hcju/proofwikips huggingface dataset:
Integral of Survival Function
Let $\left({X, \Sigma, \mu}\right)$ be a $\sigma$-finite measure space. Let $f: X \to \R_{\ge 0}$ be a positive $\Sigma$-measurable function. Let $F \left({f}\right): \R \to \R$ be the survival function of $f$. Then: :$\displaystyle \int f \, \mathrm d \mu = \int_{\left({0 \,.\,.\, \infty}\right)} F \left({f}\right) \, \mathrm d \lambda$ where $\lambda$ is Lebesgue measure.
Cardinal Zero is Less than Cardinal One
The zero cardinal $0$ is less than one: :$0 < 1$
Idempotent Non-Trivial Quasigroup is Not a Loop
Let $\left({S, \circ}\right)$ be an idempotent quasigroup whose underlying set $S$ comprises more than one element. Then $\left({S, \circ}\right)$ is not an algebra loop, that is, it has no identity element.
B-Algebra is Right Cancellable
Let $\struct {X, \circ}$ be a $B$-algebra. Then $\circ$ is right-cancellable for $X$. That is: :$\forall x, y, z \in X: x \circ z = y \circ z \implies x = y$
Right Regular Representation of 0 is Bijection in B-Algebra
Let $\left({X, \circ}\right)$ be a $B$-algebra. Then the right regular representation of $\left({X, \circ}\right)$ with respect to $0$ is a bijection.
Product of Cardinals is Commutative
Let $\mathbf a$ and $\mathbf b$ be cardinals. Then: : $\mathbf a \mathbf b = \mathbf b \mathbf a$ where $\mathbf a \mathbf b$ denotes the product of $\mathbf a$ and $\mathbf b$.
B-Algebra Identity: xy=x(0(0y))
Let $\left({X, \circ}\right)$ be a $B$-algebra. Then: :$\forall x,y \in X: x \circ y = x \circ\left({0 \circ\left({ 0 \circ y}\right)}\right)$
B-Algebra Identity: x (y z) = (x (0 z)) y
Let $\left({X, \circ}\right)$ be a $B$-algebra. Then: :$\forall x,y,z \in X: x \circ \left({y \circ z}\right) = \left({x \circ \left({0 \circ z}\right)}\right) \circ y$
B-Algebra Identity: xy = 0 iff x = y
Let $\left({X, \circ}\right)$ be a $B$-algebra. Then: :$\forall x,y \in X: x \circ y = 0 \iff x = y$
First Power of Element in B-Algebra
Let $\left({X, \circ}\right)$ be a $B$-algebra. Then: :$\forall x \in X: x^1 = x$ where $x^k$ for $k \in \N$ denotes the $k$th power of the element $x$.
B-Algebra Power Law
Let $\left({X, \circ}\right)$ be a B-algebra. Let $n, m \in \N$ such that $n \ge m$. Then: :$\forall x \in X: x^n \circ x^m = x^{n-m}$ where $x^k$ for $k \in \N$ denotes the $k$th power of the element $x$.
B-Algebra is Left Cancellable
Let $\struct {X, \circ}$ be a $B$-algebra. Then $\circ$ is a left cancellable operation.
B-Algebra is Quasigroup
Let $\left({X, \circ}\right)$ be a $B$-algebra. Then $\left({X, \circ}\right)$ is a quasigroup.
Quasigroup is not necessarily B-Algebra
Let $\struct {S, \circ}$ be a quasigroup. Then $\struct {S, \circ}$ is not necessarily a $B$-algebra.
Group is Quasigroup
Let $\left({G, \circ}\right)$ be a group. Then $\left({G, \circ}\right)$ is a quasigroup.
B-Algebra Power Law with Zero
:$\forall x \in X: n, m \in \N_{>0} \implies x^m \circ x^n = 0 \circ x^{n-m}$
Product of Powers in B-Algebra
Let $\left({X, \circ}\right)$ be a $B$-algebra. Let $x \in X$ and $m, n \in \N$. Then: :$x^m \circ x^n = \begin{cases} x^{m-n} & : m \ge n \\ 0 \circ x^{n-m} & : n > m \end{cases}$
Superset of Co-Countable Set
Every superset of a co-countable set is co-countable.
