hcju/proofwikips huggingface dataset:
Vector Product is Zero only if Factor is Zero
Let $F$ be a field whose zero is $0_F$ and whose unity is $1_F$. Let $\struct {\mathbf V, +, \circ}_F$ be a vector space over $F$, as defined by the vector space axioms. Then: :$\forall \lambda \in F: \forall \mathbf v \in \mathbf V: \lambda \circ \mathbf v = \bszero \implies \paren {\lambda = 0_F \lor \mathbf v = \mathbf 0}$ where $\bszero \in \mathbf V$ is the zero vector.
Null Ring is Ideal
Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$. Then the null ring $\struct {\set {0_R}, +, \circ}$ is an ideal of $R$.
Ring Epimorphism Preserves Subrings
Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring epimorphism. Let $S$ be a subring of $R_1$. Then $\phi \sqbrk S$ is a subring of $R_2$.
Preimage of Image of Ideal under Ring Homomorphism
Let $\phi: \left({R_1, +_1, \circ_1}\right) \to \left({R_2, +_2, \circ_2}\right)$ be a ring homomorphism. Let $K = \ker \left({\phi}\right)$, where $\ker \left({\phi}\right)$ is the kernel of $\phi$.  Let $J$ be an ideal of $R_1$. Then: :$\phi^{-1} \left({\phi \left({J}\right)}\right) = J + K$
Image of Preimage of Ideal under Ring Epimorphism
Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring epimorphism. Let $S_2$ be an ideal of $R_2$. Then: :$\phi \sqbrk {\phi^{-1} \sqbrk {S_2} } = S_2$
Vectors are Right Cancellable
Let $\struct {\mathbf V, +, \circ}_{\mathbb F}$ be a vector space over $\mathbb F$, as defined by the vector space axioms. Then every $\mathbf v \in \struct {\mathbf V, +}$ is right cancellable: :$\forall \mathbf a, \mathbf b, \mathbf c \in \mathbf V: \mathbf a + \mathbf c = \mathbf b + \mathbf c \implies \mathbf a = \mathbf b$
Vectors are Left Cancellable
Let $\left({\mathbf V, +, \circ}\right)_{\mathbb F}$ be a vector space over $\mathbb F$, as defined by the vector space axioms. Then every $\mathbf v \in \left({\mathbf V, +}\right)$ is left cancellable: :$\forall \mathbf a, \mathbf b, \mathbf c \in \mathbf V: \mathbf c + \mathbf a = \mathbf c + \mathbf b \implies \mathbf a = \mathbf b$
Additive Inverse in Vector Space is Unique
Let $\struct {\mathbf V, +, \circ}_F$ be a vector space over a field $F$, as defined by the vector space axioms. Then for every $\mathbf v \in \mathbf V$, the additive inverse of $\mathbf v$ is unique: :$\forall \mathbf v \in \mathbf V: \exists! \paren {-\mathbf v} \in \mathbf V: \mathbf v + \paren {-\mathbf v} = \mathbf 0$
Vector Inverse is Negative Vector
Let $F$ be a field whose zero is $0_F$ and whose unity is $1_F$. Let $\struct {\mathbf V, +, \circ}_F$ be a vector space over $F$, as defined by the vector space axioms. Then: :$\forall \mathbf v \in \mathbf V: -\mathbf v = -1_F \circ \mathbf v$
Non-Trivial Commutative Division Ring is Field
Let $\struct {R, +, \circ}$ be a non-trivial division ring such that $\circ$ is commutative. Then $\struct {R, +, \circ}$ is a field. Similarly, let $\struct {F, +, \circ}$ be a field. Then $\struct {F, +, \circ}$ is a non-trivial  division ring such that $\circ$ is commutative.
Division Ring has No Proper Zero Divisors
Let $\left({R, +, \circ}\right)$ be a division ring. Then $\left({R, +, \circ}\right)$ has no proper zero divisors.
Linear Transformation is Injective iff Kernel Contains Only Zero
Let $\mathbf V, \mathbf V'$ be vector spaces, with respective zeroes $\mathbf 0, \mathbf 0'$. Let $T: \mathbf V \to \mathbf V'$ be a linear transformation. Then: :$T$ is injective  $\map \ker T = \set {\mathbf 0}$ where: :$\mathbf 0$ is the zero of the domain of $T$ :$\map \ker T$ is the kernel of $T$.
