avewright/proofwiki-math huggingface dataset:
Theorem: Hyperbolic Cotangent of Complex Number/Formulation 3

:$\map \coth {a + b i} = \dfrac {\coth a + \coth a \cot^2 b} {\coth^2 a + \cot^2 b} + \dfrac {\cot b - \coth^2 a \cot b} {\coth^2 a + \cot^2 b} i$

Proof:
{{begin-eqn}}
{{eqn | l = \map \coth {a + b i}
      | r = \dfrac {1 - i \coth a \cot b} {\coth a - i \cot b}
      | c = Hyperbolic Cotangent of Complex Number: Formulation 2
}}
{{eqn | r = \dfrac {\paren {1 - i \coth a \cot b} \paren {\coth a + i \cot b} } {\paren {\coth a - i \cot b} \paren {\coth a + i \cot b} }
      | c = multiplying denominator and numerator by $\coth a + i \cot b$
}}
{{eqn | r = \dfrac {\paren {1 - i \coth a \cot b} \paren {\coth a + i \cot b} } {\coth^2 a + \cot^2 b}
      | c = Difference of Two Squares
}}
{{eqn | r = \dfrac {\coth a + i \cot b - i \coth^2 a \cot b + \coth a \cot^2 b} {\coth^2 a + \cot^2 b}
      | c = 
}}
{{eqn | r = \dfrac {\coth a + \coth a \cot^2 b} {\coth^2 a + \cot^2 b} + \dfrac {\cot b - \coth^2 a \cot b} {\coth^2 a + \cot^2 b} i
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Hyperbolic Tangent of Complex Number/Formulation 4

:$\map \tanh {a + b i} = \dfrac {\sinh 2 a + i \sin 2 b} {\cosh 2 a + \cos 2 b}$

Proof:
{{begin-eqn}}
{{eqn | l = \map \tanh {a + b i}
      | r = \dfrac {\sinh a \cos b + i \cosh a \sin b} {\cosh a \cos b + i \sinh a \sin b}
      | c = Hyperbolic Tangent of Complex Number: Formulation 1
}}
{{eqn | r = \dfrac {\paren {\sinh a \cos b + i \cosh a \sin b} \paren {\cosh a \cos b - i \sinh a \sin b} } {\paren {\cosh a \cos b + i \sinh a \sin b} \paren {\cosh a \cos b - i \sinh a \sin b} }
      | c = 
}}
{{eqn | r = \dfrac {\sinh a \cosh a \cos^2 b + i \cosh^2 a \cos b \sin b - i \sinh^2 a \cos b \sin b + \sinh a \cosh a \sin^2 b} {\cosh^2 a \cos^2 b + \sinh^2 a \sin^2 b}
      | c = 
}}
{{eqn | r = \dfrac {\sinh a \cosh a \paren {\cos^2 b + \sin^2 b} + i \paren {\cosh^2 a - \sinh^2 a} \cos b \sin b} {\cosh^2 a \cos^2 b + \sinh^2 a \sin^2 b}
      | c = 
}}
{{eqn | r = \dfrac {\sinh a \cosh a + i \paren {\cosh^2 a - \sinh^2 a} \cos b \sin b} {\cosh^2 a \cos^2 b + \sinh^2 a \sin^2 b}
      | c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = \dfrac {\sinh a \cosh a + i \cos b \sin b} {\cosh^2 a \cos^2 b + \sinh^2 a \sin^2 b}
      | c = Difference of Squares of Hyperbolic Cosine and Sine
}}
{{eqn | r = \dfrac {\sinh 2 a + i \sin 2 b} {2 \paren {\cosh^2 a \cos^2 b + \sinh^2 a \sin^2 b} }
      | c = Double Angle Formula for Hyperbolic Sine, Double Angle Formula for Sine
}}
{{eqn | r = \dfrac {\sinh 2 a + i \sin 2 b} {\cosh^2 a \paren {1 + \cos 2 b} +  \sinh^2 a \paren {1 - \cos 2 b} }
      | c = Double Angle Formula for Cosine
}}
{{eqn | r = \dfrac {\sinh 2 a + i \sin 2 b} {\cosh^2 a + \cosh^2 a \cos 2 b + \sinh^2 a - \sinh^2 a \cos 2 b}
      | c = 
}}
{{eqn | r = \dfrac {\sinh 2 a + i \sin 2 b} {\cosh 2 a + \paren {\cosh^2 a - \sinh^2 a} \cos 2 b}
      | c = Double Angle Formula for Hyperbolic Cosine
}}
{{eqn | r = \dfrac {\sinh 2 a + i \sin 2 b} {\cosh 2 a + \cos 2 b}
      | c = Difference of Squares of Hyperbolic Cosine and Sine
}}
{{end-eqn}}
{{qed}}
Theorem: Tangent of Complex Number/Formulation 4

