hcju/proofwikips huggingface dataset:
Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive
==== Formulation 1 ==== {{:Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive/Formulation 1}} ==== Formulation 2 ==== {{:Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive/Formulation 2}}
Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive/Formulation 1
:$\paren {q \lor r} \land p \dashv \vdash \paren {q \land p} \lor \paren {r \land p}$
Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive/Formulation 2
:$\vdash \paren {\paren {q \lor r} \land p} \iff \paren {\paren {q \land p} \lor \paren {r \land p} }$
Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive
==== Formulation 1 ==== {{:Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 1}} ==== Formulation 2 ==== {{:Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 2}}
Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive
==== Formulation 1 ==== {{:Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive/Formulation 1}} ==== Formulation 2 ==== {{:Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive/Formulation 2}}
Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 2
:$\vdash \paren {p \lor \paren {q \land r} } \iff \paren {\paren {p \lor q} \land \paren {p \lor r} }$
Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive/Formulation 1
:$\paren {q \land r} \lor p \dashv \vdash \paren {q \lor p} \land \paren {r \lor p}$
Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive/Formulation 2
:$\vdash \paren {\paren {q \land r} \lor p} \iff \paren {\paren {q \lor p} \land \paren {r \lor p} }$
Union of Subsets is Subset/Set of Sets
Let $T$ be a set. Let $\mathbb S$ be a set of sets. Suppose that for each $S \in \mathbb S$, $S \subseteq T$. Then: :$\displaystyle \bigcup \mathbb S \subseteq T$
Union is Smallest Superset/Set of Sets
Let $T$ be a set. Let $\mathbb S$ be a set of sets. Then: :$\displaystyle \paren {\forall X \in \mathbb S: X \subseteq T} \iff \bigcup \mathbb S \subseteq T$
Intersection of Relations is Relation
Let $S$ and $T$ be sets. Let $\mathcal F$ be a family of relations from $S$ to $T$. Let $\displaystyle \mathcal R = \bigcap \mathcal F$, the intersection of all the elements of $\mathcal F$.   Then $\mathcal R$ is a relation from $S$ to $T$. {{Expand|Binary case}}
Intersection of Subsets is Subset/Set of Sets
Let $T$ be a set. Let $\mathbb S$ be a non-empty set of sets. Suppose that for each $S \in \mathbb S$: :$S \subseteq T$ Then: :$\bigcap \mathbb S \subseteq T$
Equivalence of Definitions of Initial Topology
Let $X$ be a set. Let $I$ be an indexing set. Let $\family {\struct{Y_i, \tau_i}}_{i \mathop \in I}$ be an indexed family of topological spaces indexed by $I$. Let $\family {f_i: X \to Y_i}_{i \mathop \in I}$ be an indexed family of mappings indexed by $I$. {{TFAE|def = Initial Topology}}
Radius of Convergence from Limit of Sequence/Complex Case
Let $\xi \in \C$ be a complex number. Let $\displaystyle S \paren z = \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n$ be a (complex) power series about $\xi$. Let the sequence $\sequence {\cmod {\dfrac {a_{n + 1} } {a_n} } }_{n \mathop \in \N}$ converges. Then $R$ is given by: :$\displaystyle \dfrac 1 R = \lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n + 1} } {a_n} }$ If: :$\displaystyle \lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n + 1} } {a_n} } = 0$ then the radius of convergence is infinite, and $S \paren z$ is absolutely convergent for all $z \in \C$.
Radius of Convergence from Limit of Sequence/Real Case
Let $\xi \in \R$ be a real number. Let $\displaystyle S \paren x = \sum_{n \mathop = 0}^\infty a_n \paren {x - \xi}^n$ be a power series about $\xi$. Then the radius of convergence $R$ of $S \paren x$ is given by: : $\displaystyle \frac 1 R = \lim_{n \mathop \to \infty} \size {\frac {a_{n + 1} } {a_n} }$ if this limit exists and is nonzero. If : $\displaystyle \lim_{n \mathop \to \infty} \size {\frac {a_{n + 1} } {a_n} } = 0$ then the radius of convergence is infinite and therefore the interval of convergence is $\R$.