Group Induces B-Algebra
Let $\left({G, \circ}\right)$ be a group whose identity element is $e$. Let $*$ be the binary operation on $G$ defined as: :$\forall a, b \in G: a * b = a \circ b^{-1}$ where $b^{-1}$ is the inverse element of $b$ under the operation $\circ$. Then the algebraic structure $\left({G, *}\right)$ is a $B$-algebra.
Identity of Cardinal Product is One
Let $\mathbf a$ be a cardinal. Then: : $\mathbf 1 \mathbf a = \mathbf a$ where $\mathbf 1 \mathbf a$ denotes the product of the (cardinal) one and $\mathbf a$. That is, $\mathbf 1$ is the identity element of the product operation on cardinals.
Zero of Cardinal Product is Zero
Let $\mathbf a$ be a cardinal. Then: : $\mathbf 0 \mathbf a = \mathbf 0$ where $\mathbf 0 \mathbf a$ denotes the product of the (cardinal) zero and $\mathbf a$. That is, $\mathbf 0$ is the zero element of the product operation on cardinals.
Sum of Cardinals is Commutative
Let $\mathbf a$ and $\mathbf b$ be cardinals. Then: :$\mathbf a + \mathbf b = \mathbf b + \mathbf a$ where $\mathbf a + \mathbf b$ denotes the sum of $\mathbf a$ and $\mathbf b$.
B-Algebra is Commutative iff x(xy)=y
Let $\left({X, \circ}\right)$ be a $B$-algebra. Then $\left({X, \circ}\right)$ is commutative : :$\forall x, y \in X: x \circ \left({x \circ y}\right) = y$
Identity of Cardinal Sum is Zero
Let $\mathbf a$ be a cardinal. Then: : $\mathbf a + \mathbf 0 = \mathbf a$ where $\mathbf a + \mathbf 0$ denotes the sum of the zero cardinal and $\mathbf a$. That is, $\mathbf 0$ is the identity element of the sum operation on cardinals.
Commutative B-Algebra Implies (zy)(zx)=xy
Let $\struct {B, \circ}$ be a commutative $B$-algebra.  Then: :$\forall x, y, z \in X: \paren {z \circ y} \circ \paren {z \circ x} = x \circ y$
Cardinal Product Distributes over Cardinal Sum
Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be cardinals. Then: : $\mathbf a \left({\mathbf b + \mathbf c}\right) = \mathbf a \mathbf b + \mathbf a \mathbf c$ where: : $\mathbf a + \mathbf b$ denotes the sum of $\mathbf a$ and $\mathbf b$. : $\mathbf a \mathbf b$ denotes the product of $\mathbf a$ and $\mathbf b$.
Ordering of Cardinals Compatible with Cardinal Product
Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be cardinals. Then: : $\mathbf a \le \mathbf b \implies \mathbf a \mathbf c \le \mathbf b \mathbf c$ where $\mathbf a \mathbf c$ denotes the product of $\mathbf a$ and $\mathbf c$.
Ordering of Cardinals Compatible with Cardinal Sum
Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be cardinals. Then: :$\mathbf a \le \mathbf b \implies \mathbf a + \mathbf c \le \mathbf b + \mathbf c$ where $\mathbf a \mathbf c$ denotes the sum of $\mathbf a$ and $\mathbf c$.
Group is B-Algebra Iff All Elements Self-Inverse
Let $\left({G, \circ}\right)$ be a group whose identity element is $e$. Then $\left({G, \circ}\right)$ is also a $B$-algebra : :$\forall g \in G: g = g^{-1}$ That is,  all elements of $G$ are self-inverse.
Condition for Existence of Cardinal Sum
Let $\mathbf a$ and $\mathbf b$ be cardinals. Then: : $\mathbf a \le \mathbf b \iff \exists \mathbf c: \mathbf a + \mathbf c = \mathbf b$ where $\mathbf c$ is also a cardinal.
Cardinal One is Cancellable for Cardinal Sum
Let $\mathbf a$ and $\mathbf b$ be cardinals. Then: :$\mathbf a + \mathbf 1 = \mathbf b + \mathbf 1 \implies \mathbf a = \mathbf b$ where $\mathbf 1$ is (cardinal) one.
