hcju/proofwikiqa huggingface dataset:
Checking in turn each of the critera for equivalence: === Reflexive === We have that Equal Numbers are Congruent: :$\forall x, y, z \in \Z: x = y \implies x \equiv y \pmod z$ so it follows that: :$\forall x \in \Z: x \equiv x \pmod z$ and so congruence modulo $z$ is reflexive. {{qed|lemma}} === Symmetric === {{begin-eqn}} {{eqn | l = x       | o = \equiv       | r = y \bmod z       | c =  }} {{eqn | ll= \leadsto       | l = x - y       | r = k z       | c =  }} {{eqn | ll= \leadsto       | l = y - x       | r = \paren {-k} z       | c =  }} {{eqn | l = y       | o = \equiv       | r = x \bmod z       | c =  }} {{end-eqn}} So congruence modulo $z$ is symmetric. {{qed|lemma}} === Transitive === {{begin-eqn}} {{eqn | l = x_1       | o = \equiv       | r = x_2 \pmod z       | c =  }} {{eqn | lo= \land       | l = x_2       | o = \equiv       | r = x_3 \pmod z       | c =  }} {{eqn | ll= \leadsto       | l = \paren {x_1 - x_2}       | r = k_1 z       | c =  }} {{eqn | lo= \land       | l = \paren {x_2 - x_3}       | r = k_2 z       | c =  }} {{eqn | ll= \leadsto       | l = \paren {x_1 - x_2} + \paren {x_2 - x_3}       | r = \paren {k_1 + k_2} z       | c =  }} {{eqn | ll= \leadsto       | l = \paren {x_1 - x_3}       | r = \paren {k_1 + k_2} z       | c =  }} {{eqn | l = x_1       | o = \equiv       | r = x_3 \pmod z       | c =  }} {{end-eqn}} So congruence modulo $z$ is transitive. {{qed|lemma}} So we are justified in supposing that congruence, as we have defined it, is an equivalence. {{qed}}
First let $\exists m_1, m_2, \ldots, m_n \in \Z: \ds \sum_{k \mathop = 1}^n m_k a_k = 1$. Let $\gcd \set {a_1, a_2, \ldots, a_n} = d$. Then $\ds \sum_{k \mathop = 1}^n m_k a_k$ has $d$ as a divisor. That means $d$ is a divisor of $1$. Thus $\gcd \set {a_1, a_2, \ldots, a_n} = 1$. {{qed|lemma}} It remains to be shown that if $\gcd \set {a_1, a_2, \ldots, a_n} = 1$, then $\exists m_1, m_2, \ldots, m_n \in \Z: \ds \sum_{k \mathop = 1}^n m_k a_k = 1$. The proof proceeds by induction. For all $n \in \Z_{\ge 2}$, let $\map P n$ be the proposition: :$\gcd \set {a_1, a_2, \ldots, a_n} = 1 \iff \exists m_1, m_2, \ldots, m_n \in \Z: \ds \sum_{k \mathop = 1}^n m_k a_k = 1$ === Basis for the Induction === $\map P 2$ is the case: :$\gcd \set {a_1, a_2} = 1 \iff \exists m_1, m_2 \in \Z: m_1 a_1 + m_2 a_2 = 1$ This is demonstrated in Integer Combination of Coprime Integers. Thus $\map P 2$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $\map P r$ is true, where $r \ge 2$, then it logically follows that $\map P {r + 1}$ is true. So this is the induction hypothesis: :$\gcd \set {a_1, a_2, \ldots, a_r} = 1 \iff \exists m_1, m_2, \ldots, m_r \in \Z: \ds \sum_{k \mathop = 1}^r m_k a_k = 1$ from which it is to be shown that: :$\gcd \set {a_1, a_2, \ldots, a_r, a_{r + 1} } = 1 \iff \exists m_1, m_2, \ldots, m_r, m_{r + 1} \in \Z: \ds \sum_{k \mathop = 1}^{r + 1} m_k a_k = 1$ === Induction Step === This is the induction step: Let $\gcd \set {a_1, a_2, \ldots, a_r, a_{r + 1} } = 1$. Let $\gcd \set {a_1, a_2, \ldots, a_r} = d$. Dividing through by $d$, we have: :$\gcd \set {\dfrac {a_1} d, \dfrac {a_2} d, \ldots, \dfrac {a_d} d} = 1$ By the induction hypothesis: :$\exists m_1, m_2, \ldots, m_r \in \Z: \ds \sum_{k \mathop = 1}^r m_k a_k = 1$ whence multiplying through by $d$: :$\exists m'_1, m'_2, \ldots, m'_r \in \Z: \ds \sum_{k \mathop = 1}^r m'_k a_k = d$ where $m'_k = m_k d$ throughout. Then by hypothesis: :$\gcd \set {d, a_{r + 1} } = 1$ From the basis for the induction: :$\exists p, q \in \Z: p d + 1 a_{r + 1} = 1$ We take: :$m_1 = m'_1 p, m_2 = m'_2 p, \ldots, m_r = m'_r p$ and: :$m_{r + 1} = q$ and thus we have: :$\gcd \set {a_1, a_2, \ldots, a_r, a_{r + 1} } = 1 \iff \exists m_1, m_2, \ldots, m_r, m_{r + 1} \in \Z: \ds \sum_{k \mathop = 1}^{r + 1} m_k a_k = 1$ {{qed}}
{{begin-eqn}} {{eqn | l = \map \Beta {x, y}       | r = \dfrac {x + y} y \map \Beta {x, y + 1}       | c = Beta Function of x with y+1 by x+y over y }} {{eqn | r = \dfrac {\paren {x + y}^{\overline {m + 1} } } {y^{\overline {m + 1} } } \map \Beta {x, y + m + 1}       | c =  }} {{end-eqn}} Also: {{begin-eqn}} {{eqn | l = \map \Beta {y, m + 1}       | r = \map \Beta {y, m} \dfrac m {y + m}       | c = Beta Function of x with y+1 by x+y over y }} {{eqn | r = \dfrac {m!} {\paren {y + 1}^{\overline m} } \map \Beta {y, 1}       | c =  }} {{eqn | r = \dfrac {m!} {y^{\overline {m + 1} } }       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = \map \Beta {x + y, m + 1}       | r = \map \Beta {x + y, m} \dfrac m {x + y + m}       | c = Beta Function of x with y+1 by x+y over y }} {{eqn | r = \dfrac {m!} {\paren {x + y}^{\overline {m + 1} } }       | c =  }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = \map \Beta {x, y}       | r = \dfrac {\map \Beta {y, m + 1} } {\map \Beta {x + y, m + 1} } \map \Beta {x, y + m + 1}       | c =  }} {{eqn | r = \dfrac {\map {\Gamma_m} y} {m^y} \dfrac {m^{x + y} } {\map {\Gamma_m} {x + y} } \map \Beta {x, y + m + 1}       | c =  }} {{eqn | r = \dfrac {\map {\Gamma_m} y m^x} {\map {\Gamma_m} {x + y} } \map \Beta {x, y + m + 1}       | c =  }} {{end-eqn}} {{qed}} {{MissingLinks|Justification for certain of the steps above}}
From Beta Function of x with y+m+1: :$\map \Beta {x, y} = \dfrac {\map {\Gamma_m} y \, m^x} {\map {\Gamma_m} {x + y} } \map \Beta {x, y + m + 1}$ where $\Gamma_m$ is the partial Gamma function: :$\map {\Gamma_m} y := \dfrac {m^y m!} {y \paren {y + 1} \paren {y + 2} \dotsm \paren {y + m} }$ From Partial Gamma Function expressed as Integral: {{begin-eqn}} {{eqn | l = \map {\Gamma_m} x       | r = m^x \int_0^1 \paren {1 - t}^m t^{x - 1} \rd t       | c =  }} {{eqn | r = m^x \, \map \Beta {x, m + 1}       | c =  }} {{end-eqn}} Thus: :$\displaystyle \lim_{m \mathop \to \infty} m^x \, \map \Beta {x, m + 1} = \map \Gamma x$ As $m^x$ is monotone, it does not matter if $m$ is integer or real. Thus: :$\displaystyle \lim_{m \mathop \to \infty} \paren {m + y}^x \, \map \Beta {x, m + y + 1} = \map \Gamma x$ Hence: {{begin-eqn}} {{eqn | l = \map \Beta {x, y}       | r = \lim_{m \mathop \to \infty} \dfrac {\map {\Gamma_m} y \, m^x} {\map {\Gamma_m} {x + y} } \map \Beta {x, y + m + 1}       | c =  }} {{eqn | r = \lim_{m \mathop \to \infty} \dfrac {\map {\Gamma_m} y \, m^x} {\map {\Gamma_m} {x + y} } \frac {\map \Gamma x} {\paren {m + y}^x}       | c =  }} {{eqn | r = \lim_{m \mathop \to \infty} \dfrac {\map {\Gamma_m} y \, \map \Gamma x} {\map {\Gamma_m} {x + y} }       | c = as $\displaystyle \lim_{m \mathop \to \infty} \frac {m^x} {\paren {m + y}^x} = 1$ }} {{eqn | r = \dfrac {\map \Gamma y \, \map \Gamma x} {\map \Gamma {x + y} }       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \dbinom r k       | r = \dfrac {r!} {k! \, \left({r - k}\right)!}       | c = {{Defof|Binomial Coefficient/Integers|Binomial Coefficient|index = 1}} }} {{eqn | r = \dfrac {\Gamma \left({r + 1}\right)} {\Gamma \left({k + 1}\right) \Gamma \left({r - k + 1}\right)}       | c = Gamma Function Extends Factorial }} {{eqn | r = \dfrac {\Gamma \left({r + 2}\right)} {r + 1} \dfrac 1 {\Gamma \left({k + 1}\right) \Gamma \left({r - k + 1}\right)}       | c = Gamma Difference Equation }} {{eqn | r = \dfrac 1 {r + 1} \dfrac {\Gamma \left({r + 2}\right)} {\Gamma \left({k + 1}\right) \Gamma \left({r - k + 1}\right)}       | c = rearranging }} {{eqn | r = \dfrac 1 {r + 1} \dfrac 1 {B \left({k + 1, r - k + 1}\right)}       | c = {{Defof|Beta Function|index = 3}} }} {{end-eqn}} {{qed}}
From the definition of the binomial coefficient: :$\dbinom z w := \displaystyle \lim_{\zeta \mathop \to z} \lim_{\omega \mathop \to w} \dfrac {\map \Gamma {\zeta + 1} } {\map \Gamma {\omega + 1} \, \map \Gamma {\zeta - \omega + 1} }$ where $\Gamma$ denotes the Gamma function. {{begin-eqn}} {{eqn | l = \dbinom z w       | r = \lim_{\zeta \mathop \to z} \lim_{\omega \mathop \to w} \dfrac {\map \Gamma {\zeta + 1} } {\map \Gamma {\omega + 1} \, \map \Gamma {\zeta - \omega + 1} }       | c =  }} {{eqn | r = \lim_{\zeta \mathop \to z} \lim_{\omega \mathop \to w} \dfrac {\map \Gamma {\zeta + 1} } {\map \Gamma {\zeta - \omega + 1} \, \map \Gamma {\omega + 1} }       | c =  }} {{eqn | r = \lim_{\zeta \mathop \to z} \lim_{\omega \mathop \to w} \dfrac {\map \Gamma {\zeta + 1} } {\map \Gamma {\zeta - \omega + 1} \, \map \Gamma {\zeta - \paren {\zeta - \omega} + 1} }       | c =  }} {{eqn | r = \dbinom z {z - w}       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \dbinom z w       | r = \lim_{\zeta \mathop \to z} \lim_{\omega \mathop \to w} \dfrac {\Gamma \left({\zeta + 1}\right)} {\Gamma \left({\omega + 1}\right) \Gamma \left({\zeta - \omega + 1}\right)}       | c = {{Defof|Binomial Coefficient/Complex Numbers|Binomial Coefficient}} }} {{eqn | r = \lim_{\zeta \mathop \to z} \lim_{\omega \mathop \to w} \dfrac {\zeta \Gamma \left({\zeta}\right)} {\omega \Gamma \left({\omega}\right) \Gamma \left({\zeta - \omega + 1}\right)}       | c = Gamma Difference Equation }} {{eqn | r = \lim_{\zeta \mathop \to z} \lim_{\omega \mathop \to w} \dfrac \zeta \omega \dfrac {\Gamma \left({\left({\zeta - 1}\right) + 1}\right)} {\omega \Gamma \left({\left({\omega - 1}\right) + 1}\right) \Gamma \left({\left({\zeta - 1}\right) - \left({\omega - 1}\right) + 1}\right)}       | c = rearrangement }} {{eqn | r = \dfrac z w \dbinom {z - 1} {w - 1}       | c = {{Defof|Binomial Coefficient/Complex Numbers|Binomial Coefficient}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \binom z {w - 1} + \binom z w       | r = \lim_{\zeta \mathop \to z} \lim_{\omega \mathop \to w} \dfrac {\Gamma \left({\zeta + 1}\right)} {\Gamma \left({\omega}\right) \Gamma \left({\zeta - \omega + 2}\right)} + \lim_{\zeta \mathop \to z} \lim_{\omega \mathop \to w} \dfrac {\Gamma \left({\zeta + 1}\right)} {\Gamma \left({\omega + 1}\right) \Gamma \left({\zeta - \omega + 1}\right)}       | c = {{Defof|Binomial Coefficient/Complex Numbers|Binomial Coefficient}} }} {{eqn | r = \lim_{\zeta \mathop \to z} \lim_{\omega \mathop \to w} \left({\dfrac {\Gamma \left({\zeta + 1}\right)} {\Gamma \left({\omega}\right) \Gamma \left({\zeta - \omega + 2}\right)} + \dfrac {\Gamma \left({\zeta + 1}\right)} {\Gamma \left({\omega + 1}\right) \Gamma \left({\zeta - \omega + 1}\right)} }\right)       | c = Sum Rule for Limits of Functions }} {{eqn | r = \lim_{\zeta \mathop \to z} \lim_{\omega \mathop \to w} \left({\dfrac {\Gamma \left({\zeta + 1}\right)} {\Gamma \left({\omega}\right) \left({\zeta - \omega + 1}\right) \Gamma \left({\zeta - \omega + 1}\right)} + \dfrac {\Gamma \left({\zeta + 1}\right)} {\omega \Gamma \left({\omega}\right) \Gamma \left({\zeta - \omega + 1}\right)} }\right)       | c = Gamma Difference Equation }} {{eqn | r = \lim_{\zeta \mathop \to z} \lim_{\omega \mathop \to w} \left({\dfrac {\omega \Gamma \left({\zeta + 1}\right) +  \left({\zeta - \omega + 1}\right) \Gamma \left({\zeta + 1}\right)} {\omega \Gamma \left({\omega}\right) \left({\zeta - \omega + 1}\right) \Gamma \left({\zeta - \omega + 1}\right)} }\right)       | c = common demonimator }} {{eqn | r = \lim_{\zeta \mathop \to z} \lim_{\omega \mathop \to w} \left({\dfrac {\omega \Gamma \left({\zeta + 1}\right) +  \left({\zeta - \omega + 1}\right) \Gamma \left({\zeta + 1}\right)} {\Gamma \left({\omega + 1}\right) \Gamma \left({\zeta - \omega + 2}\right)} }\right)       | c = Gamma Difference Equation }} {{eqn | r = \lim_{\zeta \mathop \to z} \lim_{\omega \mathop \to w} \left({\dfrac {\Gamma \left({\zeta + 1}\right) \left({\zeta - \omega + 1 + \omega}\right)} {\Gamma \left({\omega + 1}\right) \Gamma \left({\zeta - \omega + 2}\right)} }\right)       | c =  }} {{eqn | r = \lim_{\zeta \mathop \to z} \lim_{\omega \mathop \to w} \left({\dfrac {\Gamma \left({\zeta + 1}\right) \left({\zeta + 1}\right)} {\Gamma \left({\omega + 1}\right) \Gamma \left({\zeta - \omega + 2}\right)} }\right)       | c =  }} {{eqn | r = \lim_{\zeta \mathop \to z} \lim_{\omega \mathop \to w} \left({\dfrac {\Gamma \left({\zeta + 2}\right)} {\Gamma \left({\omega + 1}\right) \Gamma \left({\zeta - \omega + 2}\right)} }\right)       | c = Gamma