hcju/proofwikips huggingface dataset:
Order Isomorphism between Ordinals and Proper Class
Let $\left({A, \prec}\right)$ be a strict well-ordering. Let $A$ be a proper class. Let the initial segment of $x$ be a set for every $x \in A$. Then we may make the following definitions: Set $G$ equal to the collection of ordered pairs $\left({x, y}\right)$ such that: :$y \in \left({A \setminus \operatorname{Im} \left({x}\right)}\right)$ :$\left({A \setminus \operatorname{Im}\left({x}\right)}\right) \cap A_y = \varnothing$ Use transfinite recursion to construct a mapping $F$ such that: : The domain of $F$ is $\operatorname{On}$ : For all ordinals $x$, $F \left({x}\right) = G \left({F {\restriction_x} }\right)$ Then $F: \operatorname{On} \to A$ is an order isomorphism between $\left({\operatorname{On}, \in}\right)$ and $\left({A, \prec}\right)$.
Strict Well-Ordering Isomorphic to Unique Ordinal under Unique Mapping
Let $S$ be a set. Let $\left({S, \prec}\right)$ be a strict well-ordering. Then there exists a unique ordinal $x$ and unique mapping $f$ such that $f: x \to S$ is an order isomorphism.
Unique Isomorphism between Ordinal Subset and Unique Ordinal
Let $\operatorname{On}$ be the class of ordinals. Let $S \subset \operatorname{On}$ where $S$ is a set. Then there exists a unique mapping $\phi$ and a unique ordinal $x$ such that $\phi : x \to S$ is an order isomorphism.
Lexicographic Order forms Well-Ordering on Ordered Pairs of Ordinals
The Lexicographic Order $\operatorname{Le}$ is a strict well-ordering on $\left({\operatorname{On} \times \operatorname{On}}\right)$.
Canonical Order Well-Orders Ordered Pairs of Ordinals
The canonical order, $R_0$ strictly well-orders the ordered pairs of ordinal numbers.
Initial Segment of Canonical Order is Set
Let $R_0$ denote the canonical ordering of $\left({\operatorname{On} \times \operatorname{On} }\right)$. Then, for all $\left({x, y}\right) \in \left({\operatorname{On} \times \operatorname{On} }\right)$, the $R_0$-initial segment is a set. {{explain|The $R_0$-initial segment of what?}}
Ordinal Addition is Closed
Let $\operatorname{On}$ be the class of all ordinals. Then: :$\forall x, y \in \operatorname{On}: \left({x + y}\right) \in \operatorname{On}$ That is: the sum $x+y$ is an ordinal.
Union of Ordinals is Ordinal
Let $y$ be a set. Let $F \left({z}\right)$ be a mapping such that: :$\displaystyle \forall z \in y: F \left({z}\right) \in \operatorname{On}$ Then: :$\displaystyle \bigcup_{z \mathop \in y} F \left({z}\right) \in \operatorname{On}$
Ordinal Addition by Zero
Let $x$ be an ordinal. Let $\varnothing$ be the zero ordinal. Then: :$x + \varnothing = x = \varnothing + x$ where $+$ denotes ordinal addition.
Union of Limit Ordinal
Let $x$ be a limit ordinal. Then: :$\displaystyle x = \bigcup x$ :$\displaystyle x = \bigcup_{y \mathop \in x} y$
Substitutivity of Equality
Let $x$ and $y$ be sets. Let $\map P x$ be a well-formed formula of the language of set theory. Let $\map P y$ be the same proposition $\map P x$ with some (not necessarily all) free instances of $x$ replaced with free instances of $y$. Let $=$ denote set equality. :$x = y \implies \paren {\map P x \iff \map P y}$
Membership is Left Compatible with Ordinal Addition
Let $x$, $y$, and $z$ be ordinals. Let $<$ denote membership $\in$, since $\in$ is a strict well-ordering on the ordinals. Then: :$x < y \implies \paren {z + x} < \paren {z + y}$
Ordinal Addition is Left Cancellable
Let $x, y, z$ be ordinals. Then: :$\left({z + x}\right) = \left({z + y}\right) \implies x = y$ That is, ordinal addition is left cancellable.