Mapping from Finite Set to Itself is Injection iff Surjection
Let $S$ be a finite set. Let $f: S \to S$ be a mapping. Then $f$ is injective  $f$ is surjective.
Finite Cardinals form Infinite Set
The finite cardinals form a set which is infinite.
Increasing Function has Countable Discontinuities
Let $f: \R \to \R$ be an increasing real function. Then $f$ has at most countably many discontinuities.
Integral of Increasing Function Composed with Measurable Function
Let $\left({X, \Sigma, \mu}\right)$ be a $\sigma$-finite measure space. Let $f: X \to \R_{\ge 0}$ be a positive measurable function. Let $\phi: \R_{\ge 0} \to \R_{\ge 0}$ be a continuously differentiable, increasing function such that $\phi \left({0}\right) = 0$. Then: :$\displaystyle \int \phi \circ f \rd \mu = \int_0^\infty \phi' \left({t}\right) F \left({t}\right) \rd t$ where: : $F$ is the survival function of $f$ : $\displaystyle \int_0^\infty$ denotes an improper integral.
Minkowski's Inequality for Double Integrals
Let $\struct {X, \Sigma, \mu}$ and $\struct {Y, \Sigma', \nu}$ be $\sigma$-finite measure spaces. Let $\struct {X \times Y, \Sigma \otimes \Sigma', \mu \times \nu}$ be their product measure space. Let $f: X \times Y \to \overline \R$ be a $\Sigma \otimes \Sigma'$-measurable function. Then, for all $p \in \R$ with $p \ge 1$: :$\displaystyle \paren {\int_X \paren {\int_Y \size {\map f {x, y} } \map {\rd \nu} y}^p \map {\rd \mu} x}^{1/p} \le \int_Y \paren {\int_X \size {\map f {x, y} }^p \map {\rd \mu} x}^{1/p} \map {\rd \nu} y$
Integral with respect to Pushforward Measure
Let $\struct {X, \Sigma, \mu}$ be a measure space. Let $\struct {X', \Sigma'}$ be a measurable space. Let $T: X \to X'$ be a $\Sigma \, / \, \Sigma'$-measurable mapping. Let $f: X' \to \overline \R$ be a $\map T \mu$-integrable function, where $\map T \mu$ denotes the pushforward measure of $\mu$ under $T$. Then $f \circ T: X \to \overline \R$ is $\mu$-integrable, and: :$\displaystyle \int_{X'} f \rd \map T \mu = \int_X f \circ T \rd \mu$
Integrable Function under Pushforward Measure
Let $\struct {X, \Sigma, \mu}$ be a measure space. Let $\struct {X', \Sigma'}$ be a measurable space. Let $T: X \to X'$ be a $\Sigma \, / \, \Sigma'$-measurable mapping. Let $f: X' \to \overline \R$ be a mapping. Then the following are equivalent: :$(1): \quad f$ is $\map T \mu$-integrable :$(2): \quad f \circ T$ is $\mu$-integrable where $\map T \mu$ is the pushforward measure of $\mu$ under $T$.
0 in B-Algebra is Left Cancellable Element
Let $\left({X, \circ}\right)$ be a $B$-Algebra. Then: :$\forall x, y \in X: 0 \circ x = 0 \circ y \implies x = y$
B-Algebra Identity: 0(0x)=x
Let $\struct {X, \circ}$ be a $B$-algebra.  Then: :$\forall x \in X: 0 \circ \paren {0 \circ x} = x$
B-Algebra Induces Group
Let $\struct {X, \circ}$ be a $B$-algebra. Let $*$ be the binary operation on $X$ defined as: :$\forall a, b \in X: a * b := a \circ \paren {0 \circ b}$ Then the algebraic structure $\struct {X, *}$ is a group such that: :$\forall x \in X: 0 \circ x$ is the inverse element of $x$ under $*$. That is: :$\forall a, b \in X: a * b^{-1} := a \circ b$
Convolution of Measurable Functions is Bilinear
Let $\BB^n$ be the Borel $\sigma$-algebra on $\R^n$. Let $f, f', g, g': \R^n \to \R$ be $\BB^n$-measurable functions. Then for all $\lambda \in \R$: :$\paren {\lambda f + f'} * g = \lambda \paren {f * g} + f' * g$ :$f * \paren {\lambda g + g'} = \lambda \paren {f * g} + f * g'$ provided the convolutions in these expressions exist. That is, convolution $*$ is a bilinear operation.
