\vert{f}\right\vert$ is the absolute value of $f$.
Absolute Value of Simple Function is Simple Function
Let $\struct {X, \Sigma}$ be a measurable space. Let $f: X \to \R$ be a simple function. Then $\size f: X \to \R$, the absolute value of $f$, is also a simple function.
Measurable Function Pointwise Limit of Simple Functions
Let $\left({X, \Sigma}\right)$ be a measurable space. Let $f: X \to \overline{\R}$ be a $\Sigma$-measurable function. Then there exists a sequence $\left({f_n}\right)_{n \in \N} \in \mathcal E \left({\Sigma}\right)$ of simple functions, such that: :$\forall x \in X: f \left({x}\right) = \displaystyle \lim_{n \to \infty} f_n \left({x}\right)$ That is, such that $f = \displaystyle \lim_{n \to \infty} f_n$ pointwise. The sequence $\left({f_n}\right)_{n \in \N}$ may furthermore be taken to satisfy: :$\forall n \in \N: \left\vert{f_n}\right\vert \le \left\vert{f}\right\vert$ where $\left\vert{f}\right\vert$ denotes the absolute value of $f$.
Finite Cartesian Product of Non-Empty Sets is Non-Empty
Let $S_1, S_2, \ldots, S_n$ be non-non-empty sets. Then their cartesian product $S_1 \times S_2 \times \cdots \times S_n$ is non-empty.
Equivalent Conditions for Dedekind-Infinite Set
For a set $S$, the following conditions are equivalent: :$(1): \quad$ $S$ is Dedekind-infinite. :$(2): \quad$ $S$ has a countably infinite subset. The above equivalence can be proven in Zermelo-Fraenkel set theory. If the axiom of countable choice is accepted, then it can be proven that the following condition is also equivalent to the above two: :$(3): \quad$ $S$ is infinite.
Relation Isomorphism Preserves Reflexivity
Let $\struct {S, \mathcal R_1}$ and $\struct {T, \mathcal R_2}$ be relational structures. Let $\struct {S, \mathcal R_1}$ and $\struct {T, \mathcal R_2}$ be (relationally) isomorphic. Then $\mathcal R_1$ is a reflexive relation  $\mathcal R_2$ is also a reflexive relation.
Preimage of Subset under Composite Mapping
Let $S_1, S_2, S_3$ be sets. Let $f: S_1 \to S_2$ and $g: S_2 \to S_3$ be mappings. Denote with $g \circ f: S_1 \to S_3$ the composition of $g$ and $f$. Let $S_3' \subseteq S_3$ be a subset of $S_3$. Then: :$\paren {g \circ f}^{-1} \sqbrk {S_3'} = \paren {f^{-1} \circ g^{-1} } \sqbrk {S_3'}$ where $g^{-1} \sqbrk {S_3'}$ denotes the preimage of $S_3'$ under $g$.
Infinite Set has Countably Infinite Subset/Proof 4
If the axiom of countable choice is accepted, then it can be proven that every infinite set has a countably infinite subset.
Set is Infinite iff exist Subsets of all Finite Cardinalities
A set $S$ is infinite  for all $n \in \N$, there exists a subset of $S$ whose cardinality is $n$.
Relation Isomorphism Preserves Symmetry
Let $\left({S, \mathcal R_1}\right)$ and $\left({T, \mathcal R_2}\right)$ be relational structures. Let $\left({S, \mathcal R_1}\right)$ and $\left({T, \mathcal R_2}\right)$ be (relationally) isomorphic. Then $\mathcal R_1$ is a symmetric relation  $\mathcal R_2$ is also a symmetric relation.
Set Difference with Proper Subset
Let $S$ be a set. Let $T \subsetneq S$ be a proper subset of $S$.   Let $S \setminus T$ denote the set difference between $S$ and $T$. Then: :$S \setminus T \ne \O$ where $\O$ denotes the empty set.
Order Isomorphism iff Strictly Increasing Surjection
Let $\left({S, \preceq_1}\right)$ and $\left({T, \preceq_2}\right)$ be totally ordered sets. A mapping $\phi: \left({S, \preceq_1}\right) \to \left({T, \preceq_2}\right)$ is an order isomorphism iff: : $(1): \quad \phi$ is a surjection : $(2): \quad \forall x, y \in S: x \mathop{\prec_1} y \implies \phi \left({x}\right) \mathop{\prec_2} \phi \left({y}\right)$
Relation Isomorphism Preserves Transitivity
Let $\struct {S, \RR_1}$ and $\struct {T, \RR_2}$ be relational structures. Let $\struct {S, \RR_1}$ and $\struct {T, \RR_2}$ be (relationally) isomorphic. Then $\RR_1$ is a transitive relation  $\RR_2$ is a transitive relation.