Diagonal Relation on Ring is Ordering Compatible with Ring Structure
Let $\struct {R, +, \circ, \preceq}$ be a ring whose zero is $0_R$. Then the diagonal relation $\Delta_R$ on $R$ is an ordering compatible with the ring structure of $R$.
Nagata-Smirnov Metrization Theorem
Let $T = \left({S, \tau}\right)$ be a topological space. Then $T$ is metrizable  $T$ is regular and has a basis that is countably locally finite.
Smirnov Metrization Theorem
Let $T = \left({S, \tau}\right)$ be a topological space. Then $T$ is metrizable  it is paracompact and locally metrizable.
Identity Mapping is Automorphism/Semigroups
Let $\left({S, \circ}\right)$ be a semigroup. Then $I_S: \left({S, \circ}\right) \to \left({S, \circ}\right)$ is a semigroup automorphism.
Inverse of Reflexive Relation is Reflexive
Let $\mathcal R$ be a relation on a set $S$. If $\mathcal R$ is reflexive, then so is $\mathcal R^{-1}$.
Inverse of Antireflexive Relation is Antireflexive
Let $\mathcal R$ be a relation on a set $S$. If $\mathcal R$ is antireflexive, then so is $\mathcal R^{-1}$.
Inverse of Non-Reflexive Relation is Non-Reflexive
Let $\mathcal R$ be a relation on a set $S$. If $\mathcal R$ is non-reflexive, then so is $\mathcal R^{-1}$.
Inverse of Symmetric Relation is Symmetric
Let $\mathcal R$ be a relation on a set $S$. If $\mathcal R$ is symmetric, then so is $\mathcal R^{-1}$.
Inverse of Asymmetric Relation is Asymmetric
Let $\mathcal R$ be a relation on a set $S$. If $\mathcal R$ is asymmetric, then so is $\mathcal R^{-1}$.
Inverse of Antisymmetric Relation is Antisymmetric
Let $\RR$ be a relation on a set $S$. If $\RR$ is antisymmetric, then so is $\RR^{-1}$.
Inverse of Non-Symmetric Relation is Non-Symmetric
Let $\mathcal R$ be a relation on a set $S$. If $\mathcal R$ is non-symmetric, then so is $\mathcal R^{-1}$.
Inverse of Transitive Relation is Transitive
Let $\mathcal R$ be a relation on a set $S$. Let $\mathcal R$ be transitive. Then its inverse $\mathcal R^{-1}$ is also transitive.
Inverse of Antitransitive Relation is Antitransitive
Let $\RR$ be a relation on a set $S$. If $\RR$ is antitransitive, then so is $\RR^{-1}$.
Inverse of Non-Transitive Relation is Non-Transitive
Let $\mathcal R$ be a relation on a set $S$. If $\mathcal R$ is non-transitive, then so is $\mathcal R^{-1}$.
Inverse of Ordered Ring Isomorphism is Ordered Ring Isomorphism
Let $\left({S, +, \circ, \preceq}\right)$ and $\left({T, \oplus, *, \preccurlyeq}\right)$ be ordered rings. Let $\phi: S \to T$ be an ordered ring isomorphism. Then $\phi^{-1}: T \to S$ is also an ordered ring isomorphism.
Feit-Thompson Theorem
All finite groups of odd order are solvable. That is, every non-abelian group finite simple group is of even order.
Orthogonal Group is Subgroup of General Linear Group
Let $k$ be a field. Let $\operatorname O \left({n, k}\right)$ be the $n$th orthogonal group on $k$. Let $\operatorname{GL} \left({n, k}\right)$ be the $n$th general linear group on $k$. Then $\operatorname O \left({n, k}\right)$ is a subgroup of $\operatorname{GL} \left({n, k}\right)$.
Orthogonal Group is Group
Let $k$ be a field. The $n$th orthogonal group on $k$ is a group.
Composite of Ordered Ring Isomorphisms is Ordered Ring Isomorphism
Let $\left({S_1, +_1, \circ_1, \preccurlyeq_1}\right), \left({S_2, +_2, \circ_2, \preccurlyeq_2}\right), \left({S_3, +_3, \circ_3, \preccurlyeq_3}\right)$ be ordered rings. Let $\phi: S_1 \to S_2$ and $\psi: S_2 \to S_3$ be ordered ring isomorphisms. Then the composite mapping $\psi \circ \phi: S_1 \to S_3$ is also an ordered ring isomorphism.