Convolution of Measurable Function and Measure is Bilinear
Let $\mu$ and $\nu$ be measures on the Borel $\sigma$-algebra $\mathcal B^n$ on $\R^n$. Let $f, f': \R^n \to \R$ be $\mathcal B^n$-measurable functions. Then for all $\lambda \in \R$: :$\left({\lambda f + f'}\right) * \mu = \lambda \left({f * \mu}\right) + f' * \mu$ :$f * \left({\lambda \mu + \nu}\right) = \lambda \left({f * \mu}\right) + f * \nu$ provided the convolutions in these expressions exist. That is, convolution $*$ is a bilinear operation.
Convolution of Measures is Bilinear
Let $\mu, \mu', \nu$ and $\nu'$ be measures on the Borel $\sigma$-algebra $\mathcal B^n$ on $\R^n$. Then for all $\lambda \in \R$: :$\left({\lambda \mu + \mu'}\right) * \nu = \lambda \left({\mu * \nu}\right) + \mu' * \nu$ :$\mu * \left({\lambda \nu + \nu'}\right) = \lambda \left({\mu * \nu}\right) + \mu * \nu'$ where $*$ denotes convolution. That is, convolution is a bilinear operation.
Convolution of Measures as Pushforward Measure
Let $\mu$ and $\nu$ be measures on the Borel $\sigma$-algebra $\BB^n$ on $\R^n$. Let $\alpha: \R^n \times \R^n \to \R^n, \ \map \alpha {\mathbf x, \mathbf y} = \mathbf x + \mathbf y$ be vector addition on $\R^n$. Then we have the following equality of measures on $\BB^n$: :$\mu * \nu = \map {\alpha_*} {\mu \times \nu}$ where $\mu * \nu$ is the convolution of $\mu$ and $\nu$, and $\map {\alpha_*} {\mu \times \nu}$ is the pushforward of the product measure $\mu \times \nu$ under $\alpha$.
Continuous Functions with Compact Support Dense in Lebesgue P-Space
Let $C_c \left({\R^n}\right)$ be the space of continuous functions with compact support on $\R^n$. Let $p \in \R$, $p \ge 1$, and let $\mathcal L^p \left({\lambda^n}\right)$ be Lebesgue $p$-space for Lebesgue measure $\lambda^n$. Then $C_c \left({\R^n}\right)$ is everywhere dense in $\mathcal L^p \left({\lambda^n}\right)$ with respect to the $p$-seminorm $\left\Vert{\, \cdot \,}\right\Vert_p$.
Convolution of Integrable Function with Bounded Function
Let $f: \R^n \to \R$ be a Lebesgue integrable function. Let $g: \R^n \to \R$ be an essentially bounded function under Lebesgue measure $\lambda^n$. Then the convolution $f * g$ of $f$ and $g$ is bounded and continuous. In particular, $f * g$ is again essentially bounded.
F-Sigma Sets Closed under Union
Let $T = \struct {S, \tau}$ be a topological space. Let $F, F'$ be $F_\sigma$ sets of $T$. Then their union $F \cup F'$ is also an $F_\sigma$ set of $T$.
F-Sigma Sets Closed under Intersection
Let $T = \left({S, \tau}\right)$ be a topological space. Let $F, F'$ be $F_\sigma$ sets of $T$. Then their intersection $F \cap F'$ is also a $F_\sigma$ set of $T$.
G-Delta Sets Closed under Union
Let $T = \left({S, \tau}\right)$ be a topological space. Let $G, G'$ be $G_\delta$ sets of $T$. Then their union $G \cup G'$ is also a $G_\delta$ set of $T$.
G-Delta Sets Closed under Intersection
Let $T = \struct {S, \tau}$ be a topological space. Let $G, G'$ be $G_\delta$ sets of $T$. Then their intersection $G \cap G'$ is also a $G_\delta$ set of $T$.
