hcju/proofwikips huggingface dataset:
Equivalence of Formulations of Peano's Axioms
Let $P$ be a set. Let $s: P \to P$ be a mapping. Let $0 \in P$ be a distinguished element. {{TFAE|def = Peano's Axioms|view = Peano's Axioms}}
Product of Row Sum Unity Matrices
Let $\mathbf A = \left[{a}\right]_{m n}$ be an $m \times n$ matrix. Let $\mathbf B = \left[{b}\right]_{n p}$ be an $n \times p$ matrix. Let the row sum of $\mathbf A$ and $\mathbf B$ be equal to $1$. Then the row sum of their (conventional) product is also $1$.
Restriction of Injection is Injection
Let $f: S \to T$ be an injection. Let $X \subseteq S$ be a subset of $S$. Let $f \sqbrk X$ denote the image of $X$ under $f$. Let $Y \subseteq T$ be a subset of $T$ such that $f \sqbrk X \subseteq Y$. The restriction $f \restriction_{X \times Y}$ of $f$ to $X \times Y$ is an injection from $X$ to $Y$.
Absolute Value is Many-to-One
Let $f: \R \to \R$ be the absolute value function: :$\forall x \in \R:  \map f x = \begin{cases} x & : x \ge 0 \\ -x & : x < 0 \end{cases}$ Then $f$ is a many-to-one relation.
Laws of Logarithms
Let $x, y, b \in \R_{>0}$ be (strictly) positive real numbers. Let $a \in \R$ be any real number such that $a > 0$ and $a \ne 1$. Let $\log_a$ denote the logarithm to base $a$. Then:
Subset of Preimage under Relation is Preimage of Subset
Let $\mathcal R \subseteq S \times T$ be a relation. Let $X \subseteq S, Y \subseteq T$. Then: :$X \subseteq \mathcal R^{-1} \sqbrk Y \iff \mathcal R \sqbrk X \subseteq Y$ In the language of direct image mappings, this can be written: :$X \subseteq \map {\mathcal R^\gets} Y \iff \map {\mathcal R^\to} X \subseteq Y$
Composition of Direct Image Mappings of Relations
Let $A, B, C$ be non-empty sets. Let $\mathcal R_1 \subseteq A \times B, \mathcal R_2 \subseteq B \times C$ be relations. Let: :${\mathcal R_1}^\to: \powerset A \to \powerset B$ and :${\mathcal R_2}^\to: \powerset B \to \powerset C$ be the direct image mappings of $\mathcal R_1$ and $\mathcal R_2$. Then: :$\paren {\mathcal R_2 \circ \mathcal R_1}^\to = {\mathcal R_2}^\to \circ {\mathcal R_1}^\to$
Subset equals Preimage of Image iff Mapping is Injection
Let $f: S \to T$ be a mapping. Let $f^{-1}$ denote the inverse of $f$. Then: :$\forall A \subseteq S: A = \paren {f^{-1} \circ f} \sqbrk A$  $f$ is an injection where: :$f \sqbrk A$ denotes the image of $A$ under $f$ :$f^{-1} \circ f$ denotes the composition of $f^{-1}$ and $f$. This can be expressed in the language and notation of direct image mappings and inverse image mappings as: :$\forall A \in \powerset S: A = \map {\paren {f^\gets \circ f^\to} } A$  $f$ is an injection
Subset equals Image of Preimage iff Mapping is Surjection
Let $f: S \to T$ be a mapping. Let $f^\to: \powerset S \to \powerset T$ be the direct image mapping of $f$. Similarly, let $f^\gets: \powerset T \to \powerset S$ be the inverse image mapping of $f$. Then: :$\forall B \in \powerset T: B = \map {\paren {f^\to \circ f^\gets} } B$  $f$ is a surjection.
Inverse of Injection is Many-to-One Relation
Let $f: S \to T$ be an injection. Let $f^{-1}: T \to S$ be the inverse relation of $f$. Then $f^{-1}$ is many-to-one.
Bijection iff Left and Right Cancellable
Let $f$ be a mapping. Then $f$ is a bijection  $f$ is both left cancellable and right cancellable.