Supremum Inequality for Ordinals
Let $A \subseteq \operatorname{On}$ and $B \subseteq \operatorname{On}$ be ordinals. Then: :$\displaystyle \forall x \in A: \exists y \in B: x \le y \implies \bigcup A \le \bigcup B$
Subset is Right Compatible with Ordinal Addition
Let $x, y, z$ be ordinals. Then: :$x \le y \implies \left({x + z}\right) \le \left({y + z}\right)$
Ordinal Subtraction when Possible is Unique
Let $x$ and $y$ be ordinals such that $x \le y$. Then there exists a unique ordinal $z$ such that $\paren {x + z} = y$. That is: :$x \le y \implies \exists! z \in \On: \paren {x + z} = y$
Equality of Successors
Let $x$ and $y$ be ordinals. Let $x^+$ denote the successor set of $x$. Then, $x = y \iff x^+ = y^+$
Finite Ordinal Plus Transfinite Ordinal
Let $n$ be a finite ordinal. Let $x$ be a transfinite ordinal. Then: : $n + x = x$
Class is Transitive iff Union is Subset
A class $A$ is transitive : :$\displaystyle \bigcup A \subseteq A$
Union of Successor Ordinal
Let $x$ be an ordinal. Let $x^+$ denote the successor of $x$. Then: :$\displaystyle \map \bigcup {x^+} = x$
Union Distributes over Union
Set union is distributive over itself: :$\forall A, B, C: \paren {A \cup B} \cup \paren {A \cup C} = A \cup B \cup C = \paren {A \cup C} \cup \paren {B \cup C}$ where $A, B, C$ are sets.
Subset of Empty Set
Let $A$ be a class. Then: : $A$ is a subset of the empty set $\O$ ; : $A$ is equal to the empty set: :$\displaystyle A \subseteq \O \iff A = \O$
Ordinal is Subset of Successor
Let $x$ and $y$ be ordinals. Let $x^+$ denote the successor of $x$. Then: : $x \subseteq y^+ \iff \left({x \subseteq y \lor x = y^+}\right)$
Indexed Union Equality
Let $A$, $B_x$, and $C_x$ be classes. Then: :$\displaystyle \forall x \in A: B_x = C_x \implies \bigcup_{x \mathop \in A} B_x = \bigcup_{x \mathop \in A} C_x$ {{MissingLinks|$\bigcup_{x \mathop \in A} B_x$ and $\bigcup_{x \mathop \in A} C_x$}}
Indexed Union Subset
Let $A$, $B_x$ and $C_x$ be classes. {{explain|It can be inferred from the context, but the meaning of the subscript on $B_x$ and $C_x$ needs to be explained.}} Then: :$\displaystyle \forall x \in A: B_x \subseteq C_x \implies \bigcup_{x \mathop \in A} B_x \subseteq \bigcup_{x \mathop \in A} C_x$ {{MissingLinks|$\bigcup_{x \mathop \in A} B_x$ and $\bigcup_{x \mathop \in A} C_x$}}
Ordinal is Less than Sum
Let $x$ and $y$ be ordinals. Then: :$x \le \left({x + y}\right)$ :$x \le \left({y + x}\right)$
Limit Ordinals Preserved Under Ordinal Addition
Let $x$ and $y$ be ordinals such that $x$ is a limit ordinal. Then $\left({y + x}\right)$ is a limit ordinal. That is, letting $K_{II}$ denote the class of all limit ordinals: :$\forall x \in K_{II}: \left({y + x}\right) \in K_{II}$
Successor in Limit Ordinal
Suppose that $x$ is a limit ordinal and that $y \in x$. Then $y^+ \in x$ where $y^+$ denotes the successor set of $y$: :$\forall y \in x: y^+ \in x$
Successor is Less than Successor
Let $x$ and $y$ be ordinals and let $x^+$ denote the successor set of $x$. Then, $x \in y \iff x^+ \in y^+$.
Ordinal Addition is Associative
Ordinal addition is associative, i.e.: :$\left({x + y}\right) + z = x + \left({y + z}\right)$ holds for all ordinals $x$, $y$ and $z$.
Derivative of Composite Function/Corollary
:$\dfrac {\d y} {\d x} = \dfrac {\paren {\dfrac {\d y} {\d u} } } {\paren {\dfrac {\d x} {\d u} } }$ for $\dfrac {\d x} {\d u} \ne 0$.