Existence of Radius of Convergence of Complex Power Series
Let $\xi \in \C$. Let $\displaystyle S \paren z = \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n $ be a (complex) power series about $\xi$. Then there exists a radius of convergence $R \in \overline \R$ of $S \paren z$.
Conjunction Equivalent to Negation of Implication of Negative/Formulation 1
:$p \land q \dashv \vdash \neg \left({p \implies \neg q}\right)$
Conjunction Equivalent to Negation of Implication of Negative/Formulation 2
:$\vdash \paren {p \land q} \iff \paren {\neg \paren {p \implies \neg q} }$
Conjunction Equivalent to Negation of Implication of Negative/Formulation 1/Forward Implication
:$p \land q \vdash \neg \left({p \implies \neg q}\right)$
Conjunction Equivalent to Negation of Implication of Negative/Formulation 1/Reverse Implication
:$\neg \left({p \implies \neg q}\right) \vdash p \land q$
Rule of Material Implication/Formulation 1/Forward Implication
:$p \implies q \vdash \neg p \lor q$
Rule of Material Implication/Formulation 1
:$p \implies q \dashv \vdash \neg p \lor q$
Rule of Material Implication/Formulation 1/Reverse Implication
:$\neg p \lor q \vdash p \implies q$
Rule of Material Implication/Formulation 1/Proof
:$p \implies q \dashv \vdash \neg p \lor q$
Rule of Material Implication/Formulation 2
:$\vdash \paren {p \implies q} \iff \paren {\neg p \lor q}$
Modus Tollendo Ponens/Variant/Formulation 1
:$p \lor q \dashv \vdash \neg p \implies q$
Modus Tollendo Ponens/Sequent Form
==== Case 1 ==== {{:Modus Tollendo Ponens/Sequent Form/Case 1}} ==== Case 2 ==== {{:Modus Tollendo Ponens/Sequent Form/Case 2}}
Modus Tollendo Ponens/Variant/Formulation 1/Proof
:$p \lor q \dashv \vdash \neg p \implies q$
Radius of Convergence of Power Series over Factorial/Complex Case
Let $\xi \in \C$ be a complex number. Let $\displaystyle \map f z = \sum_{n \mathop = 0}^\infty \dfrac {\paren {z - \xi}^n} {n!}$. Then $\map f z$ converges absolutely for all $z \in \C$. That is, the radius of convergence of the power series $\displaystyle \sum_{n \mathop = 0}^\infty \frac {\paren {z - \xi}^n} {n!}$ is infinite.
Radius of Convergence of Power Series over Factorial/Real Case
Let $\xi \in \R$ be a real number. Let $\displaystyle f \left({x}\right) = \sum_{n \mathop = 0}^\infty \frac {\left({x - \xi}\right)^n} {n!}$. Then $f \left({x}\right)$ converges for all $x \in \R$. That is, the interval of convergence of the power series $\displaystyle \sum_{n \mathop = 0}^\infty \frac {\left({x - \xi}\right)^n} {n!}$ is $\R$.
Sum of Logarithms/Natural Logarithm
Let $x, y \in \R$ be strictly positive real numbers. Then: :$\ln x + \ln y = \map \ln {x y}$ where $\ln$ denotes the natural logarithm.
Sum of Logarithms/General Logarithm
Let $x, y, b \in \R$ be strictly positive real numbers such that $b > 1$. Then: :$\log_b x + \log_b y = \map {\log_b} {x y}$ where $\log_b$ denotes the logarithm to base $b$.
Rule of Simplification/Sequent Form/Formulation 1/Proof
:$(1): \quad p \land q \vdash p$ :$(2): \quad p \land q \vdash q$
Rule of Simplification/Sequent Form
The Rule of Simplification can be symbolised by the sequents:
Rule of Conjunction/Sequent Form
The Rule of Conjunction can be symbolised in sequent form as follows:
Radius of Convergence of Derivative of Complex Power Series
Let $\xi \in \C$. For all $z \in \C$, define the power series: : $\displaystyle S \paren z = \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n$ and: : $\displaystyle S' \paren z = \sum_{n \mathop = 1}^\infty n a_n \paren {z - \xi}^{n - 1}$ Let $R$ be the radius of convergence of $S \paren z$, and let $R'$ be the radius of convergence of $S' \paren z$. Then $R =R'$.