Relation Induced by Mapping is Equivalence Relation
Let $f: S \to T$ be a mapping. Let $\RR_f \subseteq S \times S$ be the relation induced by $f$: :$\tuple {s_1, s_2} \in \RR_f \iff \map f {s_1} = \map f {s_2}$ Then $\RR_f$ is an equivalence relation.
Canonical Injection is Monomorphism
Let $\struct {S_1, \circ_1}$ and $\struct {S_2, \circ_2}$ be algebraic structures with identities $e_1, e_2$ respectively. The canonical injections: :$\inj_1: \struct {S_1, \circ_1} \to \struct {S_1, \circ_1} \times \struct {S_2, \circ_2}: \forall x \in S_1: \map {\inj_1} x = \tuple {x, e_2}$ :$\inj_2: \struct {S_2, \circ_2} \to \struct {S_1, \circ_1} \times \struct {S_2, \circ_2}: \forall x \in S_2: \map {\inj_2} x = \tuple {e_1, x}$ are monomorphisms.
Order Embedding is Injection
Let $\left({S, \preceq_1}\right)$ and $\left({T, \preceq_2}\right)$ be ordered sets. Let $\phi: S \to T$ be an order embedding. That is: : $\forall x, y \in S: x \preceq_1 y \iff \phi \left({x}\right) \preceq_2 \phi \left({y}\right)$ Then $\phi$ is an injection.
Order Isomorphism is Surjective Order Embedding
Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be ordered sets. Let $f: S \to T$ be a mapping. Then $f$ is an order isomorphism : :$(1): \quad f$ is a surjection :$(2): \quad \forall x, y \in S: x \preceq_1 y \iff \map f x \preceq_2 \map f y$ That is,  $f$ is an order embedding which is also a surjection.
Infimum of Power Set
Let $S$ be a set. Let $\mathcal P \left({S}\right)$ be the power set of $S$. Let $\left({\mathcal P \left({S}\right), \subseteq}\right)$ be the relational structure defined on $\mathcal P \left({S}\right)$ by the relation $\subseteq$. (From Subset Relation on Power Set is Partial Ordering, this is an ordered set.) Then the infimum of $\left({\mathcal P \left({S}\right), \subseteq}\right)$ is the empty set $\varnothing$.
Supremum of Power Set
Let $S$ be a set. Let $\mathcal P \left({S}\right)$ be the power set of $S$. Let $\left({\mathcal P \left({S}\right), \subseteq}\right)$ be the relational structure defined on $\mathcal P \left({S}\right)$ by the relation $\subseteq$. (From Subset Relation on Power Set is Partial Ordering, this is an ordered set.) Then the supremum of $\left({\mathcal P \left({S}\right), \subseteq}\right)$ is the set $S$.
Diagonal Relation is Ordering and Equivalence
Let $\left({S, \Delta_S}\right)$ be a relational structure where $\Delta_S$ is the diagonal relation, defined as: :$\forall x, y \in S: \paren {x, y} \in \Delta_S \iff x = y$ Then $\Delta_S$ is the '''only''' relation on $S$ which is both an equivalence and an ordering.
Cantor-Bernstein-Schröder Theorem/Proof 3
Let $S$ and $T$ be sets, such that: :$\exists f: S \to T$ such that $f$ is an injection :$\exists g: T \to S$ such that $g$ is an injection. Then there exists a bijection from $S$ to $T$.
Cantor-Bernstein-Schröder Theorem/Proof 2
Let $S$ and $T$ be sets, such that: :$S \preccurlyeq T$, that is: $T$ dominates $S$ :$T \preccurlyeq S$, that is: $S$ dominates $T$. Then: :$S \sim T$ that is, $S$ is equivalent to $T$.
Huygens-Steiner Theorem
Let $B$ be a body of mass $M$. Let $I_0$ be the moment of inertia of $B$ about some axis $A$ through the centre of mass of $B$. Let $I$ the moment of inertia of $B$ about another axis $A'$ parallel to $A$. Then $I_0$ and $I$ are related by: :$I = I_0 + M l^2$ where $l$ is the perpendicular distance between $A$ and $A'$.