Product of Subset with Intersection/Corollary
Let $\struct {G, \circ}$ be a group. Let $X, Y, Z \subseteq G$ such that $X$ is a singleton. Then: :$X \circ \paren {Y \cap Z} = \paren {X \circ Y} \cap \paren {X \circ Z}$ :$\paren {Y \cap Z} \circ X = \paren {Y \circ X} \cap \paren {Z \circ X}$ where $X \circ Y$ denotes the subset product of $X$ and $Y$.
Single Instruction URM Programs/Zero Function
The zero function $\Zero: \N \to \N$, defined as: :$\forall n \in \N: \map \Zero n = 0$
Single Instruction URM Programs/Successor Function
The successor function $\Succ: \N \to \N$, defined as: :$\forall n \in \N: \map \Succ n = n + 1$
Single Instruction URM Programs/Projection Function
The projection functions $\pr_j^k: \N^k \to \N$, defined as: :$\forall j \in \closedint 1 k: \forall \tuple {n_1, n_2, \ldots, n_k} \in \N^k: \map {\pr_j^k} {n_1, n_2, \ldots, n_k} = n_j$
Antiassociative Operation has no Idempotent Elements
Let $\struct {S, \circ}$ be an algebraic structure. Let $\circ$ be antiassociative on $S$. Then no element of $S$ is idempotent under $ \circ$. That is: :$\forall x \in S: x \circ x \ne x$
Antiassociative Operation is not Commutative
Let $\left({S, \circ}\right)$ be an algebraic structure. Let $\circ$ be antiassociative on $S$. Then $\circ$ is not commutative on $S$.
Antiassociative Structure of Finite Order
Let $n \in \N$ such that $n > 2$. Then there exists an algebraic structure $\left({S, \circ}\right)$ of order $n$ such that $\circ$ is antiassociative on $S$.
Example:Antiassociative Structure
Let $\left({\R_{>0}, \circ}\right)$ be an algebraic structure where $\R_{>0}$ denotes the strictly positive real numbers. Define: :$\forall x, y \in \R_{>0}: x \circ y = x y + y$ Then $\circ$ is antiassociative on $\R_{>0}$: :$\forall x, y, z \in \R_{>0}: \left({x \circ y}\right) \circ z \ne x \circ \left({y \circ z}\right)$
Reciprocal Sequence is Strictly Decreasing
The reciprocal sequence: :$\sequence {\operatorname {recip} }: \N_{>0} \to \R$: $n \mapsto \dfrac 1 n$ is strictly decreasing.
Two-Step Vector Subspace Test
Let $V$ be a vector space over a division ring $K$. Let $U \subseteq V$ be a non-empty subset of $V$ such that: :$(1): \qquad \forall u \in U, \lambda \in K: \lambda u \in U$ :$(2): \qquad \forall u, v \in U: u + v \in U$ Then $U$ is a subspace of $V$.
Reciprocal Function is Strictly Decreasing
The reciprocal function: :$\operatorname{recip}: \R \setminus \set 0 \to \R$, $x \mapsto \dfrac 1 x$ is strictly decreasing: :on the open interval $\openint 0 \to$ :on the open interval $\openint \gets 0$
Coefficients of Polynomial Product
Let $J$ be a set. Let $p_1, \ldots p_n$ be polynomial forms in the indeterminates $\set {X_j : j \in J}$ over a commutative ring $R$. Suppose that for each $i$ with $1 \le i \le n$, we have, for appropriate $a_{i, k} \in R$: :$p_i = \displaystyle \sum_{k \mathop \in Z} a_{i, k} X^k$ where $Z$ comprises the multiindices of natural numbers over $J$. Then: :$\displaystyle \prod_{i \mathop = 1}^n p_i = \displaystyle \sum_{k \mathop \in Z} b_k X^k$ where: :$\displaystyle b_k := \sum_{k_1 + \cdots + k_n \mathop = k} \paren {\prod_{i \mathop = 1}^n a_{i, k_i} }$
Subset Product of Subgroups/Necessary Condition
Let $\left({G, \circ}\right)$ be a group. Let $H, K$ be subgroups of $G$. Let $H \circ K$ be a subgroup of $G$. Then $H$ and $K$ are permutable. That is: :$H \circ K = K \circ H$ where $H \circ K$ denotes subset product.