:$\tan \paren {a + b i} = \dfrac {\sin 2 a + i \sinh 2 b} {\cos 2 a + \cosh 2 b}$

Proof:
{{begin-eqn}}
{{eqn | l = \tan \paren {a + b i}
      | r = \dfrac {\sin a \cosh b + i \cos a \sinh b} {\cos a \cosh b - i \sin a \sinh b}
      | c = Tangent of Complex Number: Formulation 1
}}
{{eqn | r = \dfrac {\paren {\sin a \cosh b + i \cos a \sinh b} \paren {\cos a \cosh b + i \sin a \sinh b} } {\paren {\cos a \cosh b - i \sin a \sinh b} \paren {\cos a \cosh b + i \sin a \sinh b} }
      | c = 
}}
{{eqn | r = \dfrac {\sin a \cos a \cosh^2 b + i \cos^2 a \cosh b \sinh b + i \sin^2 a \cosh b \sinh b - \sin a \cos b \sinh^2 b} {\cos^2 a \cosh^2 b + \sin^2 a \sinh^2 b}
      | c = 
}}
{{eqn | r = \dfrac {\sin a \cos a \paren {\cosh^2 b - \sinh^2 b} + i \paren {\cos^2 a + \sin^2 a} \cosh b \sinh b} {\cos^2 a \cosh^2 b + \sin^2 a \sinh^2 b}
      | c = 
}}
{{eqn | r = \dfrac {\sin a \cos a + i \paren {\cos^2 a + \sin^2 a} \cosh b \sinh b} {\cos^2 a \cosh^2 b + \sin^2 a \sinh^2 b}
      | c = Difference of Squares of Hyperbolic Cosine and Sine
}}
{{eqn | r = \dfrac {\sin a \cos a + i \cosh b \sinh b} {\cos^2 a \cosh^2 b + \sin^2 a \sinh^2 b}
      | c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = \dfrac {\sin 2 a + i \sinh 2 b} {2 \paren {\cos^2 a \cosh^2 b + \sin^2 a \sinh^2 b} }
      | c = Double Angle Formula for Sine, Double Angle Formula for Hyperbolic Sine
}}
{{eqn | r = \dfrac {\sin 2 a + i \sinh 2 b } {\paren {1 + \cos 2 a} \cosh^2 b + \paren {1 - \cos 2 a} \sinh^2 b}
      | c = Double Angle Formula for Cosine
}}
{{eqn | r = \dfrac {\sin 2 a + i \sinh 2 b } {\cosh^2 b + \cos 2 a \cosh^2 b + \sinh^2 b - \cos 2 a \sinh^2 b}
      | c = 
}}
{{eqn | r = \dfrac {\sin 2 a + i \sinh 2 b } {\cos 2 a \paren {\cosh^2 b - \sinh^2 b} + \cosh 2 b}
      | c = Double Angle Formula for Hyperbolic Cosine
}}
{{eqn | r = \dfrac {\sin 2 a + i \sinh 2 b } {\cos 2 a + \cosh 2 b}
      | c = Difference of Squares of Hyperbolic Cosine and Sine
}}
{{end-eqn}}
{{qed}}
Theorem: Characterisation of Cauchy Sequence in Non-Archimedean Norm

Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring with non-Archimedean norm $\norm {\, \cdot \,}$.
Let $\sequence {x_n}$ be a sequence in $R$.
Then:
:$\sequence {x_n}$ is a Cauchy sequence
{{iff}}:
:$\ds \lim_{n \mathop \to \infty} \norm {x_{n + 1} - x_n} = 0$