Pointwise Supremum of Measurable Functions is Measurable
Let $\left({X, \Sigma}\right)$ be a measurable space, and let $I$ be a countable set. Let $\left({f_i}\right)_{i \in I}$, $f_i: X \to \overline \R$ be an $I$-indexed collection of $\Sigma$-measurable functions. Then the pointwise supremum $\displaystyle \sup_{i \mathop \in I} f_i: X \to \overline \R$ is also $\Sigma$-measurable.
Pointwise Infimum of Measurable Functions is Measurable
Let $\left({X, \Sigma}\right)$ be a measurable space, and let $I$ be a countable set. Let $\left({f_i}\right)_{i \in I}$, $f_i: X \to \overline{\R}$ be an $I$-indexed collection of $\Sigma$-measurable functions. Then the pointwise infimum $\displaystyle \inf_{i \mathop \in I} f_i: X \to \overline{\R}$ is also $\Sigma$-measurable.
Pointwise Upper Limit of Measurable Functions is Measurable
Let $\left({X, \Sigma}\right)$ be a measurable space. Let $\left({f_n}\right)_{n \in \N}$, $f_n: X \to \overline{\R}$ be a sequence of $\Sigma$-measurable functions. Then the pointwise upper limit $\displaystyle \limsup_{n \to \infty} f_n: X \to \overline{\R}$ is also $\Sigma$-measurable.
Pointwise Lower Limit of Measurable Functions is Measurable
Let $\left({X, \Sigma}\right)$ be a measurable space. Let $\left({f_n}\right)_{n \mathop \in \N}$, $f_n: X \to \overline \R$ be a sequence of $\Sigma$-measurable functions. Then the pointwise lower limit: :$\displaystyle \liminf_{n \mathop \to \infty} f_n: X \to \overline \R$ is also $\Sigma$-measurable.
Pointwise Limit of Measurable Functions is Measurable
Let $\left({X, \Sigma}\right)$ be a measurable space. Let $\left({f_n}\right)_{n \in \N}$, $f_n: X \to \overline{\R}$ be a sequence of $\Sigma$-measurable functions. Then the pointwise limit $\displaystyle \lim_{n \to \infty} f_n: X \to \overline{\R}$ is also $\Sigma$-measurable.
Pointwise Sum of Measurable Functions is Measurable
Let $\left({X, \Sigma}\right)$ be a measurable space. Let $f, g: X \to \overline{\R}$ be $\Sigma$-measurable functions. Assume that the pointwise sum $f + g: X \to \overline{\R}$ is well-defined. Then $f + g$ is a $\Sigma$-measurable function.
Pointwise Difference of Measurable Functions is Measurable
Let $\struct {X, \Sigma}$ be a measurable space. Let $f, g: X \to \overline \R$ be $\Sigma$-measurable functions. Assume that the pointwise difference $f - g: X \to \overline \R$ is well-defined. Then $f - g$ is a $\Sigma$-measurable function.
Pointwise Product of Measurable Functions is Measurable
Let $\left({X, \Sigma}\right)$ be a measurable space. Let $f, g: X \to \overline{\R}$ be $\Sigma$-measurable functions. Then the pointwise product $f \cdot g: X \to \overline{\R}$ is also $\Sigma$-measurable.
Pointwise Maximum of Measurable Functions is Measurable
Let $\left({X, \Sigma}\right)$ be a measurable space. Let $f, g: X \to \overline{\R}$ be $\Sigma$-measurable functions. Then the pointwise maximum $\max \left({f, g}\right): X \to \overline{\R}$ is also $\Sigma$-measurable.
Function Measurable iff Positive and Negative Parts Measurable
Let $\left({X, \Sigma}\right)$ be a measurable space. Let $f: X \to \overline{\R}$ be an extended real-valued function. Let $f^+, f^-: X \to \overline{\R}$ be the positive and negative parts of $f$. Then $f$ is $\Sigma$-measurable iff both $f^+$ and $f^-$ are $\Sigma$-measurable.