Composite of Ordered Ring Monomorphisms is Ordered Ring Monomorphism
Let $\left({S_1, +_1, \circ_1, \preccurlyeq_1}\right), \left({S_2, +_2, \circ_2, \preccurlyeq_2}\right), \left({S_3, +_3, \circ_3, \preccurlyeq_3}\right)$ be ordered rings. Let $\phi: S_1 \to S_2$ and $\psi: S_2 \to S_3$ be ordered ring monomorphisms. Then the composite mapping $\psi \circ \phi: S_1 \to S_3$ is also an ordered ring monomorphism.
Rescaling is Linear Transformation
Let $\left({R, +, \cdot}\right)$ be a commutative ring. Let $\left({V, +, \circ}\right)_R$ be an $R$-module. Then for any $r \in R$, the rescaling: :$m_r: V \to V, v \mapsto r \circ v$ is a linear transformation.
Determinant of Rescaling Matrix
Let $R$ be a commutative ring. Let $r \in R$. Let $r \, \mathbf I_n$ be the square matrix of order $n$ defined by: :$\sqbrk {r \, \mathbf I_n}_{i j} = \begin{cases} r & : i = j \\ 0 & : i \ne j \end{cases}$ Then: :$\map \det {r \, \mathbf I_n} = r^n$ where $\det$ denotes determinant.
Inverse of Rescaling Matrix
Let $R$ be a commutative ring with unity. Let $r \in R$ be a unit in $R$. Let $r \, \mathbf I_n$ be the $n \times n$ rescaling matrix of $r$. Then $\left({r \, \mathbf I_n}\right)^{-1} = r^{-1} \, \mathbf I_n$.
Powers of Ring Elements/General Result
:$\forall m, n \in \Z: \forall x \in R: \paren {m \cdot x} \circ \paren {n \cdot x} = \paren {m n} \cdot \paren {x \circ x}$.
Characteristic of Division Ring is Zero or Prime
Let $\struct {D, +, \circ}$ be a division ring. Let $\map {\operatorname {Char} } D$ be the characteristic of $D$. Then $\map {\operatorname {Char} } D$ is either $0$ or a prime number.
Magdy's Exchanger
'''Magdy's Exchanger''' is a simple method to get a semi-algebraic formula for inverse trigonometric functions using the help of inverse hyperbolic functions that has a real logarthmic formula hence we will have an example to get the inverse sine function. First: :$\displaystyle \arcsin x  = \int \frac 1 {\sqrt{1 - x^2}} \, \mathrm d x$ Using Integration by Parts: :$\displaystyle \int \frac 1 {\sqrt{1-x^2}} \,\mathrm d x = -\frac {\sqrt {1-x^2}} x - \int \frac{\sqrt{1-x^2}} {x^2} \, \mathrm d x$ The problem is that the integration on the  will inverse itself if we solve it by parts. Then the result will be: :$0 = 0$ Magdy's method will come into effect now. As we know the inverse hyperbolic function that is most similar to the inverse sine function is the inverse sinh function. The idea is to get the last integration in the  from the inverse sinh function. So we use parts again on the derivative of inverse sinh as follows: :$\displaystyle \operatorname{arcsinh} x = \int \frac 1 {\sqrt{1 + x^2}} \, \mathrm d x$ {{begin-eqn}} {{eqn | l = \operatorname{arcsinh} x       | r = \int \frac 1 {\sqrt{1 + x^2} } \, \mathrm d x       | c = Primitive of $\dfrac 1 {\sqrt {x^2 + a^2} }$ }} {{eqn | r = \frac {\sqrt {1 + x^2} } x + \int \frac {\sqrt {1 + x^2} } {x^2} \,\mathrm d x       | c = Integration by Parts }} {{eqn | r = \ln \left({x + \sqrt{x^2 + 1} }\right)       | c = Definition of Inverse Sinh Function }} {{end-eqn}} By Primitive of Root of $\dfrac {\sqrt {x^2 + a^2} } {x^2}$: :$\displaystyle \int \frac {\sqrt {1 + x^2} } {x^2} \, \mathrm d x = \ln \left({x + \sqrt{x^2 + 1} }\right) - \frac {\sqrt {1 + x^2} } x$ Taking the derivativeof both sides: :$\displaystyle \frac{\sqrt{1 + x^2} } {x^2} = \frac {1 + x^2 + x \sqrt{x^2 + 1} } {x^3 + x^2 \sqrt {x^2 + 1} } = A$ Squaring