Equivalence of Definitions of Well-Ordering/Definition 1 implies Definition 2
Let $\left({S, \preceq}\right)$ be an ordered set such that: : $\forall T \subseteq S: \exists a \in T: \forall x \in T: a \preceq x$ That is, such that $\preceq$ is a well-ordering by definition 1. Then $\preceq$ is a total ordering.
Choice Function Exists for Well-Orderable Union of Sets
Let $\mathbb S$ be a set of sets such that: :$\forall S \in \mathbb S: S \ne \varnothing$ that is, none of the sets in $\mathbb S$ may be empty. Let the union $\bigcup \mathbb S$ be well-orderable. Then there exists a choice function $f: \mathbb S \to \bigcup \mathbb S$ defined as: :$\forall S \in \mathbb S: \exists x \in S: f \left({S}\right) = x$ Thus, if every member of $\mathbb S$ is a well-ordered, then we can create a choice function $f$ defined as: :$\forall S \in \mathbb S: f \left({S}\right) = \inf \left({S}\right)$ True, we may be making infinitely many choices, but we have a rule for doing so.
Principle of Finite Choice
Let $I$ be a non-empty finite indexing set. Let $\left\langle{S_i}\right\rangle_{i \mathop \in I}$ be an $I$-indexed family of non-empty sets. Then there exists an $I$-indexed family $\left\langle{x_i}\right\rangle_{i \mathop \in I}$ such that: :$\forall i \in I: x_i \in S_i$ That is, there exists a mapping: :$\displaystyle f: I \to \bigcup_{i \mathop \in I} S_i$ such that: :$\forall i \in I: f \left({i}\right) \in S_i$
Choice Function Exists for Set of Well-Ordered Sets
Let $\mathbb S$ be a set of sets such that: :$\forall S \in \mathbb S: S \ne \varnothing$ that is, none of the sets in $\mathbb S$ may be empty. Let every element of $\mathbb S$ be well-ordered. Then there exists a choice function $f: \mathbb S \to \bigcup \mathbb S$ satisfying: :$\forall S \in \mathbb S: \exists x \in S: f \left({S}\right) = x$
Identity of Power Set with Union
Let $S$ be a set and let $\powerset S$ be its power set. Consider the algebraic structure $\struct {\powerset S, \cup}$, where $\cup$ denotes set union. Then the empty set $\O$ serves as the identity for $\struct {\powerset S, \cup}$.
Identity of Power Set with Intersection
Let $S$ be a set and let $\powerset S$ be its power set. Consider the algebraic structure $\struct {\powerset S, \cap}$, where $\cap$ denotes set intersection. Then $S$ serves as the identity for $\struct {\powerset S, \cap}$.
Commutation with Inverse in Monoid
Let $x, y \in S$ such that $y$ is invertible. Then $x$ commutes with $y$  $x$ commutes with $y^{-1}$.
Commutation of Inverses in Monoid
Let $x, y \in S$ such that $x$ and $y$ are both invertible. Then $x$ commutes with $y$  $x^{-1}$ commutes with $y^{-1}$.
Inverse of Commuting Pair
Let $x, y \in S$ such that $x$ and $y$ are both invertible. Then $x$ commutes with $y$ : : $\struct {x \circ y}^{-1} = x^{-1} \circ y^{-1}$
Conjugate of Commuting Elements
Let $x, y \in S$ such that $x$ and $y$ are both invertible. Then $x \circ y \circ x^{-1} = y$  $x$ and $y$ commute.
Product of Commuting Elements with Inverses
Let $x, y \in S$ such that $x$ and $y$ are both invertible. Then: :$x \circ y \circ x^{-1} \circ y^{-1} = e_S = x^{-1} \circ y^{-1} \circ x \circ y$  $x$ and $y$ commute.
Exists Element Not in Set
Let $S$ be a set. Then $\exists x: x \notin S$. That is, for any set, there exists some element which is not in that set.