Factorization Lemma/Extended Real-Valued Function
Then an extended real-valued function $g: X \to \overline{\R}$ is $\sigma \left({f}\right)$-measurable : :There exists a $\Sigma$-measurable mapping $\tilde g: Y \to \overline{\R}$ such that $g = \tilde g \circ f$ where: :$\sigma \left({f}\right)$ denotes the $\sigma$-algebra generated by $f$
Factorization Lemma
Let $X$ be a set, and $\left({Y, \Sigma}\right)$ be a measurable space. Let $f: X \to Y$ be a mapping.
Piecewise Combination of Measurable Mappings is Measurable/Binary Case
Let $f, g: X \to X'$ be $\Sigma \, / \, \Sigma'$-measurable mappings. Let $E \in \Sigma$ be a measurable set. Define $h: X \to X'$ by: :$\forall x \in X: \map h x := \begin{cases} \map f x & : \text {if $x \in E$} \\ \map g x & : \text {if $x \notin E$} \end{cases}$ Then $h$ is also a $\Sigma \, / \, \Sigma'$-measurable mapping.
Piecewise Combination of Measurable Mappings is Measurable/General Case
Let $\left({E_n}\right)_{n \in \N} \in \Sigma, \displaystyle \bigcup_{n \mathop \in \N} E_n = X$ be a countable cover of $X$ by $\Sigma$-measurable sets. For each $n \in \N$, let $f_n: E_n \to X'$ be a $\Sigma_{E_n} \, / \, \Sigma'$-measurable mapping. Here, $\Sigma_{E_n}$ is the trace $\sigma$-algebra of $E_n$ in $\Sigma$. Suppose that for every $m, n \in \N$, $f_m$ and $f_n$ satisfy: :$(1): \quad f_m \restriction_{E_m \cap E_n} = f_n \restriction_{E_m \cap E_n}$ that is, $f_m$ and $f_n$ coincide whenever both are defined; here $\restriction$ denotes restriction. Define $f: X \to X'$ by: :$\displaystyle \forall n \in \N, x \in E_n: f \left({x}\right) := f_n \left({x}\right)$ Then $f$ is a $\Sigma \, / \, \Sigma'$-measurable mapping.
Piecewise Combination of Measurable Mappings is Measurable
Let $\struct {X, \Sigma}$ and $\struct {X', \Sigma'}$ be measurable spaces.
Function Simple iff Positive and Negative Parts Simple
Let $\left({X, \Sigma}\right)$ be a measurable space. Let $g: X \to \overline{\R}$ be an extended real-valued function. Then $g$ is a simple function  its positive part $g^+$ and negative part $g^-$ are simple functions.
Bounded Measurable Function Uniform Limit of Simple Functions
Let $\left({X, \Sigma}\right)$ be a measurable space. Let $f: X \to \overline{\R}$ be a bounded $\Sigma$-measurable function. Then there exists a sequence $\left({f_n}\right)_{n \in \N} \in \mathcal E \left({\Sigma}\right)$ of simple functions, such that: :$\forall \epsilon > 0: \exists n \in \N: \forall x \in X: \left\vert{f \left({x}\right) - f_n \left({x}\right)}\right\vert < \epsilon$ That is, such that $f = \displaystyle \lim_{n \to \infty} f_n$ uniformly. The sequence $\left({f_n}\right)_{n \in \N}$ may furthermore be taken to satisfy: :$\forall n \in \N: \left\vert{f_n}\right\vert \le \left\vert{f}\right\vert$ where $\left\vert{f}\right\vert$ denotes the absolute value of $f$.
Integral of Characteristic Function
Let $\left({X, \Sigma, \mu}\right)$ be a measure space. Let $E \in \Sigma$ be a measurable set, and let $\chi_E: X \to \R$ be its characteristic function. Then $I_\mu \left({\chi_E}\right) = \mu \left({E}\right)$, where $I_\mu \left({\chi_E}\right)$ is the $\mu$-integral of $\chi_E$.