Subset Product of Subgroups/Sufficient Condition
Let $\struct {G, \circ}$ be a group. Let $H, K$ be subgroups of $G$. Let $H$ and $K$ be permutable subgroups of $G$. That is, suppose: :$H \circ K = K \circ H$ where $H \circ K$ denotes subset product. Then $H \circ K$ is a subgroup of $G$.
Characteristics of Eulerian Graph/Necessary Condition
Let $G$ be a finite (undirected) graph. Let $G$ be Eulerian. Then $G$ is connected and each vertex of $G$ is even. Note that the definition of graph here includes: * Simple graph * Loop-graph * Multigraph * Loop-multigraph but does not include directed graph.
Characteristics of Eulerian Graph/Sufficient Condition/Proof 1
Let $G$ be a finite (undirected) graph which is connected Let each vertex of $G$ be even. Then $G$ is an Eulerian graph. Note that the definition of graph here includes: * Simple graph * Loop-graph * Multigraph * Loop-multigraph but does not include directed graph.
Characteristics of Eulerian Graph/Sufficient Condition/Proof 2
Let $G$ be a finite (undirected) graph which is connected. Let each vertex of $G$ be even. Then $G$ is an Eulerian graph. Note that the definition of graph here includes: * Simple graph * Loop-graph * Multigraph * Loop-multigraph but does not include directed graph.
Characteristics of Eulerian Graph/Sufficient Condition
Let $G$ be a finite (undirected) graph which is connected Let each vertex of $G$ be even. Then $G$ is an Eulerian graph. Note that the definition of graph here includes: * Simple graph * Loop-graph * Multigraph * Loop-multigraph but does not include directed graph.
Characterization of Pre-Measures
Let $X$ be a set, and let $\mathcal S \subseteq \mathcal P \left({X}\right)$ be a collection of subsets of $X$. Let $\varnothing \in \mathcal S$. Denote $\overline \R_{\ge 0}$ for the set of positive extended real numbers. A mapping $\mu: \mathcal S \to \overline \R_{\ge 0}$ is a pre-measure : :$(1):\quad \mu \left({\varnothing}\right) = 0$ :$(2):\quad \mu$ is finitely additive :$(3):\quad$ For every increasing sequence $\left({E_n}\right)_{n \in \N}$ in $\mathcal S$, if $E_n \uparrow E$ for some $E \in \mathcal S$, then: ::$\mu \left({E}\right) = \displaystyle \lim_{n \to \infty} \mu \left({E_n}\right)$ where $E_n \uparrow E$ denotes limit of increasing sequence of sets. Alternatively, and equivalently, $(3)$ may be replaced by either of: :$(3'):\quad$ For every decreasing sequence $\left({E_n}\right)_{n \in \N}$ in $\mathcal S$ for which $\mu \left({E_1}\right)$ is finite, if $E_n \downarrow E$ for some $E \in \mathcal S$, then: ::$\mu \left({E}\right) = \displaystyle \lim_{n \to \infty} \mu \left({E_n}\right)$ :$(3''):\quad$ For every decreasing sequence $\left({E_n}\right)_{n \in \N}$ in $\mathcal S$ for which $\mu \left({E_1}\right)$ is finite, if $E_n \downarrow \varnothing$, then: ::$\displaystyle \lim_{n \to \infty} \mu \left({E_n}\right) = 0$ where $E_n \downarrow E$ denotes limit of decreasing sequence of sets.
Power Set is Magma of Sets
Let $X$ be a set. Let $\family {\phi_i}: i \in I$ be an indexed family of mappings. Then $\powerset X$, the power set of $X$, is a magma of sets for $\family {\phi_i}: i \in I$ on $X$.
Existence and Uniqueness of Magma of Sets Generated by Collection of Subsets
Let $X$ be a set, and let $\Phi := \left\{{\phi_i: i \in I}\right\}$ be a collection of partial mappings with codomain $\mathcal P \left({X}\right)$, the power set of $X$. Let $\mathcal G \subseteq \mathcal P \left({X}\right)$ be a collection of subsets of $X$. Then the magma of sets generated by $\mathcal G$ exists and is unique.
