hcju/proofwikiqa huggingface dataset:
{{begin-eqn}} {{eqn	| l = \int_0^\infty \frac {x^n} {\sinh a x} \rd x 	| r = \frac 1 a \int_0^\infty \frac {\paren {\frac u a}^n} {\sinh u} \rd u 	| c = substituting $u = a x$ }} {{eqn	| r = \frac 2 {a^{n + 1} } \int_0^\infty \frac {u^n e^{-u} } {1 - e^{-2 u} } \rd u 	| c = {{Defof|Hyperbolic Sine}} }} {{eqn	| r = \frac 2 {a^{n + 1} } \int_0^\infty u^n e^{-u} \paren {\sum_{N \mathop = 0}^\infty \paren {e^{-2 u} }^N} \rd u 	| c = Sum of Infinite Geometric Sequence }} {{eqn	| r = \frac 2 {a^{n + 1} } \sum_{N \mathop = 0}^\infty \int_0^\infty u^n e^{-\paren {2 N + 1} u} \rd u 	| c = Fubini's Theorem }} {{eqn	| r = \frac 2 {a^{n + 1} } \map \Gamma {n + 1} \sum_{N \mathop = 0}^\infty \frac 1 {\paren {2 N + 1}^{n + 1} }  	| c = Laplace Transform of Real Power }} {{end-eqn}} We have:  {{begin-eqn}} {{eqn	| l = \sum_{N \mathop = 1}^\infty \frac 1 {N^{n + 1} }  	| r = \map \zeta {n + 1} 	| c = {{Defof|Riemann Zeta Function}} }} {{eqn	| r = \sum_{N \mathop = 1}^\infty \frac 1 {\paren {2 N}^{n + 1} } + \sum_{N \mathop = 0}^\infty \frac 1 {\paren {2 N + 1}^{n + 1} }  }} {{eqn	| r = \frac 1 {2^{n + 1} } \sum_{N \mathop = 1}^\infty \frac 1 {N^{n + 1} } + \sum_{N \mathop = 0}^\infty \frac 1 {\paren {2 N + 1}^{n + 1} }  }} {{eqn	| r = \frac 1 {2^{n + 1} } \map \zeta {n + 1} + \sum_{N \mathop = 0}^\infty \frac 1 {\paren {2 N + 1}^{n + 1} } 	| c = {{Defof|Riemann Zeta Function}} }} {{end-eqn}} So:  {{begin-eqn}} {{eqn	| l = \sum_{N \mathop = 0}^\infty \frac 1 {\paren {2 N + 1}^{n + 1} } 	| r = \paren {1 - \frac 1 {2^{n + 1} } } \map \zeta {n + 1} }} {{eqn	| r = \frac {2^{n + 1} - 1} {2^{n + 1} } \map \zeta {n + 1} }} {{end-eqn}} giving:  :$\displaystyle \int_0^\infty \frac {x^n} {\sinh a x} \rd x = \frac {2^{n + 1} - 1} {2^n a^{n + 1} } \map \Gamma {n + 1} \map \zeta {n + 1}$ {{qed}}
{{begin-eqn}} {{eqn	| l = \int_0^\infty \frac x {\sinh a x} \rd x 	| r = \frac {2^2 - 1} {2 a^2} \map \Gamma 2 \map \zeta 2 	| c = Definite Integral to Infinity of $\dfrac {x^n} {\sinh a x}$ }} {{eqn	| r = \frac 3 {2 a^2} \times 1! \times \frac {\pi^2} 6 	| c = Gamma Function Extends Factorial, Basel Problem }} {{eqn	| r = \frac \pi {4 a^2} }} {{end-eqn}} {{qed}}
From Identity Function is Odd Function, $\map f x$ is a odd function. By Fourier Series for Odd Function over Symmetric Range, we have: :$\displaystyle \map f x \sim \sum_{n \mathop = 1}^\infty b_n \sin \frac {n \pi x} \lambda$ where: {{begin-eqn}} {{eqn | l = b_n       | r = \frac 2 \lambda \int_0^\pi \map f x \sin \frac {n \pi x} \lambda \rd x       | c =  }} {{eqn | r = \frac 2 \lambda \int_0^\pi x \sin \frac {n \pi x} \lambda \rd x       | c = Definition of $\map f x$ }} {{eqn | r = \frac {2 \lambda} {n \pi} \paren {-1}^{n + 1}       | c = Half-Range Fourier Sine Series for Identity Function }} {{end-eqn}} Substituting for $b_n$ in $(1)$:  :$\displaystyle x = \dfrac {2 \lambda} \pi \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} } n \sin \frac {n \pi x} \lambda$ as required.  {{qed}} Category:Identity Mappings Category:Fourier Series for Identity Function 3zk880ik4ng176ofyrpm9guf5wtma7n