Rule of Material Implication
==== Formulation 1 ==== {{:Rule of Material Implication/Formulation 1}} ==== Formulation 2 ==== {{:Rule of Material Implication/Formulation 2}}
De Morgan's Laws (Set Theory)/Proof by Induction
Let $\mathbb T = \left\{{T_i: i \mathop \in I}\right\}$, where each $T_i$ is a set and $I$ is some finite indexing set. Then:
Union equals Intersection iff Sets are Equal
Let $S$ and $T$ be sets. Then: :$\paren {S \cup T = S} \land \paren {S \cap T = S} \iff S = T$ where: :$S \cup T$ denotes set union :$S \cap T$ denotes set intersection.
Intersection with Set Difference is Set Difference with Intersection
: $\left({R \setminus S}\right) \cap T = \left({R \cap T}\right) \setminus S$
Set Difference is Right Distributive over Set Intersection
:$\paren {R \cap S} \setminus T = \paren {R \setminus T} \cap \paren {S \setminus T}$
Set Intersection Distributes over Set Difference
: $\left({R \setminus S}\right) \cap T = \left({R \cap T}\right) \setminus \left({S \cap T}\right)$ : $R \cap \left({S \setminus T}\right) = \left({R \cap S}\right) \setminus \left({R \cap T}\right)$
Set Difference with Intersection is Difference
:$S \setminus \paren {S \cap T} = S \setminus T$
Set Difference of Intersection with Set is Empty Set
:$\left({S \cap T}\right) \setminus S = \varnothing$ :$\left({S \cap T}\right) \setminus T = \varnothing$
Set Difference Intersection with First Set is Set Difference
:$\left({S \setminus T}\right) \cap S = S \setminus T$
Set Difference Intersection with Second Set is Empty Set
:$\paren {S \setminus T} \cap T = \O$
Set Difference with Union is Set Difference
:$\left({S \cup T}\right) \setminus T = S \setminus T$
Set Difference Union First Set is First Set
:$\paren {S \setminus T} \cup S = S$
Set Difference Union Second Set is Union
:$\left({S \setminus T}\right) \cup T = S \cup T$
Set Difference with Union
:$R \setminus \paren {S \cup T} = \paren {R \cup T} \setminus \paren {S \cup T} = \paren {R \setminus S} \setminus T = \paren {R \setminus T} \setminus S$
Set Difference is Right Distributive over Union
:$\paren {R \cup S} \setminus T = \paren {R \setminus T} \cup \paren {S \setminus T}$
Set Difference with Set Difference is Union of Set Difference with Intersection
:$R \setminus \paren {S \setminus T} = \paren {R \setminus S} \cup \paren {R \cap T}$
Set Difference is Subset of Union of Differences
: $R \setminus S \subseteq \left({R \setminus T}\right) \cup \left({T \setminus S}\right)$
Ordered Sum of Tosets is Totally Ordered Set
Let $\left({S, \preceq_1}\right)$ and $\left({T, \preceq_2}\right)$ be tosets. Let $S + T = \left({S \cup T, \preceq}\right)$ be the ordered sum of $S$ and $T$. Then $\left({S \cup T, \preceq}\right)$ is itself a toset.
Ordered Product of Tosets is Totally Ordered Set
Let $\struct {S_1, \preceq_1}$ and $\struct {S_2, \preceq_2}$ be tosets. Let $S_1 \cdot S_2 = \struct {S_1 \times S_2, \preceq}$ be the ordered product of $S_1$ and $S_2$. Then $\struct {S_1 \times S_2, \preceq}$ is itself a toset.
Restriction of Mapping is Mapping
Let $f: S \to T$ be a mapping. Let $X \subseteq S$. Let $f \restriction_X$ be the restriction of $f$ to $X$. Then $f \restriction_X: X \to T$ is a mapping: : whose domain is $X$ : whose preimage is $X$.
Hausdorff Maximal Principle
Let $\struct {\PP, \preceq}$ be a partially ordered set. Then there exists a maximal chain in $\PP$.
Logical Consequence with Union
Let $U$ be a set of propositional formulas. Let $P$ be a propositional formula. Let $U \models P$ denote that $P$ is a semantic consequence of $U$. Then: : $U \models P$ iff: : $U \cup P \models P$
Logical Consequence Union Negation
Let $U$ be a set of propositional formulas. Let $P$ be a propositional formula. Let $U \models P$ denote that $U$ is a semantic consequence $P$. Then: : $U \models P$ iff: : $U \cup \left\{{\neg P}\right\}$ has no models.