Rule of Addition/Sequent Form
The Rule of Addition can be symbolised by the sequents: :$(1): \quad p \vdash p \lor q$ :$(2): \quad q \vdash p \lor q$
Rule of Addition/Sequent Form/Proof 2
:$(1): \quad p \vdash p \lor q$ :$(2): \quad q \vdash p \lor q$
Rule of Addition/Sequent Form/Proof 1
The Rule of Addition can be symbolised by the sequents: :$(1): \quad p \vdash p \lor q$ :$(2): \quad q \vdash p \lor q$
Rule of Addition/Sequent Form/Form 2/Proof 1
:$q \vdash p \lor q$
Proof by Cases/Sequent Form
Proof by Cases can be symbolised by the sequent: :$p \lor q, \paren {p \vdash r}, \paren {q \vdash r} \vdash r$
Modus Ponendo Ponens/Sequent Form
:$p \implies q, p \vdash q$
Logarithm of Power/Natural Logarithm
Let $x \in \R$ be a strictly positive real number. Let $a \in \R$ be a real number such that $a > 1$. Let $r \in \R$ be any real number. Let $\ln x$ be the natural logarithm of $x$. Then: :$\map \ln {x^r} = r \ln x$
Logarithm of Power/General Logarithm
Let $x \in \R$ be a strictly positive real number. Let $a \in \R$ be a real number such that $a > 1$. Let $r \in \R$ be any real number. Let $\log_a x$ be the logarithm to the base $a$ of $x$. Then: :$\map {\log_a} {x^r} = r \log_a x$
Rule of Implication/Sequent Form
The Rule of Implication can be symbolised by the sequent: :$\left({p \vdash q}\right) \vdash p \implies q$
Rule of Implication/Sequent Form/Proof 2
:$\left({p \vdash q}\right) \vdash p \implies q$
Sum of Absolutely Convergent Series
Let $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ and $\displaystyle \sum_{n \mathop = 1}^\infty b_n$ be two real or complex series that are absolutely convergent. Then the series $\displaystyle \sum_{n \mathop = 1}^\infty \paren {a_n + b_n}$ is absolutely convergent, and: :$\displaystyle \sum_{n \mathop = 1}^\infty \paren {a_n + b_n} = \displaystyle \sum_{n \mathop = 1}^\infty a_n + \sum_{n \mathop = 1}^\infty b_n$
Rule of Implication/Sequent Form/Proof 1
:$\left({p \vdash q}\right) \vdash p \implies q$
Rule of Simplification/Sequent Form/Formulation 1
:$(1): \quad p \land q \vdash p$ :$(2): \quad p \land q \vdash q$
Rule of Simplification/Sequent Form/Formulation 2
:$(1): \quad \vdash p \land q \implies p$ :$(2): \quad \vdash p \land q \implies q$
Rule of Simplification/Sequent Form/Formulation 2/Proof 1/Form 1
:$\vdash p \land q \implies p$
Rule of Simplification/Sequent Form/Formulation 2/Proof 1/Form 2
:$\vdash p \land q \implies q$
Complement of Complement in Uniquely Complemented Lattice
Let $\left({S, \wedge, \vee, \preceq}\right)$ be a uniquely complemented lattice. For each $x \in S$, let $\neg x$ be the complement of $x$. Then for each $x \in S$: :$\neg \neg x = x$
Complement of Bottom/Boolean Algebra
Let $\left({S, \vee, \wedge, \neg}\right)$ be a Boolean algebra. Then $\neg \bot = \top$.
Complement of Bottom/Bounded Lattice
Let $\left({S, \vee, \wedge, \preceq}\right)$ be a bounded lattice. Then the bottom $\bot$ has a unique complement, namely $\top$, top.