Difference Equation }} {{eqn | r = \binom {z + 1} w       | c = {{Defof|Binomial Coefficient/Complex Numbers|Binomial Coefficient}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \dbinom z t \dbinom t w       | r = \displaystyle \lim_{\zeta \mathop \to z} \lim_{\tau \mathop \to t} \dfrac {\map \Gamma {\zeta + 1} } {\map \Gamma {\tau + 1} \map \Gamma {\zeta - \tau + 1} } \displaystyle \lim_{\tau \mathop \to t} \lim_{\omega \mathop \to w} \dfrac {\map \Gamma {\tau + 1} } {\map \Gamma {\omega + 1} \map \Gamma {\tau - \omega + 1} }       | c = {{Defof|Binomial Coefficient|subdef = Complex Numbers}} }} {{eqn | r = \displaystyle \lim_{\zeta \mathop \to z} \lim_{\tau \mathop \to t} \paren {\dfrac {\map \Gamma {\zeta + 1} } {\map \Gamma {\tau + 1} \map \Gamma {\zeta - \tau + 1} } \dfrac {\map \Gamma {\tau + 1} } {\map \Gamma {\omega + 1} \map \Gamma {\tau - \omega + 1} } }       | c = Product Rule for Limits of Functions }} {{eqn | r = \displaystyle \lim_{\zeta \mathop \to z} \lim_{\tau \mathop \to t} \paren {\dfrac {\map \Gamma {\zeta + 1} } {\map \Gamma {\zeta - \tau + 1} \map \Gamma {\omega + 1} \map \Gamma {\tau - \omega + 1} } }       | c =  }} {{eqn | r = \displaystyle \lim_{\zeta \mathop \to z} \lim_{\tau \mathop \to t} \paren {\dfrac {\map \Gamma {\zeta + 1} \map \Gamma {\zeta - \omega + 1} } {\map \Gamma {\omega + 1} \map \Gamma {\zeta - \omega + 1} \map \Gamma {\zeta - \tau + 1} \map \Gamma {\tau - \omega + 1} } }       | c =  }} {{eqn | r = \binom z w \binom {z - w} {t - w}       | c = {{Defof|Binomial Coefficient|subdef = Complex Numbers}} }} {{end-eqn}} {{qed}}
By definition of the Beta function: :$\displaystyle \Beta \left({x, y}\right) := 2 \int_0^{\pi / 2} \left({\sin \theta}\right)^{2x - 1} \left({\cos \theta}\right)^{2y - 1} \rd \theta$ Thus: {{begin-eqn}} {{eqn | l = \Beta \left({\dfrac 1 2, \dfrac 1 2}\right)       | r = 2 \int_0^{\pi / 2} \left({\sin \theta}\right)^{2 \times \frac 1 2 - 1} \left({\cos \theta}\right)^{2 \times \frac 1 2 - 1} \rd \theta       | c =  }} {{eqn | r = 2 \int_0^{\pi / 2} \left({\sin \theta}\right)^0 \left({\cos \theta}\right)^0 \rd \theta       | c =  }} {{eqn | r = 2 \int_0^{\pi / 2} \rd \theta       | c =  }} {{eqn | r = 2 \left[{\theta}\right]_0^{\pi / 2}       | c =  }} {{eqn | r = 2 \left({\pi / 2 - 0}\right)       | c =  }} {{eqn | r = \pi       | c =  }} {{end-eqn}} {{qed}}
By definition of the Beta function: :$\displaystyle \Beta \left({x, y}\right) := \int_{\mathop \to 0}^{\mathop \to 1} t^{x - 1} \left({1 - t}\right)^{y - 1} \rd t$ Thus: {{begin-eqn}} {{eqn | l = \Beta \left({\dfrac 1 2, \dfrac 1 2}\right)       | r = \int_{\mathop \to 0}^{\mathop \to 1} t^{\frac 1 2 - 1} \left({1 - t}\right)^{\frac 1 2 - 1} \rd t       | c =  }} {{eqn | r = \int_{\mathop \to 0}^{\mathop \to 1} \dfrac {\rd t} {\sqrt {t \left({1 - t}\right)} }       | c =  }} {{end-eqn}} Let $t = \sin^2 \theta$. Then: :$\rd t = 2 \sin \theta \cos \theta \rd \theta$ and: {{begin-eqn}} {{eqn | l = t       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \sin^2 \theta       | r = 0       | c =  }} {{eqn | l = t       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = \sin^2 \theta       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = \theta       | r = \pi / 2       | c =  }} {{end-eqn}} and so: {{begin-eqn}} {{eqn | l = \Beta \left({\dfrac 1 2, \dfrac 1 2}\right)       | r = \int_{\mathop \to 0}^{\mathop \to \pi / 2} \frac {2 \sin \theta \cos \theta \rd \theta} {\sin \theta \sqrt {1 - \sin^2 \theta} }       | c = Integration by Substitution }} {{eqn | r = \int_{\mathop \to 0}^{\mathop \to \pi / 2} \frac {2 \sin \theta \cos \theta \rd \theta} {\sin \theta \cos \theta}       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = \int_{\mathop \to 0}^{\mathop \to \pi / 2} 2 \rd \theta       | c =  }} {{eqn | r = \left[{2 \theta}\right]_0^{\pi / 2}       | c =  }} {{eqn | r = 2 \times \pi / 2 - 0       | c =  }} {{eqn | r = \pi       | c =  }} {{end-eqn}} {{qed}}
By definition of the Beta function: :$\displaystyle \map \Beta {x, y} := \int_{\mathop \to 0}^{\mathop \to 1} t^{x - 1} \paren {1 - t}^{y - 1} \rd t$ Thus: {{begin-eqn}} {{eqn | l = \map \Beta {\dfrac 1 2, \dfrac 1 2}       | r = \int_{\mathop \to 0}^{\mathop \to 1} t^{\frac 1 2 - 1} \paren {1 - t}^{\frac 1 2 - 1} \rd t       | c =  }} {{eqn | r = \int_{\mathop \to 0}^{\mathop \to 1} \dfrac {\rd t} {t^{1/2} \paren {1 - t}^{1/2} }       | c =  }} {{end-eqn}} Let $t = u^2$. Then: :$\rd t = 2 u \rd u$ and: {{begin-eqn}} {{eqn | l = t       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = u^2       | r = 0       | c =  }} {{eqn | l = t       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = u^2       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = u       | r = 1       | c =  }} {{end-eqn}} and so: {{begin-eqn}} {{eqn | l = \map \Beta {\dfrac 1 2, \dfrac 1 2}       | r = \int_{\mathop \to 0}^{\mathop \to 1} \frac {2 u \rd u} {u \paren {1 - u^2}^{1/2} }       | c = Integration by Substitution }} {{eqn | r = 2 \int_{\mathop \to 0}^{\mathop \to 1} \frac {\rd u} {\sqrt {\paren {1 - u^2} } }       | c =  }} {{eqn | r = 2 \bigintlimits {\arcsin u} 0 1       | c = Primitive of $\dfrac 1 {\sqrt {1 - u^2} }$ }} {{eqn | r = 2 \times \pi / 2 - 0       | c =  }} {{eqn | r = \pi       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \dbinom r {1 / 2}       | r = \lim_{\rho \mathop \to r} \dfrac {\Gamma \left({\rho + 1}\right)} {\Gamma \left({\frac 1 2 + 1}\right) \Gamma \left({\rho - \frac 1 2 + 1}\right)}       | c = {{Defof|Binomial Coefficient/Complex Numbers|Binomial Coefficient}} }} {{eqn | r = \lim_{\rho \mathop \to r} \dfrac {\Gamma \left({\rho + 1}\right)} {\frac 1 2 \Gamma \left({\frac 1 2}\right) \Gamma \left({\rho + \frac 1 2}\right)}       | c = Gamma Difference Equation }} {{eqn | r = \lim_{\rho \mathop \to r} \dfrac {2 \Gamma \left({\rho + 1}\right)} {\sqrt \pi \, \Gamma \left({\rho + \frac 1 2}\right)}       | c = Gamma Function of One Half }} {{end-eqn}} Legendre's Duplication Formula gives: :$\forall z \notin \left\{{-\dfrac