{{AimForCont}} $x \in \O$ as defined here. Thus we have: :$x \in E$ and: :$x \notin E$ This is a contradiction. It follows by Proof by Contradiction that $x \notin \O$. Hence, as $x$ was arbitrary, there can be no $x$ such that $x \in \O$. Thus $\O$ is the empty set by definition. {{qed}}
{{begin-eqn}} {{eqn	| l = \map \Si {-x} 	| r = \int_0^{-x} \frac {\sin u} u \rd u 	| c = {{Defof|Sine Integral Function}} }} {{eqn	| r = -\int_0^x \frac {\map \sin {-u} } {-u} \rd u 	| c = substituting $u \mapsto -u$ }} {{eqn	| r = -\int_0^x \frac {\sin u} u \rd u 	| c = Sine Function is Odd }} {{eqn	| r = -\map \Si x 	| c = {{Defof|Sine Integral Function}} }} {{end-eqn}} {{qed}}
By Sine Integral Function is Odd, $\Si$ is an odd function.  Therefore, by Odd Function of Zero is Zero: :$\map \Si 0 = 0$ {{qed}}
By definition, $\arctan$ is the inverse of the tangent function's restriction to $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$.  By Tangent of $30 \degrees$:  :$\tan \dfrac \pi 6 = \dfrac {\sqrt 3} 3$. As $\dfrac \pi 6 \in \openint {-\dfrac \pi 2} {\dfrac \pi 2}$, we have by the definition of an inverse function:  :$\map \arctan {\dfrac {\sqrt 3} 3} = \dfrac \pi 6$ {{qed}} Category:Arctangent Function 9jygfor7luy0d130k7uap7bokq4v8hm
We have that Cardinality is Additive Function. From Additive Function is Strongly Additive: :$\card {S_1 \cup S_2} + \card {S_1 \cap S_2} = \card {S_1} + \card {S_2}$ from which the result follows. {{qed}}
This is a specific example of Cardinality of Set Union: General Case. {{qed}}
Let $a \in \Q_p$. === Case: $\norm a_p \le 1$ === Let $\norm a_p \le 1$. By definition of the $p$-adic integers: :$a \in \Z_p$. Hence: :$p^0 a \in \Z_p$. {{qed|lemma}} === Case: $\norm a_p > 1$ === Let $\norm a_p > 1$. From P-adic Number times P-adic Norm is P-adic Unit, there exists $n \in \Z$ such that $p^n a$ is a $p$-adic unit. Then: {{begin-eqn}} {{eqn | l = \norm{p^n a}_p       | o = =       | r = 1       | c = P-adic Unit has Norm Equal to One }} {{eqn | lo= \leadstoandfrom       | l = \norm{p^n}_p \norm a_p       | o = =       | r = 1       | c = Norm Axiom (N2) : Multiplicativity }} {{eqn | lo= \leadstoandfrom       | l = \norm{p^n}_p        | o = <       | r = 1       | c = As $\norm{a}_p > 1$ by assumption }} {{eqn | lo= \leadstoandfrom       | l = p^{-n}        | o = <       | r = 1       | c = {{Defof|P-adic Norm|$p$-adic norm on integers}} }} {{eqn | lo= \leadstoandfrom       | l = p^n        | o = >       | r = 1 }} {{eqn | lo= \leadstoandfrom       | l = n        | o = >       | r = 0       | c = Real Power Function on Base Greater than One is Strictly Increasing }} {{end-eqn}} {{qed}}