Intersection of Magmas of Sets is Magma of Sets
Let $X$ be a set. Let $\Phi := \set {\phi_i: i \in I}$ be a collection of partial mappings with codomain $\powerset X$, the power set of $X$. Let $\mathcal S_j$ be a magma of sets for $\Phi$, for each $j \in J$, for some index set $J$. Then: :$\mathcal S := \displaystyle \bigcap_{j \mathop \in J} \mathcal S_j$ is also a magma of sets for $\Phi$.
Sigma-Algebra as Magma of Sets
The concept of $\sigma$-algebra is an instance of a magma of sets.
Ordinal Multiplication is Closed
Let $x$ and $y$ be ordinals. Let $\operatorname{On}$ denote the ordinal class. :$\left({x \cdot y}\right) \in \operatorname{On}$
Natural Number Multiplication is Closed
Let $m$ and $n$ be natural numbers. Then: :$m \times n \in \N$ where $\times$ denotes natural number multiplication.
Ordinal Multiplication by Zero
Let $x$ be an ordinal. {{begin-eqn}} {{eqn | l = \left({x \cdot \varnothing}\right)       | r = \varnothing       | c =  }} {{eqn | l = \left({\varnothing \cdot x}\right)       | r = \varnothing       | c =  }} {{end-eqn}}
Ordinal Multiplication by One
Let $x$ be an ordinal. Let $1$ denote the successor of $\varnothing$. {{begin-eqn}} {{eqn|l = \left({x \cdot 1}\right)      |r = x      |c =  }} {{eqn|l = \left({1 \cdot x}\right)      |r = x      |c =  }} {{end-eqn}}
Ordinal Addition by One
Let $x$ be an ordinal. Let $x^+$ denote the successor of $x$. Let $1$ denote (ordinal) one, the successor of the zero ordinal $\O$. Then: :$x + 1 = x^+$ where $+$ denotes ordinal addition.
Closed Set Measurable in Borel Sigma-Algebra
Let $\left({S, \tau}\right)$ be a topological space, and let $\mathcal B \left({\tau}\right)$ be the associated Borel $\sigma$-algebra. Let $C$ be a closed set in $S$. Then $C$ is $\mathcal B \left({\tau}\right)$-measurable.
Euler Phi Function of Product with Prime/Corollary
:$d \divides n \implies \map \phi d \divides \map \phi n$ where $d \divides n$ denotes that $d$ is a divisor of $n$.
Commutative B-Algebra Induces Abelian Group
Let $\left({X, \circ }\right)$ be a commutative $B$-algebra. Let $*$ be the binary operation on $X$ defined as: :$\forall a, b \in X: a * b := a \circ \left({0 \circ b}\right)$ Then the algebraic structure $\left({X, *}\right)$ is an abelian group such that: :$\forall x \in X: 0 \circ x$ is the inverse element of $x$ under $*$. That is: :$\forall a, b \in X: a * b^{-1} := a \circ b$
Membership is Left Compatible with Ordinal Multiplication
Let $x$, $y$, and $z$ be ordinals. Then: :$\left({ x < y \land z > 0 }\right) \iff \left({ z \cdot x }\right) < \left({ z \cdot y }\right)$
Ordinal Multiplication is Left Cancellable
Let $x, y, z$ be ordinals. Let $z \ne 0$. Then: :$\displaystyle \left({z \cdot x}\right) = \left({z \cdot y}\right) \implies x = y$ That is, ordinal multiplication is left cancellable.
Subset is Right Compatible with Ordinal Multiplication
Let $x, y, z$ be ordinals. Then: :$x \le y \implies \left({x \cdot z}\right) \le \left({y \cdot z}\right)$
Ordinals have No Zero Divisors
Let $x$ and $y$ be ordinals. Then: :$\paren {x \cdot y} = 0 \iff \paren {x = 0 \lor y = 0}$
Absolute Value of Simple Function is Simple Function/Proof 2
Let $\left({X, \Sigma}\right)$ be a measurable space. Let $f: X \to \R$ be a simple function. Then $\left\vert{f}\right\vert: X \to \R$, the absolute value of $f$, is also a simple function.
Absolute Value of Simple Function is Simple Function/Proof 1
Let $\left({X, \Sigma}\right)$ be a measurable space. Let $f: X \to \R$ be a simple function. Then $\left\vert{f}\right\vert: X \to \R$, the absolute value of $f$, is also a simple function.