n 2: n \in \N}\right\}: \Gamma \left({z}\right) \Gamma \left (z + \dfrac 1 2 \right) = 2^{1 - 2 z} \sqrt \pi \ \Gamma \left({2 z}\right)$ and so: :$(1): \quad \Gamma \left ({\rho + \dfrac 1 2}\right) = \dfrac {2^{1 - 2 \rho} \sqrt \pi \ \Gamma \left({2 \rho}\right)} {\Gamma \left({\rho}\right)}$ Hence: {{begin-eqn}} {{eqn | l = \dbinom r {1 / 2}       | r = \lim_{\rho \mathop \to r} \dfrac {2 \Gamma \left({\rho + 1}\right)} {\sqrt \pi \, \dfrac {2^{1 - 2 \rho} \sqrt \pi \ \Gamma \left({2 \rho}\right)} {\Gamma \left({\rho}\right)} }       | c = from $(1)$ }} {{eqn | r = \lim_{\rho \mathop \to r} \dfrac {2 \Gamma \left({\rho + 1}\right) \Gamma \left({\rho}\right)} {\pi 2^{1 - 2 \rho} \Gamma \left({2 \rho}\right)}       | c =  }} {{eqn | r = \lim_{\rho \mathop \to r} \dfrac {2^{2 \rho - 1} 2 \Gamma \left({\rho + 1}\right) \Gamma \left({\rho}\right)} {\pi \Gamma \left({2 \rho}\right)}       | c = rearranging and simplifying }} {{eqn | r = \lim_{\rho \mathop \to r} \dfrac {2^{2 \rho} } {\pi \dfrac {\Gamma \left({2 \rho}\right)} {\Gamma \left({\rho}\right) \Gamma \left({\rho + 1}\right)} }       | c = rearranging }} {{eqn | r = \lim_{\rho \mathop \to r} \dfrac {2^{2 \rho} } {\pi \dfrac {\rho \Gamma \left({2 \rho + 1}\right)} {2 \rho \Gamma \left({\rho + 1}\right) \Gamma \left({\rho + 1}\right)} }       | c = Gamma Difference Equation twice }} {{eqn | r = \lim_{\rho \mathop \to r} \dfrac {2^{2 \rho + 1} } {\pi \dfrac {\Gamma \left({2 \rho + 1}\right)} {\Gamma \left({\rho + 1}\right) \Gamma \left({\rho + 1}\right)} }       | c = simplifying }} {{eqn | r = \dfrac {2^{2 r + 1} } {\dbinom {2 r} r \pi}       | c = {{Defof|Binomial Coefficient/Complex Numbers|Binomial Coefficient}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \lim_{r \mathop \to \infty} \frac {\dbinom r k} {r^k}       | r = \lim_{r \mathop \to \infty} \frac {\map \Gamma {r + 1} } {\map \Gamma {k + 1} \map \Gamma {r - k + 1} r^k}       | c = Gamma Function Extends Factorial }} {{eqn | r = \lim_{r \mathop \to \infty} \frac 1 {\map \Gamma {k + 1} } \frac {\sqrt {2 \pi r} \paren {r / e}^r} {\sqrt {2 \pi \paren {r - k} } \paren {\paren {r - k} / e}^{r - k} r^k}       | c = Stirling's Formula for Gamma Function; first term }} {{eqn | r = \lim_{r \mathop \to \infty} \frac 1 {\map \Gamma {k + 1} } \sqrt {\frac r {r - k} } \paren {\frac r e}^r \paren {\frac e {r - k} }^{r - k} \frac 1 {r^k}       | c =  }} {{eqn | r = \lim_{r \mathop \to \infty} \frac 1 {\map \Gamma {k + 1} } \sqrt {\frac r {r - k} } \frac 1 {e^k} \frac {r^{r - k} } {\paren {r - k}^{r - k} }       | c =  }} {{eqn | r = \lim_{r \mathop \to \infty} \frac 1 {\map \Gamma {k + 1} } \sqrt {\frac r {r - k} } \frac 1 {e^k} \frac {\paren {1 - k / r}^k} {\paren {1 - k / r}^r}       | c =  }} {{eqn | r = \frac 1 {\map \Gamma {k + 1} }       | c =  }} {{end-eqn}} The last equality is justified, since as $r \to \infty$: {{begin-eqn}} {{eqn | l = \sqrt {\frac r {r - k} }       | o = \to       | r = 1 }} {{eqn | l = \paren {1 - k / r}^k       | o = \to       | r = 1 }} {{eqn | l = \paren {1 - k / r}^r       | o = \to       | r = e^{-k}       | c = from definition of Euler's Number $e$ }} {{end-eqn}} {{qed}}