We have, by Euler's Formula: Corollary: :$\map \exp {-i a x^2} = -i \map \sin {a x^2} + \map \cos {a x^2}$ As $\map \sin {a x^2}$ and $\map \cos {a x^2}$ are both real for real $a, x$, we therefore have: {{begin-eqn}} {{eqn	| l = \int_0^\infty \map \sin {a x^2} \rd x 	| r = -\int_0^\infty \map \Im {\map \exp {-i a x^2} } \rd x }} {{eqn	| r = -\map \Im {\int_0^\infty \map \exp {-i a x^2} \rd x} }} {{eqn	| r = -\frac 1 {\sqrt a} \map \Im {\int_0^\infty \map \exp {-i t^2} \rd t} 	| c = substituting $\sqrt a x = t$ }} {{eqn	| r = -\frac 1 {\sqrt a} \map \Im {\frac 1 2 \sqrt {\frac \pi 2} \paren {1 - i} } 	| c = Definite Integral to Infinity of $\map \exp {-i x^2}$ }} {{eqn	| r = \frac 1 2 \sqrt {\frac \pi {2 a} } }} {{end-eqn}} {{qed}}
We have, by Euler's Formula: Corollary: :$\map \exp {-i a x^2} = -i \map \sin {a x^2} + \map \cos {a x^2}$ As $\map \sin {a x^2}$ and $\map \cos {a x^2}$ are both real for real $a, x$, we therefore have: {{begin-eqn}} {{eqn	| l = \int_0^\infty \map \cos {a x^2} \rd x 	| r = \int_0^\infty \map \Re {\map \exp {-i a x^2} \rd x} }} {{eqn	| r = \map \Re {\int_0^\infty \map \exp {-i a x^2} \rd x} }} {{eqn	| r = \frac 1 {\sqrt a} \map \Re {\int_0^\infty \map \exp {-i t^2} \rd t} 	| c = substituting $\sqrt a x = t$ }} {{eqn	| r = \frac 1 {\sqrt a} \map \Re {\frac 1 2 \sqrt {\frac \pi 2} \paren {1 - i} } 	| c = Definite Integral to Infinity of $\map \exp {-i x^2}$ }} {{eqn	| r = \frac 1 2 \sqrt {\frac \pi {2 a} } }} {{end-eqn}} {{qed}}
Let $C$ be the unit open disk centred at $0$.  The boundary of $C$, $\partial C$, can be parameterized by:  :$\map \gamma \theta = e^{i \theta}$ for $0 \le \theta \le 2 \pi$.  We have:  {{begin-eqn}} {{eqn	| l = \int_0^{2 \pi} \frac {\d x} {1 - 2 a \cos x + a^2} 	| r = \int_0^{2 \pi} \frac {\d x} {1 - a \paren {e^{i x} + e^{-i x} } + a^2} 	| c = Cosine Exponential Formulation }} {{eqn	| r = -\int_0^{2 \pi} \frac {e^{i x} } {a e^{2 i x} - \paren {a^2 + 1} e^{i x} + a} \rd x }} {{eqn	| r = -\frac 1 a \int_0^{2 \pi} \frac {e^{i x} } {e^{2 i x} - \paren {a + \frac 1 a} e^{i x} + \frac a a} \rd x }} {{eqn	| r = -\frac 1 a \int_0^{2 \pi} \frac {e^{i x} } {\paren {e^{i x} - \frac 1 a} \paren {e^{i x} - a} } \rd x 	| c = factorising }} {{eqn	| r = -\frac 1 {i a} \int_{\partial C} \frac 1 {\paren {z - \frac 1 a} \paren {z - a} } \rd z 	| c = {{Defof|Contour Integral}}, Derivative of Exponential Function }} {{end-eqn}} The integrand has poles:  :$z_1 = a$ and:  :$z_2 = \dfrac 1 a$ We have $0 < a < 1$, so $\dfrac 1 a > 1$.  So $z_2$ lies outside the closed disk $\size z \le 1$, and $z_1$ lies in $C$, so the only pole of concern is $z_1$.  Therefore:  {{begin-eqn}} {{eqn	| l = -\frac 1 {i a} \int_{\partial C} \frac 1 {\paren {z - \frac 1 a} \paren {z - a} } \rd z 	| r = -\frac {2 \pi} a \Res {\frac 1 {\paren {z - \frac 1 a} \paren {z - a} } } a 	| c = Residue Theorem }} {{eqn	| r = -\frac {2 \pi} a \paren {\frac 1 {2 z - \paren {a + \frac 1 a} } }_{z = a} 	| c = Residue at Simple Pole, Derivative of Power }} {{eqn	| r = -\frac {2 \pi} a \paren {\frac 1 {2 a - a - \frac 1 a} } }} {{eqn	| r = \frac {2 \pi} {1 - a^2} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn	| l = \int_0^\infty \frac {\sinh a x} {e^{b x} - 1} \rd x 	| r = \frac 1 2 \int_0^\infty \frac {e^{-b x} \paren {e^{a x} - e^{-a x} } } {1 - e^{-b x} } 	| c = {{Defof|Hyperbolic Sine}} }} {{eqn	| r = \frac 1 2 \int_0^\infty \paren {e^{\paren {a - b} x} - e^{-\paren {a + b} x} } \sum_{n \mathop = 0}^\infty e^{-n b x} 	| c = Sum of Infinite Geometric Sequence }} {{eqn	| r = \frac 1 2 \sum_{n \mathop = 0}^\infty \int_0^\infty \paren {e^{\paren {a - \paren {n + 1} b} x} - e^{-\paren {a + \paren {n + 1} b} x} } \rd x 	| c = Fubini's Theorem }} {{eqn	| r = \frac 1 2 \sum_{n \mathop = 0}^\infty \paren {\intlimits {\frac {e^{\paren {a - \paren {n + 1} b} x} } {a - \paren {n + 1} b} } 0 \infty - \intlimits {-\frac {e^{-\paren {a + \paren {n + 1} b} x} } {a + \paren {n + 1} b} } 0 \infty} 	| c = Derivative of Exponential Function: Corollary 1 }} {{end-eqn}} Note that as $b > a$, we have that $a - b < 0$. As $b > 0$, we therefore have $a - \paren {n + 1} b < 0$ for all positive integer $n$. We also have that as $a + \paren {n + 1} b > 0$, that $-\paren {a + \paren {n + 1} b} < 0$. So, by Exponential Tends to Zero and Infinity:  :$\displaystyle \lim_{x \mathop \to \infty} \frac {e^{\paren {a - \paren {n + 1} b} x} } {a - \paren {n + 1} b} = 0$ and:  :$\displaystyle \lim_{x \mathop \to \infty} \frac {e^{-\paren {a + \paren {n + 1} b} x} } {a + \paren {n + 1} b} = 0$ We therefore have:  {{begin-eqn}} {{eqn	| l = \frac 1 2 \sum_{n \mathop = 0}^\infty \paren {\intlimits {\frac {e^{\paren {a - \paren {n + 1} b} x} } {a - \paren {n + 1} b} } 0 \infty - \intlimits {-\frac {e^{-\paren {a + \paren {n + 1} b} x} } {a + \paren {n + 1} b} } 0 \infty} 	| r = \frac 1 2 \sum_{n \mathop = 0}^\infty \paren {\frac 1 {\paren {n + 1} b - a} - \frac 1 {\paren {n + 1} b + a} } 	| c = Exponential of Zero }} {{eqn	| r = \sum_{n \mathop = 0}^\infty \frac a {\paren {n + 1}^2 b^2 - a^2} 	| c = Difference of Two Squares }} {{eqn	| r = -\frac 1 b \sum_{n \mathop = 1}^\infty \frac {\paren {\frac a b} } {\paren {\frac a b}^2 - n^2} 	| c = shifting the index, extracting factors }} {{end-eqn}} By Mittag-Leffler Expansion for Cotangent Function, we have:  :$\displaystyle \pi \cot \pi z = \frac 1 z + 2 \sum_{n \mathop = 1}^\infty \frac z {z^2 - n^2}$ where $z$ is not an integer.  Setting $z = \dfrac a b$, we have:  :$\displaystyle \pi \cot \frac {\pi a} b = \frac b a + 2 \sum_{n \mathop = 1}^\infty \frac {\paren {\frac a b} } {\paren {\frac a b}^2 - n^2}$ Rearrangement gives: :$\displaystyle \frac \pi 2 \cot \frac {\pi a} b - \frac b {2 a} = \sum_{n \mathop = 1}^\infty \frac {\paren {\frac a b} } {\paren {\frac a b}^2 - n^2}$ So:  {{begin-eqn}} {{eqn	| l = \int_0^\infty \frac {\sinh a x} {e^{b x} - 1} \rd x  	| r = -\frac 1 b \sum_{n \mathop = 1}^\infty \frac {\paren {\frac a b} } {\paren {\frac a b}^2 - n^2} }} {{eqn	| r = \frac 1 {2 a} - \frac \pi {2 b} \cot \frac {a \pi} b }} {{end-eqn}} {{qed}}