Integral of Positive Simple Function is Positive Homogeneous
Let $\left({X, \Sigma, \mu}\right)$ be a measure space. Let $f: X \to \R, f \in \mathcal E^+$ be a positive simple function. Let $\lambda \in \R_{\ge 0}$ be a positive real number. Then $I_\mu \left({\lambda \cdot f}\right) = \lambda \cdot I_\mu \left({f}\right)$, where: :$\lambda \cdot f$ is the pointwise $\lambda$-multiple of $f$ :$I_\mu$ denotes $\mu$-integration This can be summarized by saying that $I_\mu$ is positive homogeneous.
Integral of Positive Simple Function is Additive
Let $\left({X, \Sigma, \mu}\right)$ be a measure space. Let $f,g: X \to \R$, $f,g \in \mathcal{E}^+$ be positive simple functions. Then $I_\mu \left({f + g}\right) = I_\mu \left({f}\right) + I_\mu \left({g}\right)$, where: :$f + g$ is the pointwise sum of $f$ and $g$ :$I_\mu$ denotes $\mu$-integration This can be summarized by saying that $I_\mu$ is additive.
Strict Ordering Preserved under Product with Invertible Element
Let $\left({S, \circ, \preceq}\right)$ be an ordered semigroup. Let $z \in S$ be invertible. Suppose that either $x \circ z \prec y \circ z$ or $z \circ x \prec z \circ y$. Then $x \prec y$.
Strict Ordering Preserved under Cancellability in Totally Ordered Semigroup
Let $\left({S, \circ, \preceq}\right)$ be a totally ordered semigroup. If either: : $x \circ z \prec y \circ z$ or : $z \circ x \prec z \circ y$ then $x \prec y$.
Relation Compatibility in Totally Ordered Semigroup
Let $\left({S, \circ, \preceq}\right)$ be an ordered semigroup such that: :$(1): \quad$ All the elements of $\left({S, \circ, \preceq}\right)$ are cancellable for $\circ$ :$(2): \quad \preceq$ is a total ordering. Then: :$\forall x, y, z \in S: x \circ z \preceq y \circ z \iff x \preceq y$
Integral of Positive Measurable Function Extends Integral of Positive Simple Function
Let $\left({X, \Sigma, \mu}\right)$ be a measure space. Let $f: X \to \R, f \in \mathcal{E}^+$ be a positive simple function. Then $\displaystyle \int f \, \mathrm d\mu = I_\mu \left({f}\right)$, where: :$\displaystyle \int \cdot \, \mathrm d\mu$ denotes the $\mu$-integral of positive measurable functions :$I_\mu$ denotes the $\mu$-integral of positive simple functions That is, $\displaystyle \int \cdot \, \mathrm d\mu \restriction_{\mathcal{E}^+} = I_\mu$, using the notion of restriction, $\restriction$.
Beppo Levi's Theorem
Let $\struct {X, \Sigma, \mu}$ be a measure space. Let $\sequence {f_n}_{n \mathop \in \N} \in \MM_{\overline \R}^+$ be an increasing sequence of positive $\Sigma$-measurable functions. Let $\displaystyle \sup_{n \mathop \in \N} f_n: X \to \overline \R$ be the pointwise supremum of $\sequence {f_n}_{n \mathop \in \N}$, where $\overline \R$ denotes the extended real numbers. Then: :$\displaystyle \int \sup_{n \mathop \in \N} f_n \rd \mu = \sup_{n \mathop \in \N} \int f_n \rd \mu$ where the supremum on the  is in the ordering on $\overline \R$.
Ordered Semigroup Isomorphism is Surjective Monomorphism
Let $\left({S, \circ, \preceq}\right)$ and $\left({T, *, \preccurlyeq}\right)$ be ordered semigroups. Let $\phi: \left({S, \circ, \preceq}\right) \to \left({T, *, \preccurlyeq}\right)$ be a mapping. Then $\phi$ is an ordered semigroup isomorphism iff: :$(1): \quad \phi$ is an ordered semigroup monomorphism :$(2): \quad \phi$ is a surjection.
Ordered Semigroup Monomorphism into Image is Isomorphism
Let $\struct {S, \circ, \preceq}$ and $\struct {T, *, \preccurlyeq}$ be ordered semigroups. Let $\phi: \struct {S, \circ, \preceq} \to \struct {T, *, \preccurlyeq}$ be an ordered semigroup monomorphism. Let $S'$ be the image of $\phi$. Then $\phi$ is an ordered semigroup isomorphism from $\struct {S, \circ, \preceq}$ into $\struct {S', * {\restriction_{S'} }, \preccurlyeq \restriction_{S'} }$. Here: :$* {\restriction_{S'}}$ denotes the restriction of $*$ to $S' \times S'$ :$\preccurlyeq \restriction_{S'}$ denotes the restriction of $\preccurlyeq$ to $S' \times S'$.