both sides: :$\dfrac {1 + x^2} {x^4} = A^2$ Now the most important step. As we want to have $1 - x^2$ in the numerator we will have the fraction on the } as: :$\dfrac {1 - x^2 + 2 x^2} {x^4}$ which does not affect the value of the fraction. Now the  is: :$\dfrac {1 - x^2} {x^4} + \dfrac 2 {x^2}$ then the wanted fraction will be alone on the : :$\dfrac {1 - x^2} {x^4} = A^2 - \dfrac 2 {x^2}$ Taking the root of both sides: :$\dfrac {\sqrt{1 - x^2} } {x^2} = \sqrt {A^2 - \dfrac 2 {x^2} }$ then integrating both sides: :$\displaystyle \int \frac {\sqrt{1 - x^2} } {x^2} \, \mathrm d x = \int \sqrt{A^2 - \frac 2 {x^2}} \, \mathrm d x$ Now the  is complicated but evaluates as follows: :$\displaystyle \int \sqrt{A^2 - \frac 2 {x^2} } \, \mathrm d x = \frac {x^2 \sqrt {\frac 1 {x^4} - \frac 1 {x^2} } \ln \left({2 \sqrt {x^2 - 1} + 2 x}\right)} {\sqrt {x^2 - 1} } - x \sqrt {\frac 1 {x^4} - \frac 1 {x^2} } + C$ Substituting the last result in the first equation of the inverse sine we get: :$\displaystyle \arcsin x = - \frac {\sqrt {1 - x^2} } x - \frac {x^2 \sqrt {\frac 1 {x^4} - \frac 1 {x^2} } \ln \left({2 \sqrt {x^2 - 1} + 2 x}\right)} {\sqrt {x^2 - 1}} +  x \sqrt {\frac 1 {x^4} - \frac 1 {x^2} } + C$ Hence we get a semi-algebraic formula for the inverse sine function.
Sigma-Algebra Generated by Complements of Generators
Let $\Sigma$ be a $\sigma$-algebra on a set $X$. Let $\mathcal G$ be a generator for $\Sigma$. Then: :$\mathcal{G}' := \left\{{X \setminus G: G \in \mathcal G}\right\}$ the set of relative complements of $\mathcal G$, is also a generator for $\Sigma$.
Primary Decomposition Theorem
Let $K$ be a field. Let $V$ be a vector space over $K$. Let $T: V \to V$ be a linear operator on $V$. Let $\map p x \in K \sqbrk x$ be a polynomial such that: :$\map \deg p \ge 1$ :$\map p T = 0$ where $0$ is the zero operator on $V$. {{explain|Link to definition of $K \sqbrk x$}} Let $\map {p_1} x, \map {p_2} x, \ldots, \map {p_r} x$ be distinct irreducible monic polynomials. Let $c \in K \setminus \set 0$ and $a_1, a_2, \ldots, a_r, r \in \Z_{\ge 1}$ be constants. We have that: :$\map p x = c \map {p_1} x^{a_1} \map {p_2} x^{a_2} \dotsm \map {p_r} x^{a_r}$ The primary decomposition theorem then states the following : :$(1): \quad \map \ker {\map {p_i} T^{a_i} }$ is a $T$-invariant subspace of $V$ for all $i = 1, 2, \dotsc, r$ :$(2): \quad \displaystyle V = \bigoplus_{i \mathop = 1}^r \map \ker {\map {p_i} T^{a_i} }$ {{explain|Link to definition of $\bigoplus$ in this context.}} {{improve|Rather than dragging the unwieldy $\map \ker {\map {p_i} T^{a_i} }$ all around the page, suggest that a symbol e.g. $\kappa_i$ be used for it instead. There are many more places where the exposition is repetitive and could benefit from being broken into more modular units. And there are still places where the logical flow is compromised by being wrapped backwards upon itself with "because" and "indeed" and nested "if-thens", although I have done my best to clean out most of these.}}
Sigma-Algebra Extended by Single Set
Let $\Sigma$ be a $\sigma$-algebra on a set $X$. Let $S \subseteq X$ be a subset of $X$. For subsets $T \subseteq X$ of $X$, denote $T^c$ for the set difference $X \setminus T$. Then: :$\sigma \left({\Sigma \cup \left\{{S}\right\}}\right) = \left\{{\left({E_1 \cap S}\right) \cup \left({E_2 \cap S^c}\right): E_1, E_2 \in \Sigma}\right\}$ where $\sigma$ denotes generated $\sigma$-algebra.