From Complement Union with Superset is Universe: Corollary: :$A \cup B = \mathbb U \iff \relcomp {\Bbb U} A \subseteq B$ and from Empty Intersection iff Subset of Complement: :$A \cap B = \O \iff B \subseteq \relcomp {\Bbb U} A$ The result follows by definition of set equality. {{qed}}
=== Necessary Condition === Let $C \subseteq A$. We have: {{begin-eqn}} {{eqn | l = C       | o = \subseteq       | r = A       | c =  }} {{eqn | ll= \leadsto       | l = A \cap C       | r = C       | c = Intersection with Subset is Subset }} {{eqn | ll= \leadsto       | l = \paren {A \cap B} \cup \paren {A \cap C}       | r = \paren {A \cap B} \cup C       | c =  }} {{eqn | ll= \leadsto       | l = A \cap \paren {B \cup C}       | r = \paren {A \cap B} \cup C       | c = Intersection Distributes over Union }} {{end-eqn}} and another way: {{begin-eqn}} {{eqn | l = C       | o = \subseteq       | r = A       | c =  }} {{eqn | ll= \leadsto       | l = A \cup C       | r = A       | c = Union with Superset is Superset }} {{eqn | ll= \leadsto       | l = \paren {A \cup C} \cap \paren {B \cup C}       | r = A \cap \paren {B \cup C}       | c =  }} {{eqn | ll= \leadsto       | l = \paren {A \cap B} \cup C       | r = A \cap \paren {B \cup C}       | c = Intersection Distributes over Union }} {{end-eqn}} Thus: :$C \subseteq A \implies \paren {A \cap B} \cup C = A \cap \paren {B \cup C}$ {{qed|lemma}} === Sufficient Condition === Let $\paren {A \cap B} \cup C = A \cap \paren {B \cup C}$. {{AimForCont}} $\exists x \in C: x \notin A$. From Set is Subset of Union: :$C \subseteq \paren {A \cap B} \cup C$ and so by definition of subset: :$x \in \paren {A \cap B} \cup C$ From Intersection is Subset: :$A \cap \paren {B \cup C} \subseteq A$ and so (indirectly) by definition of subset: :$x \notin A \cap \paren {B \cup C}$ Hence by definition of subset: :$\paren {A \cap B} \cup C \nsubseteq A \cap \paren {B \cup C}$ that is: :$\paren {A \cap B} \cup C \ne A \cap \paren {B \cup C}$ This contradicts our initial assertion. Hence by Proof by Contradiction: :$\neg \exists x \in C: x \notin A$ From Assertion of Universality: :$\forall x \in C: x \in A$ and so $C \subseteq A$ by definition of subset. {{qed}}
{{begin-eqn}} {{eqn | l = \paren {A \cap C} \cup \paren {B \cap \map \complement C}       | r = \O       | c =  }} {{eqn | ll= \leadstoandfrom       | l = A \cap C       | r = \O       | c = Union is Empty iff Sets are Empty }} {{eqn | lo= \land       | l = B \cap \map \complement C       | r = \O       | c =  }} {{eqn | ll= \leadstoandfrom       | l = C       | o = \subseteq       | r = \map \complement A       | c = Empty Intersection iff Subset of Complement }} {{eqn | lo= \land       | l = B       | o = \subseteq       | r = C       | c = Intersection with Complement is Empty iff Subset }} {{eqn | ll= \leadstoandfrom       | l = B       | o = \subseteq       | r = C \subseteq \map \complement A       | c =  }} {{end-eqn}} {{qed}}
