hcju/proofwikiqa huggingface dataset:
The proof proceeds by induction. For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition: :$\displaystyle {n + 1 \brace 2} = 2^n - 1$ === Basis for the Induction === $\map P 0$ is the case: {{begin-eqn}} {{eqn | l = {1 \brace 2}       | r = \delta_{1 2}       | c = Stirling Number of the Second Kind of 1 }} {{eqn | r = 0       | c = {{Defof|Kronecker Delta}} }} {{eqn | r = 2^0 - 1       | c = {{Defof|Kronecker Delta}} }} {{end-eqn}} So the result holds for $\map P 0$. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :$\displaystyle {k + 1 \brace 2} = 2^k - 1$ from which it is to be shown that: :$\displaystyle {k + 2 \brace 2} = 2^{k + 1} - 1$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = {k + 2 \brace 2}       | r = 2 \times {k + 1 \brace 2} + {k + 1 \brace 1}       | c = {{Defof|Stirling Numbers of the Second Kind}} }} {{eqn | r = 2 \times {k + 1 \brace 2} + 1       | c = Stirling Number of the Second Kind of n+1 with 1 }} {{eqn | r = 2 \times {2^k - 1} + 1       | c = Induction Hypothesis }} {{eqn | r = 2^{k + 1} - 2 + 1       | c =  }} {{eqn | r = 2^{k + 1} - 1       | c =  }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \Z_{\ge 0}: {n + 1 \brace 2} = 2^n - 1$ {{qed}}
The proof proceeds by induction. For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition: :$\displaystyle \forall m \in \Z_{\ge 0}: \sum_k \left[{n \atop k}\right] \binom k m = \left[{n + 1 \atop m + 1}\right]$ === Basis for the Induction === $P \left({0}\right)$ is the case: {{begin-eqn}} {{eqn | l = \sum_k \left[{0 \atop k}\right] \binom k m       | r = \sum_k \delta_{0 k} \binom k m       | c = Unsigned Stirling Number of the First Kind of 0 }} {{eqn | r = \binom 0 m       | c = all terms vanish except for $k = 0$ }} {{eqn | r = \delta_{0 m}       | c = Zero Choose n }} {{eqn | r = \delta_{1 \left({m + 1}\right)}       | c = {{Defof|Kronecker Delta}} }} {{eqn | r = \left[{1 \atop m + 1}\right]       | c = Unsigned Stirling Number of the First Kind of 1 }} {{eqn | r = \left[{0 + 1 \atop m + 1}\right]       | c =  }} {{end-eqn}} So $P \left({0}\right)$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({r}\right)$ is true, where $r \ge 1$, then it logically follows that $P \left({r + 1}\right)$ is true. So this is the induction hypothesis: :$\displaystyle \sum_k \left[{r \atop k}\right] \binom k m = \left[{r + 1 \atop m + 1}\right]$ from which it is to be shown that: :$\displaystyle \sum_k \left[{r + 1 \atop k}\right] \binom k m = \left[{r + 2 \atop m + 1}\right]$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | o =        | r = \sum_k \left[{r + 1 \atop k}\right] \binom k m       | c =  }} {{eqn | r = \sum_k \left({r \left[{r \atop k}\right] + \left[{r \atop k - 1}\right]}\right) \binom k m       | c = {{Defof|Unsigned Stirling Numbers of the First Kind}} }} {{eqn | r = r \sum_k \left[{r \atop k}\right] \binom k m + \sum_k \left[{r \atop k - 1}\right] \binom k m       | c =  }} {{eqn | r = r \left[{r + 1 \atop m + 1}\right] + \sum_k \left[{r \atop k - 1}\right] \binom k m       | c = Induction Hypothesis }} {{eqn | r = r \left[{r + 1 \atop m + 