Subring Module/Special Case
Let $S$ be a subring of the ring $\struct {R, +, \circ}$. Let $\circ_S$ be the restriction of $\circ$ to $S \times R$. Then $\struct {R, +, \circ_S}_S$ is an $S$-module. If $\struct {R, +, \circ}$ has a unity, $1_R$, and $1_R \in S$, then $\struct {R, +, \circ_S}_S$ is a unitary $S$-module.
Trivial Module is Module
Let $\struct {G, +_G}$ be an abelian group whose identity is $e_G$. Let $\struct {R, +_R, \circ_R}$ be a ring. Let $\struct {G, +_G, \circ}_R$ be the trivial $R$-module, such that: :$\forall \lambda \in R: \forall x \in G: \lambda \circ x = e_G$ Then $\struct {G, +_G, \circ}_R$ is a module.
Null Module is Module
Let $\left({R, +_R, \circ_R}\right)$ be a ring. Let $G$ be the trivial group. Let $\left({G, +_G, \circ}\right)_R$ be the null module. Then $\left({G, +_G, \circ}\right)_R$ is a module.
Submodule of Module of Polynomial Functions
Let $K$ be a commutative ring with unity. Let $P \left({K}\right)$ be the set of all polynomial functions on $K$. Consider the set $P_m \left({K}\right)$ of all the polynomial functions: :$\displaystyle \sum_{k \mathop = 0}^{m-1} \alpha_k {I_K}^k$ for some $m \in \N^*$ where: :$\left \langle {\alpha_k} \right \rangle_{k \in \left[{0 \,.\,.\, m-1}\right]}$ is any sequence of $m$ terms of $K$. Then $P_m \left({K}\right)$ is a submodule of $P \left({K}\right)$.
Condition on Equality of Generated Sigma-Algebras
Let $X$ be a set, and let $\mathcal G$, $\mathcal H$ be sets of subsets of $X$. Suppose that: :$\mathcal G \subseteq \mathcal H \subseteq \sigma \left({\mathcal G}\right)$ where $\sigma$ denotes generated $\sigma$-algebra. Then: :$\sigma \left({\mathcal G}\right) = \sigma \left({\mathcal H}\right)$
Sigma-Algebra is Monotone Class
Let $\Sigma$ be a $\sigma$-algebra on a set $X$. Then $\Sigma$ is also a monotone class.
Trace Sigma-Algebra of Generated Sigma-Algebra
Let $X$ be a Set, and let $\GG \subseteq \powerset X$ be a collection of subsets of $X$. Let $A \subseteq X$ be a subset of $X$. Then the following equality holds: :$A \cap \map \sigma \GG = \map \sigma {A \cap \GG}$ where :$\map \sigma \GG$ denotes the smallest $\sigma$-algebra on $X$ that contains $\GG$ :$\map \sigma {A \cap \GG}$ denotes the smallest $\sigma$-algebra on $A$ that contains ${A \cap \GG}$ :$A \cap \map \sigma \GG$ denotes the trace $\sigma$-algebra on $A$ :$A \cap \GG$ is a shorthand for $\set {A \cap G: G \in \GG}$
Intersection is Subset/General Result
Let $S$ be a set. Let $\mathcal P \left({S}\right)$ be the power set of $S$. Let $\mathbb S \subseteq \mathcal P \left({S}\right)$. Then: : $\displaystyle \forall T \in \mathbb S: \bigcap \mathbb S \subseteq T$ === Family of Sets === In the context of a family of sets, the result can be presented as follows: {{:Intersection is Subset/Family of Sets}}
Union of Relative Complements of Nested Subsets
Let $R \subseteq S \subseteq T$ be sets with the indicated inclusions. Then: :$\complement_T \left({S}\right) \cup \complement_S \left({R}\right) = \complement_T \left({R}\right)$ where $\complement$ denotes relative complement. Phrased via Set Difference as Intersection with Relative Complement: :$\left({T \setminus S}\right) \cup \left({S \setminus R}\right) = T \setminus R$ where $\setminus$ denotes set difference.