Complement of Top/Boolean Algebra
Let $\struct {S, \vee, \wedge, \neg}$ be a Boolean algebra. Then $\neg \top = \bot$.
Duality Principle (Boolean Algebras)
Let $\struct {S, \vee, \wedge}$ be a Boolean algebra. Then any theorem in $\struct {S, \vee, \wedge}$ remains valid if both $\vee$ and $\wedge$ are interchanged, and also $\bot$ and $\top$ are interchanged throughout the whole theorem.
Identities of Boolean Algebra also Zeroes
Let $\struct {S, \vee, \wedge, \neg}$ be a Boolean algebra, defined as in Definition 1. Let the identity for $\vee$ be $\bot$ and the identity for $\wedge$ be $\top$. Then: :$(1): \quad \forall x \in S: x \vee \top = \top$ :$(2): \quad \forall x \in S: x \wedge \bot = \bot$ That is, $\bot$ is a zero element for $\wedge$, and $\top$ is a zero element for $\vee$.
Complement of Complement (Boolean Algebras)
Let $\left({S, \vee, \wedge, \neg}\right)$ be a Boolean algebra. Then for all $a \in S$: :$\neg (\neg a) = a$
De Morgan's Laws (Boolean Algebras)
:$\neg \paren {a \vee b} = \neg a \wedge \neg b$ :$\neg \paren {a \wedge b} = \neg a \vee \neg b$
Complement in Boolean Algebra is Unique
Let $\left({S, \vee, \wedge}\right)$ be a Boolean algebra. Then for all $a \in S$, there is a unique $b \in S$ such that: :$a \wedge b = \bot, a \vee b = \top$ i.e., a valid choice for $\neg a$ as in axiom $(BA \ 4)$ for Boolean algebras.
Cancellation of Join in Boolean Algebra
Let $\left({S, \vee, \wedge, \neg}\right)$ be a Boolean algebra. Let $a, b, c \in S$, and suppose that: {{begin-eqn}} {{eqn | l = a \vee c       | r = b \vee c }} {{eqn | l = a \vee \neg c       | r = b \vee \neg c }} {{end-eqn}} Then $a = b$.
Cancellation of Meet in Boolean Algebra
Let $\struct {S, \vee, \wedge, \neg}$ be a Boolean algebra. Let $a, b, c \in S$. Let: {{begin-eqn}} {{eqn | l = a \wedge c       | r = b \wedge c }} {{eqn | l = a \wedge \neg c       | r = b \wedge \neg c }} {{end-eqn}} Then: : $a = b$
Operations of Boolean Algebra are Associative
Let $\struct {S, \vee, \wedge, \neg}$ be a Boolean algebra, defined as in Definition 1. Then: :$\forall a, b, c \in S: \paren {a \wedge b} \wedge c = a \wedge \paren {b \wedge c}$ :$\forall a, b, c \in S: \paren {a \vee b} \vee c = a \vee \paren {b \vee c}$ That is, both $\vee$ and $\wedge$ are associative operations.
Absorption Laws (Boolean Algebras)
Let $\struct {S, \vee, \wedge}$ be a Boolean algebra, defined as in Definition 1. Then for all $a, b \in S$: :$a = a \vee \paren {a \wedge b}$ :$a = a \wedge \paren {a \vee b}$ That is, $\vee$ absorbs $\wedge$, and $\wedge$ absorbs $\vee$.