Union of Singleton
Consider the set of sets $\mathbb S$ such that $\mathbb S$ consists of just one set $S$. Then the union of $\mathbb S$ is $S$: :$\displaystyle \mathbb S = \set S \implies \bigcup \mathbb S = S$
Intersection of Singleton
Consider the set of sets $\mathbb S$ such that $\mathbb S$ consists of just one set $S$. Then the intersection of $\mathbb S$ is $S$: :$\displaystyle \mathbb S = \set S \implies \bigcap \mathbb S = S$
Union of Empty Set
Consider the set of sets $\mathbb S$ such that $\mathbb S$ is the empty set $\O$. Then the union of $\mathbb S$ is $\O$: :$\mathbb S = \O \implies \displaystyle \bigcup \mathbb S = \O$
Intersection of Empty Set
Consider the set of sets $\mathbb S$ such that $\mathbb S$ is the empty set $\O$. Then the intersection of $\mathbb S$ is $\mathbb U$: :$\mathbb S = \O \implies \displaystyle \bigcap \mathbb S = \mathbb U$ where $\mathbb U$ is the universe. A paradoxical result.
Inverse of Identity Mapping
Let $S$ be a set. Let $I_S: S \to S$ be the identity mapping on $S$. Then the inverse of $I_S$ is itself: :$\paren {I_S}^{-1} = I_S$
Factoring Mapping into Quotient and Injection
Let $f: S \to T$ be a mapping. Then $f$ can be uniquely '''factored into''' a quotient mapping, followed by an injection. Thus: :$f = h \circ q_{\RR_f}$ where: :$q_{\RR_f}: S \to S / \RR_f: \map {q_{\RR_f} } s = \eqclass s {\RR_f}$ :$h: S / \RR_f \to T: \map h {\eqclass s {\RR_f} } = \map f s$ :$\eqclass s {\RR_f}$ denotes the equivalence class of $s$ with respect to the equivalence relation $\RR$ induced on $S$ by $f$. This can be illustrated using a commutative diagram as follows: ::$\begin{xy}\xymatrix@L+2mu@+1em{  S \ar[r]^*{q_{\RR_f} }      \ar@{-->}[rd]_*{f = h \circ q_{\RR_f} } &  S / \RR_f \ar[d]^*{h} \\ &  T }\end{xy}$
Subsets in Increasing Union
Let $S_0, S_1, S_2, \ldots, S_i, \ldots$ be a nested sequence of sets, that is: :$S_0 \subseteq S_1 \subseteq S_2 \subseteq \ldots \subseteq S_i \subseteq \ldots$ Let $S$ be the increasing union of $S_0, S_1, S_2, \ldots, S_i, \ldots$: :$\displaystyle S = \bigcup_{i \mathop \in \N} S_i$ Then: :$\forall s \in S: \exists k \in \N: \forall j \ge k: s \in S_j$
Mapping on Increasing Union
Let $S_0, S_1, S_2, \ldots, S_i, \ldots$ be sets such that: :$S_0 \subseteq S_1 \subseteq S_2 \subseteq \ldots \subseteq S_i \subseteq \ldots$ that is, each set is contained in the next as a subset. Let $S$ be the increasing union of $S_0, S_1, S_2, \ldots, S_i, \ldots$: :$\displaystyle S = \bigcup_{i \mathop \in \N} S_i$ For each $i \in \N$, let $f_i: S_i \to T$ be a mapping such that: :$\forall j < i: f_i \restriction_{S_j} = f_j$ where $f_i \restriction_{S_j}$ denotes the restriction of $f_i$ to $S_j$. Then there is a '''unique''' mapping: :$f: S \to T$ which extends each $f_i$ to $S$.
Domain of Injection Not Larger than Codomain
Let $f: S \rightarrowtail T$ be an injection. Then: :$\left|{S}\right| \le \left|{T}\right|$ where $\left|{S}\right|$ denotes the cardinality of $S$.