Proof:
=== Necessary Condition ===
{{:Characterisation of Cauchy Sequence in Non-Archimedean Norm/Necessary Condition}}{{qed|lemma}}
Theorem: Characterisation of Cauchy Sequence in Non-Archimedean Norm/Necessary Condition

Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring with non-Archimedean norm $\norm {\,\cdot\,}$.
Let $\sequence {x_n}$ be a Cauchy sequence in $R$.
Then:
:$\lim_{n \mathop \to \infty} \norm {x_{n + 1} - x_n} = 0$

Proof:
Let $\epsilon > 0$ be given.
By the definition of a Cauchy sequence:
:$\exists N: \forall n, m > N: \norm {x_n - x_m} < \epsilon$
So
:$\exists N: \forall n > N: \norm {x_{n + 1} - x_n} < \epsilon$
Hence the result follows:
:$\lim_{n \mathop \to \infty} \norm {x_{n + 1} - x_n} = 0$.
Theorem: Characterisation of Cauchy Sequence in Non-Archimedean Norm/Sufficient Condition

Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring with non-Archimedean norm $\norm {\,\cdot\,}$.
Let $\sequence {x_n}$ be a sequence in $R$.
Let $\ds \lim_{n \mathop \to \infty} \norm {x_{n + 1} - x_n} = 0$.
Then:
:$\sequence {x_n}$ is a Cauchy sequence.

Proof:
Let $\epsilon > 0$ be given.
By assumption $\exists N \in \N$ such that:
:$(1) \quad  \forall n > N: \norm {x_{n + 1} - x_n} < 0$
Suppose $n, m > N$, and $m = n + r > n$.
Then:
{{begin-eqn}}
{{eqn | l = \norm {x_m - x_n}
      | r = \norm {x_{n + r} - x_{n + r - 1} + x_{n + r - 1} - x_{n + r - 2} + \dotsb + x_{n + 1} - x_n}
}}
{{eqn | r = \max \set {\norm {x_{n + r} - x_{n + r - 1} }, \norm {x_{n + r - 1} - x_{n + r - 2} }, \dotsc, \norm {x_{n + 1} - x_n} }
      | c = as $\norm {\,\cdot\,}$ is non-Archimedean
}}
{{eqn | r = \norm {x_{n + s} - x_{n + s - 1} }
      | c = for some $s$: $0 < s \le r$
}}
{{eqn | o = <
      | r = \epsilon
      | c = by $(1)$
}}
{{end-eqn}}
It follows that:
::$\sequence {x_n}$ is a Cauchy sequence.
Theorem: Power Series of Sine of Odd Theta

Let $r \in \R$ such that $\size r < 1$.
Let $\theta \in \R$ such that $\theta \ne m \pi$ for any $m \in \Z$.
Then:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop \ge 0} \map \sin {2 k + 1} \theta r^k
      | r = \sin \theta + r \sin 3 \theta + r^2 \sin 5 \theta + \cdots
      | c = 
}}
{{eqn | r = \dfrac {\paren {1 + r} \sin \theta} {1 - 2 r \cos 2 \theta + r^2}
      | c = 
}}
{{end-eqn}}