Convergence in Norm Implies Convergence in Measure
Let $\struct {X, \Sigma, \mu}$ be a measure space, and let $p \in \R, p \ge 1$. Let $\sequence {f_n}_{n \mathop \in \N}, f_n : X \to \R$ be a sequence of $p$-integrable functions. Also, let $f: X \to \R$ be a $p$-integrable function. Suppose that $f_n$ converges in norm to $f$ (in the $p$-norm). Then $f_n$ converges in measure to $f$ (in $\mu$). That is: :$\displaystyle \operatorname {\mathcal L^{\textit p}-\!\lim\,} \limits_{n \mathop \to \infty} f_n = f \implies \operatorname {\mu-\!\lim\,} \limits_{n \mathop \to \infty} f_n = f$
Pointwise Convergence Implies Convergence in Measure
Let $\struct {X, \Sigma, \mu}$ be a measure space. Let $\sequence {f_n}_{n \mathop \in \N}, f_n : X \to \R$ be a sequence of measurable functions. Also, let $f: X \to \R$ be a measurable function. Suppose that $f$ is the pointwise limit of the $f_n$ $\mu$-almost everywhere. Then $f_n$ converges in measure to $f$ (in $\mu$). That is: :$\displaystyle \lim_{n \mathop \to \infty} \map {f_n} x \stackrel {a.e.} {=} \map f x \implies \operatorname {\mu-\!\lim\,} \limits_{n \mathop \to \infty} f_n = f$
Convergence in Sigma-Finite Measure
Let $\left({X, \Sigma, \mu}\right)$ be a $\sigma$-finite measure space. Let $\left({f_n}\right)_{n \in \N}, f_n: X \to \R$ be a sequence of measurable functions. Also, let $f, g: X \to \R$ be measurable functions. Suppose that $f_n$ converges in measure to both $f$ and $g$ (in $\mu$). Then $f$ and $g$ are equal $\mu$-almost everywhere.
Vitali's Theorem
Let $\left({X, \Sigma, \mu}\right)$ be a $\sigma$-finite measure space, and let $p \in \R, p \ge 1$. Let $\left({f_n}\right)_{n \in \N}, f_n: X \to \R$ be a sequence of $p$-integrable functions. Also, let $f: X \to \R$ be a measurable function. Suppose that $\displaystyle \operatorname{\mu-\!\lim\,} \limits_{n \to \infty} f_n = f$, i.e. $f_n$ converges in measure to $f$. Then the following are equivalent: :$(1): \quad \displaystyle \lim_{n \to \infty} \left\Vert{f_n - f}\right\Vert_p = 0$, where $\left\Vert{\cdot}\right\Vert_p$ is the $p$-seminorm (i.e. $\displaystyle \operatorname{\mathcal L^{\textit p}-\!\lim\,} \limits_{n \to \infty} f_n = f$) :$(2): \left({\left\vert{f_n}\right\vert^p}\right)_{n \in \N}$ is a uniformly integrable collection :$(3): \displaystyle \lim_{n \to \infty} \int \left\vert{f_n}\right\vert^p \, \mathrm d \mu = \int \left\vert{f}\right\vert^p \, \mathrm d \mu$ If $\left({X, \Sigma, \mu}\right)$ is not $\sigma$-finite, $(1)$ and $(3)$ must be replaced by, respectively: :$(1'): \quad \left({f_n}\right)_{n \in \N}$ converges in $\mathcal L^p$ :$(3'): \quad \displaystyle \lim_{n \to \infty} \int \left\vert{f_n}\right\vert^p \, \mathrm d \mu$ exists in $\R$
Trivial Subgroup is Subgroup
Let $\struct {G, \circ}$ be a group whose identity is $e$. Then the trivial subgroup $\struct {\set e, \circ}$ is indeed a subgroup of $\struct {G, \circ}$.
Trivial Subgroup is Normal
Let $\struct {G, \circ}$ be a group whose identity is $e$. Then the trivial subgroup $\struct {\set e, \circ}$ of $G$ is a normal subgroup in $G$.
Group is Normal in Itself
Let $\struct {G, \circ}$ be a group. Then $\struct {G, \circ}$ is a normal subgroup of itself.