Difference of Absolutely Convergent Series
Let $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ and $\displaystyle \sum_{n \mathop = 1}^\infty b_n$ be two real or complex series that are absolutely convergent. Then the series $\displaystyle \sum_{n \mathop = 1}^\infty \paren {a_n - b_n}$ is absolutely convergent, and: :$\displaystyle \sum_{n \mathop = 1}^\infty \paren {a_n - b_n} = \sum_{n \mathop = 1}^\infty a_n - \sum_{n \mathop = 1}^\infty b_n$
True Statement is implied by Every Statement/Formulation 1
:$p \vdash q \implies p$
True Statement is implied by Every Statement/Formulation 2
:$\vdash q \implies \paren {p \implies q}$
True Statement is implied by Every Statement
==== Formulation 1 ==== {{:True Statement is implied by Every Statement/Formulation 1}} ==== Formulation 2 ==== {{:True Statement is implied by Every Statement/Formulation 2}}
False Statement implies Every Statement
==== Formulation 1 ==== {{:False Statement implies Every Statement/Formulation 1}} ==== Formulation 2 ==== {{:False Statement implies Every Statement/Formulation 2}}
False Statement implies Every Statement/Formulation 1
: $\neg p \vdash p \implies q$
False Statement implies Every Statement/Formulation 1/Proof 2
:$\neg p \vdash p \implies q$
False Statement implies Every Statement/Formulation 2
:$\vdash \neg p \implies \paren {p \implies q}$
Modus Tollendo Ponens/Sequent Form/Case 1
:$p \lor q, \neg p \vdash q$
Modus Tollendo Ponens/Sequent Form/Case 2
:$p \lor q, \neg q \vdash p$
Two is Boolean Algebra
Let $\mathbf 2$ denote two. Then $\mathbf 2$ is a Boolean algebra.
Two-Valued Functions form Boolean Algebra
Let $\mathbf 2$ be the Boolean algebra two, and let $X$ be a set. Let $\mathbf 2^X$ be the set of all mappings $p: X \to \mathbf 2$. Define the operations $\vee$, $\wedge$ and $\neg$ on $\mathbf 2^X$ in pointwise fashion thus: :$\vee: \mathbf 2^X \times \mathbf 2^X \to \mathbf 2^X, \left({p \vee q}\right) (x) := p (x) \vee q (x)$ :$\wedge: \mathbf 2^X \times \mathbf 2^X \to \mathbf 2^X, \left({p \wedge q}\right) (x) := p (x) \wedge q (x)$ :$\neg: \mathbf 2^X \to \mathbf 2^X, \left({\neg p}\right) (x) := \neg p (x)$ Furthermore, write $\bot$ and $\top$ for the constant mappings with these values, viz: :$\bot: X \to \mathbf 2, \bot (x) := \bot$ :$\top: X \to \mathbf 2, \top (x) := \top$ Then $\left({\mathbf 2^X, \vee, \wedge, \neg}\right)$ is a Boolean algebra, with $\bot$ and $\top$ as identities for $\vee$ and $\wedge$, respectively.
Power Set is Boolean Ring
Let $S$ be a set, and let $\mathcal P \left({S}\right)$ be its power set. Denote with $*$ and $\cap$ symmetric difference and intersection, respectively. Then $\left({S, *, \cap}\right)$ is a Boolean ring.
Power Set and Two-Valued Functions are Isomorphic Boolean Rings
Let $S$ be a set. Let $\mathbf 2$ be the Boolean ring two. Let $\mathcal P \left({S}\right)$ be the power set of $S$; by Power Set is Boolean Ring, it is a Boolean ring. Let $\mathbf 2^S$ be the set of all mappings $f: S \to \mathbf 2$; by Two-Valued Functions form Boolean Ring, it is also a Boolean ring. Let $\chi_{\left({\cdot}\right)}: \mathcal P \left({S}\right) \to \mathbf 2^S$ be the characteristic function operation. Then $\chi_{\left({\cdot}\right)}$ is a ring isomorphism.