From Definite Integral from $0$ to $2 \pi$ of $\dfrac 1 {a + b \sin x}$, we have:  :$\displaystyle \int_0^{2 \pi} \frac {\d x} {a + b \sin x} = \frac {2 \pi} {\sqrt {a^2 - b^2} }$ We have:  {{begin-eqn}} {{eqn	| l = \frac \partial {\partial a} \int_0^{2 \pi} \frac {\d x} {a + b \sin x} 	| r = \int_0^{2 \pi} \frac \partial {\partial a} \paren {\frac 1 {a + b \sin x} } \rd x 	| c = Definite Integral of Partial Derivative  }} {{eqn	| r = -\int_0^{2 \pi} \frac {\d x} {\paren {a + b \sin x}^2} 	| c = Quotient Rule for Derivatives }} {{end-eqn}} and:  {{begin-eqn}} {{eqn	| l = \frac \partial {\partial a} \paren {\frac {2 \pi} {\sqrt {a^2 - b^2} } } 	| r = 2 \pi \paren {\frac {-2 a} {2 \sqrt {a^2 - b^2} } } \paren {\frac 1 {a^2 - b^2} } 	| c = Quotient Rule for Derivatives }} {{eqn	| r = -\frac {2 \pi a} {\paren {a^2 - b^2}^{3/2} }  }} {{end-eqn}} giving: :$\displaystyle \int_0^{2 \pi} \frac {\d x} {\paren {a + b \cos x}^2} = \frac {2 \pi a} {\paren {a^2 - b^2}^{3/2} }$ {{qed}}
{{begin-eqn}} {{eqn	| l = \gamma 	| r = \lim_{n \mathop \to \infty} \paren {\sum_{k \mathop = 1}^n \frac 1 k - \ln n} 	| c = {{Defof|Euler-Mascheroni Constant}} }} {{eqn	| r = \lim_{n \mathop \to \infty} \paren {\sum_{k \mathop = 1}^n \paren {\int_0^1 x^{k - 1} \rd x } + \int_0^1 \frac {1 - x^{n - 1} } {\ln x} \rd x} 	| c = Primitive of Power, Definite Integral from $0$ to $1$ of $\dfrac {x^m - x^n} {\ln x}$ }} {{eqn	| r = \lim_{n \mathop \to \infty} \paren {\int_0^1 \paren {\sum_{k = 1}^n x^{k - 1} } \rd x + \int_0^1 \frac {1 - x^{n - 1} } {\ln x} \rd x} 	| c = Linear Combination of Definite Integrals }} {{eqn	| r = \lim_{n \mathop \to \infty} \paren {\int_0^1 \paren {\frac {1 - x^n} {1 - x} + \frac {1 - x^{n - 1} } {\ln x} } \rd x} 	| c = Sum of Infinite Geometric Sequence }} {{eqn	| r = \int_0^1 \lim_{n \mathop \to \infty} \paren {\frac {1 - x^n} {1 - x} + \frac {1 - x^{n - 1} } {\ln x} } \rd x 	| c = Lebesgue's Dominated Convergence Theorem }} {{eqn	| r = \int_0^1 \paren {\frac 1 {1 - x} + \frac 1 {\ln x} } \rd x 	| c = Sequence of Powers of Number less than One }} {{end-eqn}} Let:  :$x = e^{-t}$ By Derivative of Exponential Function: Corollary 1: :$\dfrac {\d x} {\d t} = -e^{-t}$ By Exponential of Zero, we have:  :as $x \to 1$, $t \to 0$. By Exponential Tends to Zero and Infinity, we have:  :as $x \to 0$, $t \to \infty$. We therefore have:  {{begin-eqn}} {{eqn	| l = \int_0^1 \paren {\frac 1 {1 - x} + \frac 1 {\ln x} } \rd x 	| r = -\int_\infty^0 e^{-t} \paren {\frac 1 {1 - e^{-t} } + \frac 1 {\map \ln {e^{-t} } } } \rd t 	| c = substituting $x = e^{-t}$ }} {{eqn	| r = \int_0^\infty e^{-t} \paren {\frac {e^t} {e^t - 1} - \frac 1 t} \rd t 	| c = Reversal of Limits of Definite Integral, {{Defof|Natural Logarithm}} }} {{eqn	| r = \int_0^\infty \paren {\frac 1 {e^t - 1} - \frac {e^{-t} } t} \rd t }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = A \uparrow A       | r = \relcomp {\Bbb U} {A \cap A}        | c = Definition of $\uparrow$ }} {{eqn | r = \relcomp {\Bbb U} A       | c = Intersection is Idempotent }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = A \cap B       | r = \relcomp {\Bbb U} {\relcomp {\Bbb U} {A \cap B} }       | c = Relative Complement of Relative Complement }} {{eqn | r = \relcomp {\Bbb U} {A \uparrow B}       | c = Intersection Complement of Set with Itself is Complement }} {{eqn | r = \paren {A \uparrow B} \uparrow \paren {A \uparrow B}       | c = Intersection Complement of Set with Itself is Complement }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = A \cup B       | r = \relcomp {\Bbb U} {\relcomp {\Bbb U} A \cap \relcomp {\Bbb U} B}       | c = De Morgan's Laws : Complement of Union }} {{eqn | r = \relcomp {\Bbb U} {\paren {A \uparrow A} \cap \paren {B \uparrow B} }       | c = Intersection Complement of Set with Itself is Complement }} {{eqn | r = \paren {A \uparrow A} \uparrow \paren {B \uparrow B}       | c = Intersection Complement of Set with Itself is Complement }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = x       | o = \in       | r = S       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \set x       | o = \subseteq       | r = S       | c = Singleton of Element is Subset }} {{eqn | ll= \leadstoandfrom       | l = \set x       | o = \in       | r = \powerset S       | c = {{Defof|Power Set}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = S       | o = \subseteq       | r = T       | c = }} {{eqn | ll= \leadsto       | l = S       | o = \in       | r = \powerset T       | c = {{Defof|Power Set}} }} {{eqn | ll= \leadsto       | l = \set S       | o = \in       | r = \powerset {\powerset T}       | c = Element in Set iff Singleton in Powerset }} {{end-eqn}} {{qed}}
We have: :$\O \in \set \O$ and so: :$\neg \paren {\forall x: x \notin \O}$ The result follows by definition of the empty set. {{qed}}
By the Kuratowski formalization of $\tuple {a, b}$: :$\tuple {a, b} = \set {\set a, \set {a, b} }$ and by Equality of Ordered Pairs: :$\tuple {a, b} = \tuple {b, a} \iff a = b$ But $a \ne b$ and so: :$\tuple {a, b} \ne \tuple {b, a}$ {{qed}}
From Definite Integral of Even Function: :$\displaystyle \frac 1 2 \int_{-\infty}^\infty \frac {\cos m x} {x^2 + a^2} \rd x = \int_0^\infty \frac {\cos m x} {x^2 + a^2} \rd x$ Let $R$ be a positive real number with $R > a$. Let $C_1$ be the straight line segment from $-R$ to $R$. Let $C_2$ be the arc of the circle of radius $R$ centred at the origin connecting $R$ and $-R$ counter-clockwise.  Let $\Gamma = C_1 \cup C_2$.  Let:  :$\map f z = \dfrac {e^{i m z} } {z^2 + a^2}$ From Euler's Formula, we have:  :$\map f z = \dfrac {\cos m z} {z^2 + a^2} + i \dfrac {\sin m z} {z^2 + a^2}$ So:  {{begin-eqn}} {{eqn	| l = \int_{-\infty}^\infty \map f x \rd x 	| r = \int_{-\infty}^\infty \frac {\cos m x} {x^2 + a^2} \rd x + i \int_{-\infty}^\infty \frac {\sin m x} {x^2 + a^2} \rd x 	| c = Linear Combination of Definite Integrals }} {{eqn	| r = \int_{-\infty}^\infty \frac {\cos m x} {x^2 + a^2} \rd x 	| c = Definite Integral of Odd Function }} {{end-eqn}} Note that the integrand is meromorphic with simple poles where $z^2 + a^2 = 0$.  