Proof:
From Euler's Formula:
:$\map \exp {i \theta} = \cos \theta + i \sin \theta$
Hence:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^\infty \map \sin {2 k + 1} \theta r^k
      | r = \map \Im {\sum_{k \mathop = 0}^\infty r^k \map \exp {\paren {2 k + 1} i \theta} }
      | c = 
}}
{{eqn | r = \map \Im {\map \exp {i \theta} \sum_{k \mathop = 0}^\infty \paren {r \map \exp {2 i \theta} }^k}
      | c = as $\map \Im {\exp \paren {i \times 0 \times x} } = \map \Im 1 = 0$
}}
{{eqn | r = \map \Im {\frac {\map \exp {i \theta} } {1 - r \map \exp {2 i \theta} } }
      | c = Sum of Infinite Geometric Sequence: valid because $\size r < 1$
}}
{{eqn | r = \map \Im {\frac {\map \exp {i \theta} \paren {1 - r \map \exp {-2 i \theta} } } {\paren {1 - r \map \exp {2 i \theta} } \paren {1 - r \map \exp {-2 i \theta} } } }
      | c = multiplying denominator and numerator by $1 - r \map \exp {- 2 i \theta}$
}}
{{eqn | r = \map \Im {\frac {\map \exp {i \theta} \paren {1 - r \map \exp {-2 i \theta} } } {1 - r \paren {\map \exp {-2 i \theta} + \map \exp {2 i \theta} } + r^2} }
      | c = expanding
}}
{{eqn | r = \map \Im {\frac {\map \exp {i \theta} \paren {1 - r \map \exp {-2 i \theta} } } {1 - 2 r \cos 2 \theta + r^2} }
      | c = Euler's Cosine Identity
}}
{{eqn | r = \map \Im {\frac {\paren {\cos \theta + i \sin \theta} \paren {1 - r \paren {\cos 2 \theta - i \sin 2 \theta} } } {1 - 2 r \cos 2 \theta + r^2} }
      | c = Euler's Formula, {{Corollary|Euler's Formula}}
}}
{{eqn | r = \map \Im {\frac {\cos \theta + i \sin \theta - r \cos \theta \cos 2 \theta + i r \cos \theta \sin 2 \theta - i r \sin \theta \cos 2 \theta - r \sin \theta \sin 2 \theta} {1 - 2 r \cos 2 \theta + r^2} }
      | c = expanding
}}
{{eqn | r = \frac {\sin \theta + r \cos \theta \sin 2 \theta - r \sin \theta \cos 2 \theta} {1 - 2 r \cos 2 \theta + r^2}
      | c = taking imaginary part
}}
{{eqn | r = \frac {\sin \theta + r \paren {\sin 2 \theta \cos \theta - \cos 2 \theta \sin \theta} } {1 - 2 r \cos 2 \theta + r^2}
      | c = 
}}
{{eqn | r = \frac {\sin \theta + r \map \sin {2 \theta - \theta} } {1 - 2 r \cos 2 \theta + r^2}
      | c = Sine of Difference
}}
{{eqn | r = \dfrac {\paren {1 + r} \sin \theta} {1 - 2 r \cos 2 \theta + r^2}
      | c = after simplification
}}
{{end-eqn}}
{{qed}}
Theorem: Sum of Infinite Series of Product of nth Power of cos 2 theta by 2n+1th Multiple of Sine

Let $\theta \in \R$ such that $\theta \ne m \pi$ for any $m \in \Z$.
Then:
{{begin-eqn}}
{{eqn | l = \sum_{n \mathop = 0}^\infty \paren {\cos 2 \theta}^n \sin \paren {2 n + 1} \theta
      | r = \sin \theta + \cos 2 \theta \sin 3 \theta + \paren {\cos 2 \theta}^2 \sin 5 \theta + \paren {\cos 2 \theta}^3 \sin 7 \theta + \cdots
      | c = 
}}
{{eqn | r = \dfrac {\csc \theta} 2
      | c = 
}}
{{end-eqn}}

Proof:
Let $\theta \ne \dfrac {m \pi} 2$ for any $m \in \Z$.
Then $\size {\cos 2 \theta} < 1$.
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop \ge 0} \sin \paren {2 k + 1} \theta r^k
      | r = \dfrac {\paren {1 + r} \sin \theta} {1 - 2 r \cos 2 \theta + r^2}
      | c = Power Series of Sine of Odd Theta: $\size r < 1$
}}
{{eqn | ll= \leadsto
      | l = \sum_{k \mathop = 0}^\infty \paren {\cos 2 \theta}^k \sin \paren {2 k + 1} \theta
      | r = \dfrac {\paren {1 + \cos 2 \theta} \sin \theta} {1 - 2 \cos 2 \theta \cos 2 \theta + \paren {\cos 2 \theta}^2}
      | c = setting $r = \cos 2 \theta$
}}
{{eqn | r = \dfrac {\paren {1 + \cos 2 \theta} \sin \theta} {1 - \paren {\cos 2 \theta}^2}
      | c = simplifying
}}
{{eqn | r = \dfrac {\paren {1 + 2 \cos^2 \theta - 1} \sin \theta} {1 - \paren {2 \cos^2 \theta - 1}^2}
      | c = Double Angle Formula for Cosine
}}
{{eqn | r = \dfrac {2 \cos^2 \theta \sin \theta} {1 - \paren {4 \cos^4 \theta - 4 \cos^2 \theta + 1} }
      | c = simplifying and expanding
}}
{{eqn | r = \dfrac {2 \cos^2 \theta \sin \theta} {4 \cos^2 \theta \paren {1 - \cos^2 \theta} }
      | c = simplifying
}}
{{eqn | r = \dfrac {2 \cos^2 \theta \sin \theta} {4 \cos^2 \theta \sin^2 \theta}
      | c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = \dfrac 1 {2 \sin \theta}
      | c = simplifying
}}
{{eqn | r = \dfrac {\csc \theta} 2
      | c = {{Defof|Real Cosecant Function}}
}}
{{end-eqn}}
{{qed}}
Theorem: Modulus of Limit/Normed Division Ring

Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.
Let $\sequence {x_n}$ be a convergent sequence in $R$ to the limit $l$.
That is, let $\ds \lim_{n \mathop \to \infty} x_n = l$.
Then
:$\ds \lim_{n \mathop \to \infty} \norm {x_n} = \norm l$

Proof:
By the Reverse Triangle Inequality, we have:
:$\cmod {\norm {x_n} - \norm l} \le \norm {x_n - l}$
Hence by the Squeeze Theorem and Convergent Sequence Minus Limit, $\norm {x_n} \to \norm l$ as $n \to \infty$.
{{Qed}}
Theorem: Three Points in Ultrametric Space have Two Equal Distances

Let $\struct {X, d}$ be an ultrametric space.
Let $x, y, z \in X$ with $\map d {x, z} \ne \map d {y, z}$.
Then:
:$\map d {x, y} = \max \set {\map d {x, z}, \map d {y, z} }$

Proof:
{{WLOG}}, let $\map d {x, z} > \map d {y, z}$.
Then:
{{begin-eqn}}
{{eqn | l = \map d {x, y}
      | o = \le
      | r = \max \set {\map d {x, z}, \map d {y, z} }
      | c = {{Defof|Non-Archimedean Metric}}
}}
{{eqn | r = \map d {x, z}
      | c = since $\map d {x, z} > \map d {y, z}$
}}
{{end-eqn}}
On the other hand:
{{begin-eqn}}
{{eqn | l = \map d {x, z}
      | o = \le
      | r = \max \set {\map d {x, y}, \map d {y, z} }
      | c = {{Defof|Non-Archimedean Metric}}
}}
{{eqn | r = \map d {x, y}
      | c = since $\map d {x, z} > \map d {y, z}$
}}
{{end-eqn}}
Putting the two inequalities together then:
:$\map d {x, y} = \map d {x, z} = \max \set {\map d {x, z}, \map d {y, z} }$
{{qed}}
Theorem: Three Points in Ultrametric Space have Two Equal Distances/Corollary

Let $\struct {X, d}$ be an ultrametric space.
Let $x, y, z \in X$.
Then:
:at least two of the distances $\map d {x, y}$, $\map d {x, z}$ and $\map d {y, z}$ are equal.

Proof:
Either:
:$\map d {x, z} = \map d {y, z}$
or:
:$\map d {x, z} \ne \map d {y, z}$
By Three Points in Ultrametric Space have Two Equal Distances:
:$\map d {x, z} = \map d {y, z}$ or $\map d {x, y} = \max \set {\map d {x, z}, \map d {y, z} }$
In either case two of the distances are equal.
{{Qed}}
Theorem: Non-Archimedean Norm iff Non-Archimedean Metric/Necessary Condition

Let $\struct {R, \norm {\,\cdot\,}}$ be a non-Archimedean normed division ring.
Let $d$ be the metric induced by $\norm {\,\cdot\,}$.
Then $d$ is a non-Archimedean metric.