It can be assumed that both $x$ and $y$ are integers. {{begin-eqn}} {{eqn | l = \dbinom {x + y} y       | r = \dfrac {\paren {x + y}!} {x! \, y!}       | c =  }} {{eqn | ll= \leadsto       | l = \displaystyle \lim_{x, y \mathop \to \infty} \dbinom {x + y} y       | r = \dfrac {\sqrt {2 \pi \paren {x + y} } \paren {\dfrac {x + y} e}^{\paren {x + y} } } {\sqrt {2 \pi x} \paren {\dfrac x e}^x \, \sqrt {2 \pi y} \paren {\dfrac y e}^y}       | c = Stirling's Formula }} {{eqn | r = \dfrac 1 {\sqrt {2 \pi} } \dfrac {\paren {\frac 1 e}^{x + y} } {\paren {\frac 1 e}^x \paren {\frac 1 e}^y} \sqrt {\dfrac {x + y} {x y} } \dfrac {\paren {x + y}^x \paren {x + y}^y} {x^x \, y^y}       | c = rearrangement }} {{eqn | r = \dfrac 1 {\sqrt {2 \pi} } \sqrt {\frac 1 x + \frac 1 y} \paren {1 + \dfrac y x}^x \paren {1 + \dfrac x y}^y       | c = simplifying }} {{eqn | r = \sqrt {\dfrac 1 {2 \pi} \paren {\frac 1 x + \frac 1 y} } \paren {1 + \dfrac y x}^x \paren {1 + \dfrac x y}^y       | c =  }} {{end-eqn}} {{qed}}
From Approximation to x+y Choose y: :$\displaystyle \lim_{x, y \mathop \to \infty} \dbinom {x + y} y = \sqrt {\dfrac 1 {2 \pi} \paren {\frac 1 x + \frac 1 y} } \paren {1 + \dfrac y x}^x \paren {1 + \dfrac x y}^y$ Thus: {{begin-eqn}} {{eqn | l = \lim_{n \mathop \to \infty} \dbinom {2 n} n       | r = \sqrt {\dfrac 1 {2 \pi} \paren {\frac 1 n + \frac 1 n} } \paren {1 + \dfrac n n}^n \paren {1 + \dfrac n n}^n       | c =  }} {{eqn | r = \sqrt {\dfrac 1 {2 \pi} \paren {\frac 2 n} } \paren {1 + 1}^n \paren {1 + 1}^n       | c =  }} {{eqn | r = \sqrt {\dfrac 1 {n \pi} } 2^{2 n}       | c =  }} {{eqn | r = \dfrac {4^n} {\sqrt {n \pi} }       | c =  }} {{end-eqn}} {{qed}}
We have that $\phi$ is a provable consequence of $\mathcal F$. Hence it is a theorem of $\mathscr P (\mathcal F)$, the proof system obtained from $\mathscr P$ by adding all of $\mathcal F$ as axioms. Now in the formal proof of $\phi$ in $\mathscr P (\mathcal F)$, both axioms and rules of inference are used. Each rule of inference of $\mathscr P (\mathcal F)$ is also a rule of inference of $\mathscr P$. Similarly, by construction, each axiom of $\mathscr P (\mathcal F)$ is either an axiom of $\mathscr P$ or an element of $\mathcal F$. But the elements of $\mathcal F$ are theorems of $\mathscr P$, each of which thus has a formal proof in $\mathscr P$. By definition, the rules of inference of a proof system do not distinguish between theorems and axioms. Therefore the formal proofs of the following can be combined: :$\vdash_{\mathscr P} \mathcal F$ :$\vdash_{\mathscr P (\mathcal F)} \phi$ by letting the latter follow the former. This yields a new formal proof, which is entirely formulated in $\mathscr P$. This is the desired formal proof of $\phi$ from $\mathscr P$, and we conclude: :$\vdash_{\mathscr P} \phi$ {{qed}} Category:Formal Systems hxt2j7f4evu5cyiu6zmbl4f6gmmagd2