Order Completion Unique up to Isomorphism
Let $\left({S, \preceq_S}\right)$ be an ordered set. Suppose that both $\left({T, \preceq_T}\right)$ and $\left({T', \preceq_{T'}}\right)$ are order completions for $\left({S, \preceq_S}\right)$. Then there exists a unique order isomorphism $\psi: T \to T'$. In particular, $\left({T, \preceq_T}\right)$ and $\left({T', \preceq_{T'}}\right)$ are isomorphic.
Intersection of Strict Upper Closures in Toset
Let $\left({S, \preceq}\right)$ be a totally ordered set. Let $a, b \in S$. Then: :$a^\succ \cap b^\succ = \left({\max \left({a, b}\right)}\right)^\succ$ where: :$a^\succ$ denotes strict upper closure of $a$ :$\max$ denotes the max operation.
Intersection of Weak Upper Closures in Toset
Let $\left({S, \preccurlyeq}\right)$ be a totally ordered set. Let $a, b \in S$. Then: :$a^\succcurlyeq \cap b^\succcurlyeq = \left({\max \left({a, b}\right)}\right)^\succcurlyeq$ where: : $a^\succcurlyeq$ denotes weak upper closure of $a$ : $\max$ denotes the max operation.
Intersection of Weak Lower Closures in Toset
Let $\struct {S, \preccurlyeq}$ be a totally ordered set. Let $a, b \in S$. Then: :$a^\preccurlyeq \cap b^\preccurlyeq = \paren {\min \set {a, b} }^\preccurlyeq$ where: :$a^\preccurlyeq$ denotes the weak lower closure of $a$ :$\min$ denotes the min operation.
Naturally Ordered Semigroup Exists
There exists a Naturally Ordered Semigroup.
Strict Upper Closure in Restricted Ordering
Let $\left({S, \preceq}\right)$ be an ordered set. Let $T \subseteq S$ be a subset of $S$, and let $\preceq \restriction_T$ be the restricted ordering on $T$. Then for all $t \in T$: :$t^{\succ T} = T \cap t^{\succ S}$ where: : $t^{\succ T}$ is the strict upper closure of $t$ in $\left({T, \preceq \restriction_T}\right)$ : $t^{\succ S}$ is the strict upper closure of $t$ in $\left({S, \preceq}\right)$.
Weak Upper Closure in Restricted Ordering
Let $\left({S, \preccurlyeq}\right)$ be an ordered set. Let $T \subseteq S$ be a subset of $S$. Let $\preccurlyeq \restriction_T$ be the restricted ordering on $T$. Then for all $t \in T$: :$t^{\succcurlyeq T} = T \cap t^{\succcurlyeq S}$ where: : $t^{\succcurlyeq T}$ is the weak upper closure of $t$ in $\left({T, \preccurlyeq \restriction_T}\right)$ : $t^{\succcurlyeq S}$ is the weak upper closure of $t$ in $\left({S, \preccurlyeq}\right)$.
Strict Lower Closure in Restricted Ordering
Let $\left({S, \preceq}\right)$ be an ordered set. Let $T \subseteq S$ be a subset of $S$, and let $\preceq \restriction_T$ be the restricted ordering on $T$. Then for all $t \in T$: :$t^{\prec T} = T \cap t^{\prec S}$ where: : $t^{\prec T}$ is the strict lower closure of $t$ in $\left({T, \preceq \restriction_T}\right)$ : $t^{\prec S}$ is the strict lower closure of $t$ in $\left({S, \preceq}\right)$.
Weak Lower Closure in Restricted Ordering
Let $\left({S, \preccurlyeq}\right)$ be an ordered set. Let $T \subseteq S$ be a subset of $S$. Let $\preccurlyeq \restriction_T$ be the restricted ordering on $T$. Then for all $t \in T$: :$t^{\preccurlyeq T} = T \cap t^{\preccurlyeq S}$ where: :$t^{\preccurlyeq T}$ is the weak lower closure of $t$ in $\left({T, \preccurlyeq \restriction_T}\right)$ :$t^{\preccurlyeq S}$ is the weak lower closure of $t$ in $\left({S, \preccurlyeq}\right)$.