Homomorphic Image of R-Module is R-Module
Let $\left({R, +_R, \times_R}\right)$ be a ring. Let $\left({G, +_G, \circ_G}\right)_R$ be an $R$-module. Let $\left({H, +_H, \circ_H}\right)_R$ be an $R$-algebraic structure. Let $\phi: G \to H$ be a homomorphism. Then the homomorphic image of $\phi$ is an $R$-module.
Evaluation Linear Transformation is Linear Transformation
Let $R$ be a commutative ring. Let $G$ be an $R$-module. Let $G^*$ be the algebraic dual of $G$. Let $G^{**}$ be the algebraic dual of $G^*$. Let the mapping $J: G \to G^{**}$ be the evaluation linear transformation from $G$ into $G^{**}$. For each $x \in G$,  $x^\wedge: G^* \to R$ is defined as: :$\forall t' \in G^*: \map {x^\wedge} {t'} = \map {t'} x$ Let the mapping $J: G \to G^{**}$ be the evaluation linear transformation from $G$ into $G^{**}$ defined as: :$\forall x \in G: \map J x = x^\wedge$ where for each $x \in G$, $x^\wedge: G^* \to R$ is defined as: :$\forall t' \in G^*: \map {x^\wedge} {t'} = \map {t'} x$   Then: :$(1): \quad x^\wedge \in G^{**}$ :$(2): \quad J$ is a linear transformation.
Evaluation Isomorphism is Isomorphism
Let $R$ be a commutative ring. Let $G$ be a unitary $R$-module whose dimension is finite. Then the evaluation linear transformation $J: G \to G^{**}$ is an isomorphism.
Zero Matrix is Identity for Hadamard Product
Let $\struct {S, \cdot}$ be a monoid whose identity is $e$. Let $\map {\MM_S} {m, n}$ be an $m \times n$ matrix space over $S$. Let $\mathbf e = \sqbrk e_{m n}$ be the zero matrix of $\map {\MM_S} {m, n}$. Then $\mathbf e$ is the identity element for Hadamard product.
Sum of Ideals is Ideal/General Result
Let $J_1, J_2, \ldots, J_n$ be ideals of a ring $\struct {R, +, \circ}$. Then: : $J = J_1 + J_2 + \cdots + J_n$ is an ideal of $R$. where $J_1 + J_2 + \cdots + J_n$ is as defined in subset product.
Integer is Expressible as Product of Primes
Let $n$ be an integer such that $n > 1$. Then $n$ can be expressed as the product of one or more primes.
Expression for Integer as Product of Primes is Unique
Let $n$ be an integer such that $n > 1$. Then the expression for $n$ as the product of one or more primes is unique up to the order in which they appear.
Modulo Addition is Closed/Integers
Let $m \in \Z$ be an integer. Then addition modulo $m$ on the set of integers modulo $m$ is closed: :$\forall \eqclass x m, \eqclass y m \in \Z_m: \eqclass x m +_m \eqclass y m \in \Z_m$.
Set is Subset of Union/General Result
Let $S$ be a set. Let $\mathcal P \left({S}\right)$ be the power set of $S$. Let $\mathbb S \subseteq \mathcal P \left({S}\right)$. Then: : $\forall T \in \mathbb S: T \subseteq \bigcup \mathbb S$
Intersection Distributes over Union/General Result
Let $S$ and $T$ be sets. Let $\powerset T$ be the power set of $T$. Let $\mathbb T$ be a subset of $\powerset T$. Then: :$\displaystyle S \cap \bigcup \mathbb T = \bigcup_{X \mathop \in \mathbb T} \paren {S \cap X}$
Union Distributes over Intersection/General Result
Let $S$ and $T$ be sets. Let $\powerset T$ be the power set of $T$. Let $\mathbb T$ be a subset of $\powerset T$. Then: :$\displaystyle S \cup \bigcap \mathbb T = \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$
Scalar Multiple of Simple Function is Simple Function
Let $\struct {X, \Sigma}$ be a measurable space. Let $f: X \to \R$ be a simple function, and let $\lambda \in \R$. Then the pointwise scalar multiple $\lambda f: X \to \R$ of $f$ is also a simple function.