Proof:
Let $x, y, z \in R$.
{{begin-eqn}}
{{eqn | l = \map d {x, y}
      | r = \norm {x - y}
      | c = {{Defof|Metric Induced by Norm on Division Ring|Metric Induced by $\norm {\,\cdot\,}$}}
}}
{{eqn | r = \norm {\paren {x - z} + \paren {z - y} }
}}
{{eqn | r = \max \set {\norm {x - z}, \norm {z - y} }
      | o = \le
      | c = {{Defof|Non-Archimedean Division Ring Norm}}
}}
{{eqn | r = \max \set {\map d {x, z}, \map d {z, y} }
      | c = {{Defof|Metric Induced by Norm on Division Ring|Metric Induced by $\norm {\,\cdot\,}$}}
}}
{{end-eqn}}
Theorem: Non-Archimedean Norm iff Non-Archimedean Metric/Sufficient Condition

Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring with zero $0$.
Let $d$ be the metric induced by $\norm {\, \cdot \,}$.
Let $d$ be non-Archimedean.
Then:
:$\norm {\, \cdot \,}$ is a non-Archimedean norm.

Proof:
Let $x, y \in R$.
{{begin-eqn}}
{{eqn | l = \norm {x + y}
      | r = \norm {x - \paren {-y} }
}}
{{eqn | r = \map d {x, - y}
      | c = {{Defof|Metric Induced by Norm on Division Ring|Metric Induced by $\norm {\, \cdot \,}$}}
}}
{{eqn | r = \max \set {\map d {x, 0}, \map d {0, -y} }
      | o = \le
      | c = {{Defof|Non-Archimedean Metric}}
}}
{{eqn | r = \max \set {\norm {x - 0 }, \norm {0 - \paren {-y} } }
      | c = {{Defof|Metric Induced by Norm on Division Ring|Metric Induced by $\norm {\, \cdot \,}$}}
}}
{{eqn | r = \max \set {\norm x, \norm y}
}}
{{end-eqn}}
Theorem: Non-Archimedean Norm iff Non-Archimedean Metric

Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring with zero $0$.
Let $d$ be the metric induced by $\norm {\,\cdot\,}$.
Then:
:$\norm {\, \cdot \,}$ is a non-Archimedean norm {{iff}} $d$ is a non-Archimedean metric.

Proof:
=== Necessary Condition ===
{{:Non-Archimedean Norm iff Non-Archimedean Metric/Necessary Condition}}{{qed|lemma}}
Theorem: Quotients of 3 Unequal Numbers are Unequal

Let $x, y, z \in \R_{\ne 0}$ be non-zero real numbers which are not all equal.
Then $\dfrac x y, \dfrac y z, \dfrac z x$ are also not all equal.

Proof:
{{AimForCont}} $\dfrac x y = \dfrac y z = \dfrac z x$.
{{begin-eqn}}
{{eqn | o = 
      | r = \dfrac x y = \dfrac y z = \dfrac z x
      | c = 
}}
{{eqn | ll= \leadsto
      | o = 
      | r = x^2 z = y^2 x = z^2 y
      | c = multiplying top and bottom by $x y z$
}}
{{eqn | ll= \leadsto
      | o = 
      | r = x z = y^2, x y = z^2, y z = x^2
      | c = 
}}
{{eqn | ll= \leadsto
      | o = 
      | r = x y z = y^3, x y z = z^3, z y z = x^3
      | c = 
}}
{{eqn | ll= \leadsto
      | o = 
      | r = x^3 = y^3 = z^3
      | c = 
}}
{{eqn | ll= \leadsto
      | o = 
      | r = x = y = z
      | c = 
}}
{{end-eqn}}
This contradicts the assertion that $x, y, z$ are all unequal.
Hence the result by Proof by Contradiction.
{{qed}}
Theorem: Union of Power Sets not always Equal to Powerset of Union

The union of the power sets of two sets $S$ and $T$ is not necessarily equal to the power set of their union.