Limit Point is Limit of Convergent Sequence
Let $M = \left({X, d}\right)$ be a metric space. Let $E \subseteq X$ be a subset of $X$. Let $p$ be a limit point of $E$. Then there exists a sequence $\left\langle{x_n}\right\rangle \subseteq E$ which converges to a limit: :$\displaystyle \lim_{n \mathop \to \infty} x_n = p$ where $\displaystyle \lim_{n \mathop \to \infty} x_n$ is the limit of the sequence $\left\langle{x_n}\right\rangle$.
F-Sigma Sets form Lattice
Let $T = \left({S, \tau}\right)$ be a topological space. Let $\mathcal F$ be the collection of all $F_\sigma$ sets of $T$. Then $\left({\mathcal F, \subseteq}\right)$ is a lattice, where $\subseteq$ denotes the subset relation.
G-Delta Sets form Lattice
Let $T = \left({S, \tau}\right)$ be a topological space. Let $\mathcal G$ be the collection of all $G_\delta$ sets of $T$. Then $\left({\mathcal G, \subseteq}\right)$ is a lattice, where $\subseteq$ denotes the subset relation.
Bounded Sequence in Euclidean Space has Convergent Subsequence
Let $\left\langle{x_i}\right\rangle_{i \in \N}$ be a bounded sequence in the Euclidean space $\R^n$. Then some subsequence of $\left\langle{x_i}\right\rangle_{i \in \N}$ converges to a limit.
Exponential Growth Equation/Special Case
All solutions of the differential equation $y' = y$ take the form $y = C e^x$.
Existence-Uniqueness Theorem for First-Order Differential Equation
Let $P$ and $Q$ be continuous functions on some open interval $I \subseteq \R$. Let $a \in I$. Let $b \in \R$. There is a unique function $f(x)=y$ on $I$ that satisfies the differential equation  :$y' + P(x)y = Q(x)$ along with the initial condition  :$f(a)=b$ This function is :$\displaystyle f(x) = be^{-A(x)} + e^{-A(x)}\int_a^x Q(t) e^{A(t)} dt$ where $\displaystyle A(x) = \int_a^x P(t)dt$
Proper Well-Ordering determines Smallest Elements
Let $S$ be a class. Let $\preceq$ be a proper well-ordering on $S$. Let $B$ be a nonempty subclass of $S$. Then $B$ has a $\preceq$-smallest element.
Subset of Ordinals has Minimal Element
Let $A$ be an ordinal (we shall allow $A$ to be a proper class). Let $B$ be a nonempty subset of $A$. Then $B$ has an $\Epsilon$-minimal element. {{explain|$\Epsilon$}} That is: : $\exists x \in B: B \cap x = \varnothing$
Union of Subset of Ordinals is Ordinal
{{explain|If $A$ is a proper class, then its union may be the class of all ordinals, which we do not consider an ordinal.}} Let $A$ be a class of ordinals. That is, $A \subseteq \On$, where $\On$ denotes the ordinal class. Then $\bigcup A$ is an ordinal.
Union of Ordinals is Least Upper Bound
Let $A \subset \operatorname{On}$. That is, let $A$ be a class of ordinals (every member of $A$ is an ordinal). Then $\bigcup A$, the union of $A$, is the least upper bound of $A$: :$\displaystyle \forall x \in A: x \le A$ :$\displaystyle \forall y \in A: y \le x \implies \bigcup A \le x$
Ordinal is Less than Successor
Let $x$ be an ordinal. Let $x^+$ denote the successor of $x$. Then: :$x \in x^+$ :$x \subset x^+$
No Natural Number between Number and Successor
Let $\N$ be the natural numbers. Let $n \in \N$. Then no natural number $m$ exists strictly between $n$ and its successor: :$\neg \exists m \in \N: \paren {n < m < n^+}$ That is: :If $m \le n \le m^+$, then $m = n$ or $m = n^+$.
No Largest Ordinal
Let $a$ be a set of ordinals. Then: :$\forall x \in a: x \prec \left({\bigcup a}\right)^+$
Minimal Infinite Successor Set is Ordinal
Let $\omega$ denote the minimal infinite successor set. Then $\omega$ is an ordinal.