Image of Composite Mapping
Let $f: S \to T$ and $g: R \to S$ be mappings. Then: :$\operatorname{Im} \left({f \circ g}\right) = f \left[{\operatorname{Im} \left({g}\right)}\right]$ where: : $f \circ g$ is the composition of $g$ and $f$ : $\operatorname{Im}$ denotes image : $f \left[{\cdot}\right]$ denotes taking image of a subset under $f$.
Scalar Multiple of Integrable Function is Integrable Function
Let $\struct {X, \Sigma, \mu}$ be a measure space. Let $f: X \to \overline \R$ be a $\mu$-integrable function, and let $\lambda \in \R$. Then $\lambda f: X \to \overline \R$, the pointwise $\lambda$-multiple of $f$, is also $\mu$-integrable. That is, the space of integrable functions $\LL^1_{\overline \R}$ is closed under pointwise $\R$-scalar multiplication.
Left Distributive Law for Natural Numbers
The operation of multiplication is left distributive over addition on the set of natural numbers $\N$: :$\forall x, y, n \in \N_{> 0}: n \times \paren {x + y} = \paren {n \times x} + \paren {n \times y}$
Right Distributive Law for Natural Numbers
The operation of multiplication is right distributive over addition on the set of natural numbers $\N_{> 0}$: :$\forall x, y, n \in \N_{> 0}: \paren {x + y} \times n = \paren {x \times n} + \paren {y \times n}$
Standard Machinery
Let $\left({X, \Sigma, \mu}\right)$ be a measure space. Let $\mathcal{L}^1_{\overline \R} \left({\mu}\right)$ be the space of $\mu$-integrable functions. Let $P \left({f_1, \ldots, f_n}\right)$ be a proposition, where the variables $f_i$ denote $\mu$-measurable functions $f_i: X \to \overline{\R}$. Let every occurrence of an $f_i$ be of the form: :$\displaystyle \int \Phi \left({f_i}\right) \, \mathrm d\mu$ for a suitable index set $I$ and multilinear mapping $\Phi: \mathcal{L}^1_{\overline \R} \left({\mu}\right)^I \to \mathcal{L}^1_{\overline \R} \left({\mu}\right)$. Denote with $\chi \left({\Sigma}\right)$ the set of characteristic functions of elements of $\Sigma$, i.e.: :$\chi \left({\Sigma}\right) := \left\{{\chi_E: X \to \R: E \in \Sigma}\right\}$ Then the following are equivalent: :$(A): \quad \forall f_1, \ldots, f_n \in \chi \left({\Sigma}\right): P \left({f_1, \ldots, f_n}\right)$ :$(B): \quad \forall f_1, \ldots, f_n \in \mathcal{L}^1_{\overline \R} \left({\mu}\right): P \left({f_1, \ldots, f_n}\right)$ that is, it suffices to verify $P$ holds for characteristic functions. {{WIP|Not completely valid as stated; need to restrict to $E$ with $\mu(E)<+\infty$ and impose more conditions on $P$}}
Third Principle of Mathematical Induction
Let $\map P n$ be a propositional function depending on $n \in \N$. If: :$(1): \quad \map P n$ is true for all $n \le d$ for some $d \in \N$ :$(2): \quad \forall m \in \N: \paren {\forall k \in \N, m \le k < m + d: \map P k} \implies \map P {m + d}$ then $\map P n$ is true for all $n \in \N$.
Cross-Relation is Equivalence Relation
{{Cross-Relation Context Definition}} Then $\boxtimes$ is an equivalence relation on $\struct {S_1 \times S_2, \oplus}$.
Cross-Relation on Natural Numbers is Equivalence Relation
Let $\struct {\N, +}$ be the semigroup of natural numbers under addition. Let $\struct {\N \times \N, \oplus}$ be the (external) direct product of $\struct {\N, +}$ with itself, where $\oplus$ is the operation on $\N \times \N$ induced by $+$ on $\N$. The relation $\boxtimes$ defined on $\N \times \N$ by: :$\tuple {x_1, y_1} \boxtimes \tuple {x_2, y_2} \iff x_1 + y_2 = x_2 + y_1$ is an equivalence relation on $\struct {\N \times \N, \oplus}$.