Peirce's Law/Formulation 1
: $\left({p \implies q}\right) \implies p \vdash p$
Peirce's Law/Formulation 2/Proof 2
: $\vdash \left({\left({p \implies q}\right) \implies p}\right) \implies p$
Peirce's Law/Strong Form/Formulation 1
:$\paren {\paren {p \implies q} \implies p} \dashv \vdash p$
Conjunction with Negative Equivalent to Negation of Implication
==== Formulation 1 ==== {{:Conjunction with Negative Equivalent to Negation of Implication/Formulation 1}} ==== Formulation 2 ==== {{:Conjunction with Negative Equivalent to Negation of Implication/Formulation 2}}
Conjunction with Negative Equivalent to Negation of Implication/Formulation 1
:$p \land \neg q \dashv \vdash \neg \left({p \implies q}\right)$
Conjunction with Negative Equivalent to Negation of Implication/Formulation 1/Forward Implication
:$p \land \neg q \vdash \neg \paren {p \implies q}$
Conjunction with Negative Equivalent to Negation of Implication/Formulation 1/Reverse Implication
:$\neg \left({p \implies q}\right) \vdash p \land \neg q$
Conjunction with Negative Equivalent to Negation of Implication/Formulation 1/Proof
:$p \land \neg q \dashv \vdash \neg \left({p \implies q}\right)$
Conjunction with Negative Equivalent to Negation of Implication/Formulation 2
:$\vdash \paren {p \land \neg q} \iff \paren {\neg \paren {p \implies q} }$
Reflexive Circular Relation is Equivalence
Let $\mathcal R \subseteq S \times S$ be a reflexive and circular relation in $S$. Then $\mathcal R$ is an equivalence relation.
Inverse of Algebraic Structure Isomorphism is Isomorphism/General Result
Let $\phi: \left({S, \circ_1, \circ_2, \ldots, \circ_n}\right)$ and $\left({T, *_1, *_2, \ldots, *_n}\right)$ be algebraic structures. Let $\phi: \left({S, \circ_1, \circ_2, \ldots, \circ_n}\right) \to \left({T, *_1, *_2, \ldots, *_n}\right)$ be a mapping. Then: : $\phi: \left({S, \circ_1, \circ_2, \ldots, \circ_n}\right) \to \left({T, *_1, *_2, \ldots, *_n}\right)$ is an isomorphism : : $\phi^{-1}: \left({T, *_1, *_2, \ldots, *_n}\right) \to \left({S, \circ_1, \circ_2, \ldots, \circ_n}\right)$ is also an isomorphism.
Fundamental Theorem of Calculus for Complex Riemann Integrals
Let $\closedint a b$ be a closed real interval. Let $F, f: \closedint a b \to \C$ be complex functions. Suppose that $F$ is a primitive of $f$. Then the complex Riemann integral of $f$ satisfies: :$\displaystyle \int_a^b \map f t \rd t = \map F b - \map F a$
Fundamental Theorem of Calculus for Contour Integrals
Let $F, f: D \to \C$ be complex functions, where $D$ is a connected domain. Let $C$ be a contour that is a concatenation of the directed smooth curves $C_1, \ldots, C_n$. Let $C_k$ be parameterized by the smooth path $\gamma_k: \closedint {a_k} {b_k} \to D$ for all $k \in \set {1, \ldots, n}$. Suppose that $F$ is a primitive of $f$. If $C$ has start point $z$ and end point $w$, then: :$\displaystyle \int_C \map f z \rd z = \map F w - \map F z$ If $C$ is a closed contour, then: :$\displaystyle \oint_C \map f z \rd z = 0$
Power Series is Taylor Series
Let $\displaystyle f \left({z}\right) = \sum_{n \mathop = 0}^\infty a_n \left({z - \xi}\right)^n$ be a complex power series about $\xi \in \C$. Let $R$ be the radius of convergence of $f$. Then, $f$ is of differentiability class $C^\infty$. For all $n \in \N$: :$a_n = \dfrac{f^{\left({n}\right) } \left({\xi}\right) }{ n! }$ Hence, $f$ is equal to its Taylor series expansion about $\xi$: :$\displaystyle \forall z \in \C, \left\vert{z - \xi}\right\vert < R: \quad f \left({z}\right) = \sum_{n \mathop = 0}^\infty \dfrac{\left({z - \xi}\right)^n}{n!} f^{\left({n}\right) } \left({\xi}\right)$
Image of Complex Exponential Function
The image of the complex exponential function is $\C \setminus \left\{ {0}\right\}$.
Reciprocal of Complex Exponential
:$\dfrac 1 {\map \exp z} = \map \exp {-z}$
Power Function Strictly Preserves Ordering in Ordered Group
Let $n \in \N_{>0}$ be a strictly positive integer.