No Infinitely Descending Membership Chains
Let $\omega$ denote the minimal infinite successor set. Let $F$ be a function whose domain is $\omega$. Then: :$\exists n \in \omega: \map F {n^+} \notin \map F n$
Transfinite Recursion/Uniqueness of Transfinite Recursion
Let $f$ be a mapping with a domain $y$ where $y$ is an ordinal. Let $f$ satisfy the condition that: : $\forall x \in y: f \left({x}\right) = G \left({f \restriction x}\right)$ where $f \restriction x$ denotes the restriction of $f$ to $x$. {{explain|What is $G$?}} Let $g$ be a mapping with a domain $z$ where $z$ is an ordinal. Let $g$ satisfy the condition that: : $\forall x \in z: g \left({x}\right) = G \left({g \restriction x}\right)$ Let $y \subseteq z$. Then: : $\forall x \in y: f \left({x}\right) = g \left({x}\right)$
Transfinite Induction/Schema 1
Let $P \left({x}\right)$ be a property  Suppose that: : If $P \left({x}\right)$ holds for all ordinals $x$ less than $y$, then $P \left({y}\right)$ also holds. Then $P \left({x}\right)$ holds for all ordinals $x$.
Transfinite Induction/Principle 2
Let $A$ be a class satisfying the following conditions: * $\varnothing \in A$ * $\forall x \in A: x^+ \in A$ * If $y$ is a limit ordinal, then $\left({ \forall x < y: x \in A }\right) \implies y \in A$ where $x^+$ denotes the successor of $x$. Then $\operatorname{On} \subseteq A$.
Transfinite Induction/Schema 2
Let $\phi \left({x}\right)$ be a property satisfying the following conditions: :$(1): \quad \phi \left({\varnothing}\right)$ is true :$(2): \quad$ If $x$ is an ordinal, then $\phi \left({x}\right) \implies \phi \left({x^+}\right)$ :$(3): \quad$ If $y$ is a limit ordinal, then $\left({\forall x < y: \phi \left({x}\right)}\right) \implies \phi \left({y}\right)$ where $x^+$ denotes the successor of $x$. Then, $\phi \left({x}\right)$ is true for all ordinals $x$.
Transfinite Recursion/Theorem 2
Let $\Dom x$ denote the domain of $x$. Let $\Img x$ denote the image of the mapping $x$. {{explain|We infer that $x$ is a mapping, but what is its context?}} Let $G$ be a class of ordered pairs $\tuple {x, y}$ satisfying at least one of the following conditions: :$(1): \quad x = \O$ and $y = a$ {{explain|What is $a$?}} :$(2): \quad \exists \beta: \Dom x = \beta^+$ and $y = \map H {\map x {\bigcup \Dom x} }$ {{explain|What is $H$?}} :$(3): \quad \Dom x$ is a limit ordinal and $y = \bigcup \Rng x$. {{explain|Is this invoking well-founded recursion?}} Let $\map F \alpha = \map G {F \restriction \alpha}$ for all ordinals $\alpha$. Then: :$F$ is a mapping and the domain of $F$ is the ordinals, $\On$. :$\map F \O = a$ :$\map F {\beta^+} = \map H {\map F \beta}$ :For limit ordinals $\beta$, $\displaystyle \map F \beta = \bigcup_{\gamma \mathop \in \beta} \map F \gamma$ :$F$ is unique. ::That is, if there is another function $A$ satisfying the above three properties, then $A = F$.