1}\right] + \sum_k \left[{r \atop k - 1}\right] \left({\binom {k - 1} m + \binom {k - 1} {m - 1} }\right)       | c = Pascal's Rule }} {{eqn | r = r \left[{r + 1 \atop m + 1}\right] + \sum_k \left[{r \atop k - 1}\right] \binom {k - 1} m + \sum_k \left[{r \atop k - 1}\right] \binom {k - 1} {m - 1}       | c =  }} {{eqn | r = r \left[{r + 1 \atop m + 1}\right] + \sum_k \left[{r \atop k}\right] \binom k m + \sum_k \left[{r \atop k}\right] \binom k {m - 1}       | c = Translation of Index Variable of Summation }} {{eqn | r = r \left[{r + 1 \atop m + 1}\right] + \left[{r + 1 \atop m + 1}\right] + \left[{r + 1 \atop m}\right]       | c = Induction Hypothesis }} {{eqn | r = \left({r + 1}\right) \left[{r + 1 \atop m + 1}\right] + \left[{r + 1 \atop m}\right]       | c =  }} {{eqn | r = \left[{r + 2 \atop m + 1}\right]       | c = {{Defof|Unsigned Stirling Numbers of the First Kind}} }} {{end-eqn}} So $P \left({r}\right) \implies P \left({r + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall m, n \in \Z_{\ge 0}: \sum_k \left[{n \atop k}\right] \binom k m = \left[{n + 1 \atop m + 1}\right]$ {{qed}}
Define relation $\mathcal R$ on $S$: :$\forall x, y \in S: \left({x, y}\right) \in \mathcal R \iff x \in f\left({y}\right)$ By definition of $\mathcal R$-segment: :$\forall x \in S: f\left({x}\right) = x^{\mathcal R}$ We will prove that :$\mathcal R$ is an approximating relation. Let $x \in S$. Suppose the case holds that: :$x \preceq \sup I$ Thus: {{begin-eqn}} {{eqn | l = \sup \left({x^{\mathcal R} }\right)       | r = \sup f \left({x}\right) }} {{eqn | r = \sup \left\{ {x \wedge d: d \in I}\right\}       | c = definition of $f$ }} {{eqn | r = x \wedge \sup I       | c = {{Defof|Meet-Continuous Lattice}} }} {{eqn | r = x       | c = Preceding iff Meet equals Less Operand }} {{end-eqn}} Suppose the case holds that :$x \npreceq \sup I$ Thus {{begin-eqn}} {{eqn | l = \sup \left({x^{\mathcal R} }\right)       | r = \sup f \left({x}\right) }} {{eqn | r = \sup \left({x^\preceq}\right)       | c = definition of $f$ }} {{eqn | r = x       | c = Supremum of Lower Closure of Element }} {{end-eqn}} {{qed}}
Let $x \in S$. Define $\mathcal R := \mathord\preceq$. By definitions of lower closure of element and $\mathcal R$-segment: :$x^\preceq = x^{\mathcal R}$ where: :$x^\preceq$ denotes the lower closure of $x$ :$x^{\mathcal R}$ denotes the $\mathcal R$-segment of $x$ Thus by Supremum of Lower Closure of Element: :$x = \sup \left({x^\preceq}\right) = \sup \left({ x^{\mathcal R} }\right)$ Hence $\preceq$ is an approximating relation on $S$. {{qed}}
Proceed by the Principle of Structural Induction on the definition of term, applied to $\beta$. If $\beta = y$ for some variable $y$, then: :$\map \beta {x \gets \tau} = \begin{cases} \tau & : \text {if $y = x$} \\ y &: \text {otherwise} \end{cases}$ In either case, $\map \beta {x \gets \tau}$ is a term. If $\beta = \map f {\tau_1, \ldots, \tau_n}$ and the induction hypothesis holds for $\tau_1, \ldots, \tau_n$, then: :$\map \beta {x \gets \tau} = \map f {\map {\tau_1} {x \gets \tau}, \ldots, \map {\tau_n} {x \gets \tau} }$ By the induction hypothesis, each $\map {\tau_i} {x \gets \tau}$ is a term. Hence so is $\map \beta {x \gets \tau}$. The result follows by the Principle of Structural Induction. {{qed}}