Relativisation is Standard Model
Let $P$ be a well-formed formula. Let $A$ be a finite set such that $x \in A$  $x$ is a free variable in $P$. Then: :$\displaystyle A \subseteq B \implies \left({B \models P \iff P^B}\right)$ {{explain|Definition of $P^B$}}
Induction on Well-Formed Formulas
Let $\mathcal F$ be a formal language with a bottom-up grammar. Let $\Phi$ be a proposition about the well-formed formulas of $\mathcal F$. Suppose that $\Phi$ is true for all letters of $\mathcal F$. Suppose further that every rule of formation preserves $\Phi$, i.e. when fed well-formed formulas satisfying $\Phi$, it yields new well-formed formulas satisfying $\Phi$. Then all well-formed formulas of $\mathcal F$ satisfy $\Phi$.
Substitution of Elements
Let $a$, $b$, and $x$ be sets. :$a = b \implies \left({a \in x \iff b \in x}\right)$
Limit Ordinals Preserved Under Ordinal Multiplication
Let $x$ and $y$ be ordinals. Let $x$ be non-empty. Let $y$ be a limit ordinal. It follows that the ordinal product $\left({x \times y}\right)$ is a limit ordinal.
Ordinal is Less than Ordinal times Limit
Let $y$ be a limit ordinal. Let $x$ and $z$ be ordinals. Then: :$z < \left({x \times y}\right) \iff \exists w < y: z < \left({x \times w}\right)$
Epsilon Induction
Let $A$ be a class. Let $\Bbb U$ denote the universe. :$\left({\forall x: \left({x \subseteq A \implies x \in A}\right)}\right) \implies A = \Bbb U$
Class Equality is Reflexive
Let $A$ be a class. Then: :$A = A$ where $=$ denotes class equality.
Class Equality is Transitive
Let $A$, $B$, and $C$ be classes. Let $=$ denote class equality. Then :$\left({ A = B \land B = C }\right) \implies B = A$
Substitutivity of Class Equality
Let $A$ and $B$ be classes. Let $P \left({A}\right)$ be a well-formed formula of the language of set theory. Let $P \left({B}\right)$ be the same proposition $P \left({A}\right)$ with all instances of $A$ replaced with instances of $B$. Let $=$ denote class equality. :$A = B \implies \left({ P \left({A}\right) \iff P \left({B}\right) }\right)$
Class Equal to All its Elements
Let $A$ be a class. Then: :$A = \left\{{x : x \in A }\right\}$
Class Member of Class Builder
Let $A$ be a class. Let $x$ be a set. Let $P \left({x}\right)$ be a well-formed formula in the language of set theory. Let $P \left({A}\right)$ denote the formula $P\left({x}\right)$ with all free instances of $x$ replaced with instances of $A$. Let $\left\{{x: P \left({x}\right)}\right\}$ be a class specified using class builder notation. Then: :$A \in \left\{{x : P \left({x}\right)}\right\} \iff \left({\exists x: x = A \land P \left({A}\right)}\right)$
Class Equality Extension of Set Equality
Let $=_1$ denote set equality. Let $=_2$ denote class equality. Let $x$ and $y$ be sets. Then $x =_1 y$ iff $x =_2 y$.
B-Algebra Induced by Group Induced by B-Algebra
Let $\left({S, *}\right)$ be a $B$-algebra. Let $\left({S, \circ}\right)$ be the group described on $B$-Algebra Induces Group. Let $\left({S, *'}\right)$ be the $B$-algebra described on Group Induces $B$-Algebra. Then $\left({S, *'}\right) = \left({S, *}\right)$.
Union of Doubleton
Let $\set {x, y}$ be a doubleton. Then: :$\displaystyle \bigcup \set {x, y} = x \cup y$
Union of Small Classes is Small
Let $x$ and $y$ be small classes. Then $x \cup y$ is also small.
Set Difference is Set
Let $x$ be a small class. Let $A$ be a class. Let $\map \MM B$ denote that $B$ is small. Then:  :$\map \MM {x \setminus A}$
Nonempty Class has Members
Let $A$ be a class. Then: :$A \ne \O \iff \exists x: x \in A$
Class is Not Element of Itself
Let $A$ be a class. Then $A$ is not an element of itself: :$A \notin A$