Transfinite Recursion/Corollary
Let $x$ be an ordinal. Let $G$ be a mapping There exists a unique mapping $f$ that satisfies the following properties: : The domain of $f$ is $x$ : $\forall y \in x: f \left({y}\right) = G \left({f \restriction y}\right)$
Inverse of Product/Monoid/General Result
Let $\left({S, \circ}\right)$ be a monoid whose identity is $e$. Let $a_1, a_2, \ldots, a_n \in S$ be invertible for $\circ$, with inverses $a_1^{-1}, a_2^{-1}, \ldots, a_n^{-1}$. Then $a_1 \circ a_2 \circ \cdots \circ a_n$ is invertible for $\circ$, and: :$\forall n \in \N_{> 0}: \left({a_1 \circ a_2 \circ \cdots \circ a_n}\right)^{-1} = a_n^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1}$
Inverse of Group Product
Let $\struct {G, \circ}$ be a group whose identity is $e$. Let $a, b \in G$, with inverses $a^{-1}, b^{-1}$. Then: :$\paren {a \circ b}^{-1} = b^{-1} \circ a^{-1}$
Inverse of Group Product/General Result
Let $\struct {G, \circ}$ be a group whose identity is $e$. Let $a_1, a_2, \ldots, a_n \in G$, with inverses $a_1^{-1}, a_2^{-1}, \ldots, a_n^{-1}$. Then: : $\paren {a_1 \circ a_2 \circ \cdots \circ a_n}^{-1} = a_n^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1}$
Well-Ordered Transitive Subset is Equal or Equal to Initial Segment
Let $\left({\prec, A}\right)$ be a well-ordered set. For every $x \in A$, let every $\prec$-initial segment $A_x$ be a set. Let $B$ be a subclass of $A$ such that :$\forall x \in A: \forall y \in B: \left({x \prec y \implies x \in B}\right)$. That is, $B$ must be $\prec$-transitive. Then: :$A = B$ or: :$\exists x \in A: B = A_x$
Inverse of Inverse/General Algebraic Structure
Let $\struct {S, \circ}$ be an algebraic structure with an identity element $e$. Let $x \in S$ be invertible, and let $y$ be an inverse of $x$. Then $x$ is also an inverse of $y$.
Inverse of Group Inverse
Let $\struct {G, \circ}$ be a group. Let $g \in G$, with inverse $g^{-1}$. Then: :$\paren {g^{-1} }^{-1} = g$
Condition for Injective Mapping on Ordinals
Let $F$ be a mapping satisfying the following properties: :$(1): \quad$ The domain of $F$ is $\On$, the ordinal class :$(2): \quad$ For all ordinals $x$, $\map F x = \map G {F \restriction x}$ :$(3): \quad$ For all ordinals $x$, $\map G {F \restriction x} \in \paren {A \setminus \Img x}$ where $\Img x$ is the image of $x$ under $F$. Let $\Img F$ denote the image of $F$. Then the following properties hold: :$(1): \quad \Img F \subseteq A$ :$(2): \quad F$ is injective :$(3): \quad A$ is a proper class. Note that only the third property of $F$ is the most important. For any function $G$, a function $F$ can be constructed satisfying the first two using transfinite recursion. {{explain|Exactly which of the above statements is the "third" is unclear. This statement needs to be made precise. It also needs to be explained. Also, it needs to be put in a separate section as it is peripheral to the statement of the theorem.}}
Maximal Injective Mapping from Ordinals to a Set
Let $F$ be a mapping satisfying the following properties: {{explain|What is $G$?}} : The domain of $F$ is $\operatorname{On}$, the ordinal class : For all ordinals $x$, $F \left({x}\right) = G \left({F \restriction x}\right)$. : For all ordinals $x$, if $(A \setminus \operatorname{Im} \left({x}\right) ) \ne \varnothing$, then $G  \left({F \restriction x}\right) \in (A \setminus \operatorname{Im} \left({x}\right))$ where $\operatorname{Im} \left({x}\right)$ is the image of the subset $x$ under $F$. : $A$ is a set. Then there exists an ordinal $y$ satisfying the following properties: : $\forall x \in y: \left({A \setminus \operatorname{Im} \left({x}\right)}\right) \ne \varnothing$ : $\operatorname {Im} \left({y}\right) = A$ : $F \restriction y$ is an injective mapping. Note that the first third and fourth properties of $F$ are the most important.  For any mapping $G$, a mapping $F$ can be constructed satisfying the first two properties using transfinite recursion.
Power of Element/Semigroup
Let $\struct {S, \oplus}$ be a magma. Let $a \in S$. Let $n \in \N_{>0}$. Let $\tuple {a_1, a_2, \ldots, a_n}$ be the ordered $n$-tuple defined by $a_k = a$ for each $k \in \N_n$. Then: :$\displaystyle \bigoplus_{k \mathop = 1}^n a_k = \oplus^n a$ where: :$\displaystyle \bigoplus_{k \mathop = 1}^n a_k$ is the composite of $\tuple {a_1, a_2, \ldots, a_n}$ for $\oplus$ :$\oplus^n a$ is the $n$th power of $a$ under $\oplus$.