Proof:
Proof by Counterexample:
Let $S = \set {1, 2, 3}, T = \set {2, 3, 4}, X = \set {1, 2, 3, 4}$.
{{begin-eqn}}
{{eqn | l = S \cup T
      | r = \set {1, 2, 3, 4}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = X
      | o = \subseteq
      | r = S \cup T
      | c = 
}}
{{eqn | ll= \leadsto
      | l = X
      | o = \in
      | r = \powerset {S \cup T}
      | c = 
}}
{{end-eqn}}
But note that $X \nsubseteq S \land X \nsubseteq T$.
Thus:
{{begin-eqn}}
{{eqn | l = X
      | o = \nsubseteq
      | r = S \land X \nsubseteq T
      | c = 
}}
{{eqn | ll= \leadsto
      | l = X
      | o = \notin
      | r = \powerset S \land X \notin \powerset T
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \neg (X
      | o = \in
      | r = \powerset S \lor X \in \powerset T)
      | c = 
}}
{{eqn | ll= \leadsto
      | l = X
      | o = \notin
      | r = \powerset S \cup \powerset T
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \powerset {S \cup T}
      | o = \nsubseteq
      | r = \powerset S \cup \powerset T
      | c = 
}}
{{end-eqn}}
So:
:$\powerset {S \cup T} \ne \powerset S \cup \powerset T$
{{qed}}
Theorem: Limit of Intersection of Closed Intervals from Zero to Positive Integer Reciprocal

For all (strictly) positive integers $n \in \Z_{>0}$, let $A_n$ be the closed real interval:
:$A_n = \closedint 0 {\dfrac 1 n}$
Let $A \subseteq \R$ be the subset of the real numbers defined as:
:$A = \ds \lim_{n \mathop \to \infty} \bigcap A_n$
Then:
:$A = \set 0$

Proof:
First it is noted that:
:$\forall x \in \R_{<0}: x \notin A$
and that by definition of closed real interval:
:$\forall n \in \Z_{>0}: 0 \in A_n$
and so by definition of intersection:
:$0 \in A$
It remains to demonstrate that:
:$\forall x \in \R_{>0}: x \notin A$
{{AimForCont}} $\exists x \in \R_{>0}: x \in A$.
By the Axiom of Archimedes:
:$\exists N \in \Z: N > \dfrac 1 x$
and so from Reciprocal Function is Strictly Decreasing:
:$\exists N \in \Z: \dfrac 1 N < x$
Thus:
:$x \notin A_N$
and so by definition of intersection:
:$x \notin A$
This contradicts our supposition that $x \in A$.
Hence the only element of $A$ is $0$, and so:
:$A = \set 0$
{{qed}}
Theorem: Total Number of Set Partitions

Let $S$ be a finite set of cardinality $n$.
Then the number of different partitions of $S$ is $B_n$, where $B_n$ is the $n$th Bell number.

Proof:
The number of different partitions of $S$ is '''defined''' as $B_n$.
From Bell Number as Summation over Lower Index of Stirling Numbers of the Second Kind:
:$B_n = \ds \sum_{k \mathop = 0}^n {n \brace k}$
where $\ds {n \brace k}$ denotes a Stirling number of the second kind.
From Number of Set Partitions by Number of Components, $\ds {n \brace k}$ is the number of partitions of $S$ into $k$ components.
Hence the result.
{{qed}}
Theorem: Combination Theorem for Sequences/Normed Division Ring/Inverse Rule

Suppose $l \ne 0$.
Then:
:$\exists k \in \N : \forall n \in \N: x_{k + n} \ne 0$
and the subsequence $\sequence { x_{k+n}^{-1} }$ is well-defined and convergent with:
:$\ds \lim_{n \mathop \to \infty} {x_{k + n} }^{-1} = l^{-1}$.

Proof:
Since $\sequence {x_n}$ converges to $l \ne 0$, by Sequence Converges to Within Half Limit then:
:$\exists k \in \N: \forall n \in \N: \dfrac {\norm l} 2 < \norm {x_{k + n} }$
By {{Norm-axiom-mult|1}}:
:$\forall n \in \N : x_{k + n} \ne 0$
Let $\sequence {y_n}$ be the subsequence of $\sequence {x_n}$ where $y_n = x_{k + n}$. 
By Limit of Subsequence equals Limit of Sequence, $\sequence {y_n}$ is convergent with:
:$\ds \lim_{n \mathop \to \infty} y_n = l$
It also follows that: 
:$\forall n \in \N : y_n \ne 0$
So $\sequence { {y_n}^{-1} }$ is well-defined and:
:$\sequence { {y_n}^{-1} } = \sequence { {x_{k + n} }^{-1} }$
Theorem: Bell Number as Summation over Lower Index of Stirling Numbers of the Second Kind

Let $B_n$ be the Bell number for $n \in \Z_{\ge 0}$.
Then:
:$B_n = \ds \sum_{k \mathop = 0}^n {n \brace k}$
where $\ds {n \brace k}$ denotes a Stirling number of the second kind.