Futurama Theorem
Let $A_{n - 2} \subset A_n$ be a subgroup of the alternating group on $n$ letters $A_n$. For any element $x \in A_{n - 2}$, let $x = x_1 x_2 \dots x_k$, where $x_i \in H$ is a transposition. {{explain|What is $H$?}} Then there exists $y$ which can be represented as a series of transpositions $y_1 y_2 \dots y_j \in A_n$ such that: :$(1): \quad y x = z$, where $z$ contains no transpositions from $H$ :$(2): \quad y_a \ne x_b$ for all $a, b$.
Pointwise Addition on Continuous Real-Valued Functions forms Group
Let $C$ be the set of all continuous real functions on the set of real numbers $\R$. Let $f, g \in C$. Let $f + g$ be the pointwise sum of $f$ and $g$: :$\forall x \in R: \map {\paren {f + g} } x = \map f x + \map g x$ Then $\struct {C, +}$, the algebraic structure on $C$ induced by $+$, forms a group.
General Intersection Property of Topological Space
Let $\struct {S, \tau}$ be a topological space. Let $S_1, S_2, \ldots, S_n$ be open sets of $\struct {S, \tau}$. Then: :$\displaystyle \bigcap_{i \mathop = 1}^n S_i$ is also an open set of $\struct {S, \tau}$. That is, the intersection of any finite number of open sets of a topology is also in $\tau$. Conversely, if the intersection of any finite number of open sets of a topology is also in $\tau$, then: :$(1): \quad$ The intersection of any two elements of $\tau$ is an element of $\tau$ :$(2): \quad S$ is itself an element of $\tau$.
Empty Set is Element of Topology
Let $\struct {X, \tau}$ be a topological space. Then $\O$ is an open set of $\struct {X, \tau}$.
Coarseness Relation on Topologies is Partial Ordering
Let $S$ be a set. Let $\mathbb T$ be the set of all topologies on $S$. Let $\le$ be the relation on $\mathbb T$ defined as: :$\forall \tau_1, \tau_2 \in \mathbb T: \tau_1 \le \tau_2$  $\tau_1$ is coarser than $\tau_2$. Then $\le$ is a partial ordering on $\mathbb T$.
Topology Defined by Closed Sets
Let $X$ be any set and let $\tau$ be a collection of subsets of $X$. Then $\tau$ is a topology on $X$ : :$(1): \quad$ Any intersection of arbitrarily many closed sets of $X$ under $\tau$ is a closed set of $X$ under $\tau$ :$(2): \quad$ The union of any finite number of closed sets of $X$ under $\tau$ is a closed set of $X$ under $\tau$ :$(3): \quad X$ and $\O$ are both closed sets of $X$ under $\tau$.
Complement of F-Sigma Set is G-Delta Set
Let $T = \struct {S, \tau}$ be a topological space. Let $X$ be an $F_\sigma$ set of $T$. Then its complement $S \setminus X$ is a $G_\delta$ set of $S$.
Set is Open iff Neighborhood of all its Points
Let $T = \struct {S, \tau}$ be a topological space. Let $V \subseteq S$ be a subset of $T$. Then: :$V$ is an open set of $T$ : :$V$ is a neighborhood of all the points in $V$.
Relationship between Limit Point Types
Let $T = \struct {X, \tau}$ be a topological space. Let $A \subseteq X$. Let: :$C$ be the set of condensation points of $A$ :$W$ be the set of $\omega$-accumulation points of $A$ :$L$ be the set of limit points of $A$ :$D$ be the set of adherent points of $A$. Then: :$C \subseteq W \subseteq L \subseteq D$ That is: :Every condensation point is an $\omega$-accumulation point :Every $\omega$-accumulation point is a limit point :Every limit point is an adherent point. In general, the inclusions do not hold in the other direction.
Limit Point of Sequence is Accumulation Point
Let $T = \struct {S, \tau}$ be a topological space. Let $A \subseteq S$. Let $\sequence {x_n}$ be a sequence in $A$. Let $\alpha$ be a limit point of $\sequence {x_n}$. Then $\alpha$ is also an accumulation point of $\sequence {x_n}$.