Period of Complex Exponential Function
:$\map \exp {z + 2 k \pi i} = \map \exp z$
Zero Staircase Integral Condition for Primitive
Let $f: D \to \C$ be a continuous complex function, where $D$ is a connected domain. Let $z_0 \in D$. Suppose that $\displaystyle \oint_C \map f z \rd z = 0$ for all closed staircase contours $C$ in $D$. Then $f$ has a primitive $F: D \to \C$ defined by: :$\displaystyle \map F w = \int_{C_w} \map f z \rd z$ where $C_w$ is any staircase contour in $D$ with start point $z_0$ and end point $w$.
Derivative of Complex Polynomial
Let $a_n \in \C$ for $n \in \left\{ {0, 1, \ldots, N}\right\}$, where $N \in \N$. Let $f: \C \to \C$ be a complex polynomial defined by $\displaystyle f \left({z}\right) = \sum_{n \mathop = 0}^N a_n z^n$. Then $f$ is complex differentiable and its derivative is: :$\displaystyle f' \left({z}\right) = \sum_{n \mathop = 1}^N n a_n z^{n-1}$
Nested Sphere Theorem
Let $M = \left({A, d}\right)$ be a complete metric space. Let $\left\langle{S_n}\right\rangle$ be a sequence of closed balls in $M$ defined by: : $S_n = B^-_{\rho_n} \left({x_n}\right)$ where $\rho_n \to 0$ as $n \to \infty$ and: : $S_1 \supseteq S_2 \supseteq \cdots \supseteq S_n \supseteq \cdots$ Then there exists $x \in A$ such that: :$\displaystyle \bigcap_{n \mathop = 1}^\infty S_n = \left\{{x}\right\}$
Principle of Dilemma/Formulation 1/Forward Implication
: $\left({p \implies q}\right) \land \left({\neg p \implies q}\right) \vdash q$
Principle of Dilemma/Formulation 1/Forward Implication/Proof 3
: $\left({p \implies q}\right) \land \left({\neg p \implies q}\right) \vdash q$
Principle of Dilemma/Formulation 1/Reverse Implication
: $q \vdash \left({p \implies q}\right) \land \left({\neg p \implies q}\right)$
Principle of Dilemma/Formulation 2/Reverse Implication
:$\vdash q \implies \paren {\paren {p \implies q} \land \paren {\neg p \implies q} }$
Properties of Complex Exponential Function
Let $z \in \C$ be a complex number. Let $\exp z$ be the exponential of $z$. Then:
Open Domain is Connected iff it is Path-Connected
Let $D \subseteq \C$ be a open subset of the set of complex numbers. Then $D$ is connected  $D$ is path-connected.
Proof by Contradiction/Variant 2/Formulation 1
:$p \implies q, p \implies \neg q \vdash \neg p$
Proof by Contradiction/Variant 2/Formulation 2
:$\vdash \paren {\paren {p \implies q} \land \paren {p \implies \neg q} } \implies \neg p$
Proof by Contradiction/Variant 3/Formulation 1
:$p \implies \neg p \vdash \neg p$
Proof by Contradiction/Variant 3/Formulation 2
:$\vdash \paren {p \implies \neg p} \implies \neg p$
Primitive of Function on Connected Domain
Let $f: D \to \C$ be a continuous complex function, where $D$ is a connected domain. Then the following three conditions are equivalent: :$(1): \quad$ $f$ has a primitive. :$(2): \quad$ For any two contours $C_1, C_2$ in $D$ with identical start points $z_1 \in D$ and end points $z_2 \in D$, we have: ::$\displaystyle \int_{C_1} \map f z \rd z = \int_{C_2} \map f z \rd z$ :$(3): \quad$ For all closed contours $C$ in $D$, we have: ::$\displaystyle \oint_C \map f z \rd z = 0$ If the conditions hold, we can choose any $z_0 \in D$ and define a primitive $F: D \to \C$ of $f$ by: :$\displaystyle \map F w = \int_{C_w} \map f z \rd z$ where $C_w$ is any contour in $D$ with start point $z_0$ and end point $w$.