Proof:
By definition of Bell numbers:
:$B_n$ is the number of partitions of a (finite) set whose cardinality is $n$.
First consider the case where $n > 0$.
From Number of Set Partitions by Number of Components, the number of partitions of $S$ into $k$ components is $\ds {n \brace k}$.
Thus the total number of all partitions of $S$ is the sum of $\ds {n \brace k}$ where $k$ runs from $1$ to $n$.
Hence:
:$B_n = \ds \sum_{k \mathop = 1}^n {n \brace k}$
But $\ds {n \brace 0} = 0$ for all $n \in \Z_{> 0}$, and so:
:$B_n = \ds \sum_{k \mathop = 0}^n {n \brace k}$
Now consider the edge case where $n = 0$, that is, $S = \O = \set{}$.
There is by conventional definition one partition of $\O$, that is: $\set \O$
In this case, we have:
:$\ds {0 \brace 0} = 1$
and so:
:$B_0 = \ds \sum_{k \mathop = 0}^0 {n \brace k} = {0 \brace 0} = 1$
{{qed}}
Category:Bell Numbers
Category:Stirling Numbers
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Theorem: Bell Number as Summation over Lower Index of Stirling Numbers of the Second Kind/Corollary

:$B_n = \ds \sum_{k \mathop = 1}^n {n \brace k}$

Proof:
From Bell Number as Summation over Lower Index of Stirling Numbers of the Second Kind, we have that:
:$B_n = \ds \sum_{k \mathop = 0}^n {n \brace k}$
But when $n > 0$:
:$\ds {n \brace 0} = 0$
Hence the result.
{{qed}}
Category:Bell Numbers
Category:Stirling Numbers
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Theorem: Equivalence of Well-Ordering Principle and Induction/Proof/PFI implies PCI

The Principle of Finite Induction implies the Principle of Complete Finite Induction.
That is:
:Principle of Finite Induction: Given a subset $S \subseteq \N$ of the natural numbers which has these properties:
::$0 \in S$
::$n \in S \implies n + 1 \in S$
:then $S = \N$.
implies:
:Principle of Complete Finite Induction: Given a subset $S \subseteq \N$ of the natural numbers which has these properties:
::$0 \in S$
::$\set {0, 1, \ldots, n} \subseteq S \implies n + 1 \in S$
:then $S = \N$.

Proof:
Let us assume that the '''PFI''' is true.
Let $S \subseteq \N$ which satisfy:
:$(A): \quad 0 \in S$
:$(B): \quad \set {0, 1, \ldots, n} \subseteq S \implies n + 1 \in S$.
We want to show that $S = \N$, that is, the '''PCI''' is true.
Let $P \paren n$ be the propositional function:
:$P \paren n \iff \set {0, 1, \ldots, n} \subseteq S$
We define the set $S'$ as:
:$S' = \set {n \in \N: P \paren n \text { is true} }$
$P \paren 0$ is true by $(A)$, so $0 \in S'$.
Assume $P \paren k$ is true where $k \ge 0$.
So $k \in S'$, and by hypothesis:
:$\set {0, 1, \ldots, k} \subseteq S$
So by $(B)$:
:$k + 1 \in S$
Thus:
:$\set {0, 1, \ldots, k, k + 1} \subseteq S$.
That last statement means $P \paren {k + 1}$ is true.
This means $k + 1 \in S'$.
Thus we have satisfied the conditions:
:$0 \in S'$
:$n \in S' \implies n + 1 \in S'$
That is, $S' = \N$, and $P \paren n$ holds for all $n \in \N$.
Hence, by definition:
:$S = \N$
So '''PFI''' gives that $S = \N$.
Therefore '''PFI''' implies '''PCI'''.