Filter on Set is Proper Filter
Let $S$ be a set. Let $\mathcal P \left({S}\right)$ denote the power set of $S$. Let $\left({\mathcal P \left({S}\right), \subseteq}\right)$ be the poset defined on $\mathcal P \left({S}\right)$ by the subset relation. Let $\mathcal F$ be a filter on $S$. Then $\mathcal F$ is a proper filter on $\left({\mathcal P \left({S}\right), \subseteq}\right)$.
Peirce's Law is Equivalent to Law of Excluded Middle
Peirce's Law: :$\paren {p \implies q} \implies p \vdash p$ is logically equivalent to the Law of Excluded Middle: :$\vdash p \lor \neg p$ That is, Peirce's Law holds  the Law of Excluded Middle holds.
Implication is Left Distributive over Conjunction
The implication operator is left distributive over the conjunction operator:
Existence of Negation Normal Form of Statement
Any propositional formula can be expressed in negation normal form (NNF).
Existence of Conjunctive Normal Form of Statement
Any propositional formula can be expressed in conjunctive normal form (CNF).
Existence of Disjunctive Normal Form of Statement
Any propositional formula can be expressed in disjunctive normal form (DNF).
Existence of Base-N Representation
{{refactor|lemmata and sub proofs}} {{tidy|Could be improved in presentational style}} Given a number $x \in \hointr 0 1$, there exists a representation of that number in a base-$p$ positional system. Specifically, there exists a sequence $\sequence {a_n}$ such that: :$0 \le a_n < p$, and :$\displaystyle \sum_{n \mathop = 1}^\infty \frac {a_n} {p^n}$ converges to $x$. Unless $\sequence {a_n}$ terminates (that is $a_n = 0$ for all sufficiently large $n$), then this representation is unique. If $\sequence {a_n}$ does terminate, then there is exactly one other sequence which satisfies the criteria of the theorem.
Symmetry Group of Equilateral Triangle is Group
The symmetry group of the equilateral triangle is a group.
Hilbert's Basis Theorem
Let $A$ be a Noetherian ring. Let $A \sqbrk x$ be the ring of polynomial forms over $A$ in the single indeterminate $x$. Then $A \sqbrk x$ is also a Noetherian ring.
Existence of Cyclic Group of Order n
Let $n \in \Z_{>0}$. Then there exists a cyclic group of order $n$ which is unique up to isomorphism.
Quotient Group is Group
Let $G$ be a group. Let $N$ be a normal subgroup of $G$. Then the quotient group $G / N$ is indeed a group.
Principal Ideal is Ideal
Let $\struct {R, +, \circ}$ be a ring with unity. Let $a \in R$. Let $\ideal a$ be the principal ideal of $R$ generated by $a$. Then $\ideal a$ is an ideal of $R$. Also, if $J$ is an ideal of $R$ and $a \in J$, then $\ideal a \subseteq J$. That is, $\ideal a$ is the smallest ideal of $R$ to which $a$ belongs. {{refactor|There are a number of results that this takes in, originally taken from Whitelaw: :$(1): \quad \forall a \in R: \ideal a$ is an ideal of $R$ :$(2): \quad \forall a \in R: a \in \ideal a$ :$(3): \quad \forall a \in R:$ if $J$ is an ideal of $R$, and $a \in J$, then $\ideal a \subseteq J$. :::That is, $\ideal a$ is the smallest ideal of $R$ containing $a$. }}
Bézout's Theorem
Let $X$ and $Y$ be two plane projective curves defined over a field $F$ that do not have a common component. (This condition is true if both $X$ and $Y$ are defined by different irreducible polynomials. In particular, it holds for a pair of "generic" curves.) {{refactor|Link to a proof of the above, and (preferably) move this statement to the proof itself.}} Then the total number of intersection points of $X$ and $Y$ with coordinates in an algebraically closed field $E$ which contains $F$, counted with their multiplicities, is equal to the product of the degrees of $X$ and $Y$.
Whitney Immersion Theorem
Let $m > 1$ be a natural number. Every smooth $m$-dimensional manifold can be immersed in Euclidean $\left({2m-1}\right)$-space.