That is, at $z = a i$ and $z = -a i$.  As our semi-circular contour lies in the upper half-plane, the only pole of concern is $a i$. As $R > a$, these poles do not lie on $C_2$, but are enclosed by the curve $\Gamma$. We have:  {{begin-eqn}} {{eqn	| l = \int_C \map f z \rd z 	| r = \int_{C_1} \frac {e^{i m z} } {z^2 + a^2} \rd z + \int_{C_2} \frac {e^{i m z} } {z^2 + a^2} \rd z 	| c = Contour Integral of Concatenation of Contours }} {{eqn	| r = \int_{-R}^R \frac {e^{i m x} } {x^2 + a^2} \rd x + \int_{C_2} \frac {e^{i m z} } {z^2 + a^2} \rd z 	| c = {{Defof|Contour Integral}} }} {{end-eqn}} The integral over $C_2$ can be shown to vanish as $R \to \infty$: {{begin-eqn}} {{eqn	| l = \size {\int_{C_2} \frac {e^{i m z} } {z^2 + a^2} \rd z} 	| o = \le 	| r = \frac \pi m \max_{0 \le \theta \le \pi} \size {\frac 1 {R^2 e^{2 i \theta} + 1} } 	| c = Jordan's Lemma }} {{eqn	| r = \frac \pi m \paren {\frac 1 {R^2 - 1} }  }} {{eqn	| o = \to 	| r = 0 	| c = as $R \to \infty$ }} {{end-eqn}} Taking $R \to \infty$, we have:  {{begin-eqn}} {{eqn	| l = \int_{-\infty}^\infty \frac {e^{i m x} } {x^2 + a^2} \rd x 	| r = \int_C \frac {e^{i m z} } {z^2 + a^2} \rd z }} {{eqn	| r = 2 \pi i \Res {\frac {e^{i m z} } {z^2 + a^2} } {a i} 	| c = Residue Theorem }} {{eqn	| r = 2 \pi i \paren {\frac {e^{i m z} } {2 z} }_{z = a i} 	| c = Residue at Simple Pole }} {{eqn	| r = 2 \pi i \frac {e^{-m a} } {2 a i} }} {{eqn	| r = \frac \pi a e^{-m a} }} {{end-eqn}} So:  :$\displaystyle \int_{-\infty}^\infty \frac {\cos m x} {x^2 + a^2} \rd x = \frac \pi a e^{-m a}$ giving: :$\displaystyle \int_0^\infty \frac {\cos m x} {x^2 + a^2} \rd x = \frac \pi {2 a} e^{-m a}$ {{qed}}
From Corollary to Normed Division Ring Operations are Continuous: :$\struct {R, +, *, \tau_{_R} }$ is a topological division ring. From Sum Rule for Continuous Mappings to Topological Division Ring: :$f + g: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is a continuous mapping. {{qed}}
From Corollary to Normed Division Ring Operations are Continuous: :$\struct {R, +, *, \tau_{_R} }$ is a topological division ring. From Multiple Rule for Continuous Mappings to Topological Division Ring: :$\lambda * f, f * \lambda: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ are continuous mappings. {{qed}}
From Corollary to Normed Division Ring Operations are Continuous: :$\struct {R, +, *, \tau_{_R} }$ is a topological division ring. From Product Rule for Continuous Mappings to Topological Division Ring: :$f * g: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is a continuous mapping. {{qed}}
From Corollary to Normed Division Ring Operations are Continuous: :$\struct {R, +, *, \tau_{_R} }$ is a topological division ring. From Inverse Rule for Continuous Mappings to Topological Division Ring: :$g^{-1}: \struct {U, \tau_{_U} } \to \struct {R, \tau_{_R} }$ is a continuous mapping. {{qed}}