Natural Numbers under Multiplication form Ordered Commutative Semigroup
Let $\N$ be the natural numbers. Let $\times$ be multiplication. Let $\le$ be the ordering on $\N$. Then $\left({\N, \times, \le}\right)$ is an ordered commutative semigroup.
Invertible Elements under Natural Number Multiplication
Let $\N$ be the natural numbers. Let $\times$ denote multiplication. Then the only invertible element of $\N$ for $\times$ is $1$.
Graph containing Closed Walk of Odd Length also contains Odd Cycle
Let $G$ be a graph. {{explain|This proof works for a simple graph, but the theorem may hold for loop graphs and/or multigraphs. Clarification needed as to what applies.}} Let $G$ have a closed walk of odd length. Then $G$ has an odd cycle.
Homomorphism of Powers/Naturally Ordered Semigroup
Let $\struct {S, \circ, \preceq}$ be a naturally ordered semigroup. For a given $a \in T_1$, let $\map {\odot^n} a$ be the $n$th power of $a$ in $T_1$. For a given $a \in T_2$, let $\map {\oplus^n} a$ be the $n$th power of $a$ in $T_2$. Then: :$\forall a \in T_1: \forall n \in \struct {S^*, \circ, \preceq}: \map \phi {\map {\odot^n} a} = \map {\oplus^n} {\map \phi a}$ where $S^* = S \setminus \set 0$.
Homomorphism of Powers/Natural Numbers
Let $n \in \N$. Let $\odot^n$ and $\oplus^n$ be the $n$th powers of $\odot$ and $\oplus$, respectively. Then: :$\forall a \in T_1: \forall n \in \N: \map \phi {\map {\odot^n} a} = \map {\oplus^n} {\map \phi a}$
Homomorphism of Powers/Integers
Let $\struct {T_1, \odot}$ and $\struct {T_2, \oplus}$ be monoids. Let $\phi: \struct {T_1, \odot} \to \struct {T_2, \oplus}$ be a (semigroup) homomorphism. Let $a$ be an invertible element of $T_1$. Let $n \in \Z$. Let $\odot^n$ and $\oplus^n$ be as defined as in Index Laws for Monoids. Then: :$\forall n \in \Z: \map \phi {\map {\odot^n} a} = \map {\oplus^n} {\map \phi a}$
Right Operation is Left Distributive over All Operations
Let $\struct {S, \circ, \rightarrow}$ be an algebraic structure where: :$\rightarrow$ is the right operation :$\circ$ is any arbitrary binary operation. Then $\rightarrow$ is left distributive over $\circ$.
Left Operation is Right Distributive over All Operations
Let $\struct {S, \circ, \leftarrow}$ be an algebraic structure where: :$\leftarrow$ is the left operation :$\circ$ is any arbitrary binary operation. Then $\leftarrow$ is right distributive over $\circ$.
Right Operation is Distributive over Idempotent Operation
Let $\struct {S, \circ, \rightarrow}$ be an algebraic structure where: :$\rightarrow$ is the right operation :$\circ$ is any arbitrary binary operation. Then: :$\rightarrow$ is distributive over $\circ$  :$\circ$ is idempotent.
Left Operation is Distributive over Idempotent Operation
Let $\struct {S, \circ, \leftarrow}$ be an algebraic structure where: :$\leftarrow$ is the left operation :$\circ$ is any arbitrary binary operation. Then: :$\leftarrow$ is distributive over $\circ$  :$\circ$ is idempotent.
Integral of Positive Measurable Function as Limit of Integrals of Positive Simple Functions
Let $\left({X, \Sigma, \mu}\right)$ be a measure space. Let $f: X \to \overline{\R} \in \mathcal{M}_{\overline{\R}}^+$ be a positive $\Sigma$-measurable function. Let $\left({f_n}\right)_{n \in \N} \in \mathcal{E}^+$, $f_n: X \to \R$ be a sequence of positive simple functions such that: :$\displaystyle \lim_{n \to \infty} f_n = f$ where $\lim$ denotes a pointwise limit. Then: :$\displaystyle \int f \, \mathrm d\mu = \lim_{n \to \infty} \int f_n \, \mathrm d\mu$ where the integral signs denote $\mu$-integration.