Proceed by the Principle of Structural Induction on the definition of term, applied to $\beta$. If $\beta = y$ for some variable $y$, then: :$\beta \left({x \gets \tau}\right) = \begin{cases} \tau &: \text{if $y = x$} \\ y &: \text{otherwise} \end{cases}$ In the first case: {{begin-eqn}} {{eqn|l = \mathop{ \operatorname{val}_{\mathcal A} \left({\beta \left({x \gets \tau}\right) }\right) } \left[{\sigma}\right]      |r = \mathop{ \operatorname{val}_{\mathcal A} \left({\tau }\right) } \left[{\sigma}\right] }} {{eqn|r = a      |c = Definition of $a$ }} {{eqn|r = \left({\sigma + \left({x / a}\right)}\right) \left({x}\right)      |c = Definition of Extension of Assignment }} {{eqn|r = \mathop{ \operatorname{val}_{\mathcal A} \left({\beta }\right) } \left[{\sigma + \left({x / a}\right)}\right]      |c = Definition of value under $\sigma + \left({x / a}\right)$ }} {{end-eqn}} In the second case: {{begin-eqn}} {{eqn|l = \mathop{ \operatorname{val}_{\mathcal A} \left({\beta \left({x \gets \tau}\right) }\right) } \left[{\sigma}\right]      |r = \mathop{ \operatorname{val}_{\mathcal A} \left({y}\right) } \left[{\sigma}\right] }} {{eqn|r = \sigma \left({y}\right)      |c = Definition of value under $\sigma$ }} {{eqn|r = \left({\sigma + \left({x / a}\right)}\right) \left({y}\right)      |c = Definition of Extension of Assignment }} {{eqn|r = \mathop{ \operatorname{val}_{\mathcal A} \left({\beta }\right) } \left[{\sigma + \left({x / a}\right)}\right]      |c = Definition of value under $\sigma + \left({x / a}\right)$ }} {{end-eqn}} as desired. If $\beta = f \left({\tau_1, \ldots, \tau_n}\right)$ and the induction hypothesis holds for $\tau_1, \ldots, \tau_n$, then: :$\beta \left({x \gets \tau}\right) = f \left({ \tau_1 \left({x \gets \tau}\right), \ldots, \tau_n \left({x \gets \tau}\right) }\right)$ Now: {{begin-eqn}} {{eqn|l = \mathop{ \operatorname{val}_{\mathcal A} \left({\beta \left({x \gets \tau}\right) }\right) } \left[{\sigma}\right]      |r = \mathop{ \operatorname{val}_{\mathcal A} \left({f \left({ \tau_1 \left({x \gets \tau}\right), \ldots, \tau_n \left({x \gets \tau}\right) }\right) }\right) } \left[{\sigma}\right] }} {{eqn|r = f_{\mathcal A} \left({ \mathop{ \operatorname{val}_{\mathcal A} \left({\tau_1 \left({x \gets \tau}\right)}\right) } \left[{\sigma}\right], \ldots, \mathop{ \operatorname{val}_{\mathcal A} \left({\tau_n \left({x \gets \tau}\right)}\right) } \left[{\sigma}\right] }\right)      |c = Definition of value under $\sigma$ }} {{eqn|r = f_{\mathcal A} \left({ \mathop{ \operatorname{val}_{\mathcal A} \left({\tau_1}\right) } \left[{\sigma + \left({x / a}\right)}\right], \ldots, \mathop{ \operatorname{val}_{\mathcal A} \left({\tau_n}\right) } \left[{\sigma + \left({x / a}\right)}\right] }\right)      |c = Induction Hypothesis }} {{eqn|r = \mathop{ \operatorname{val}_{\mathcal A} \left({\beta}\right) } \left[{\sigma + \left({x / a}\right)}\right]      |c = Definition of value under $\sigma + \left({x / a}\right)$ }} {{end-eqn}} as desired. The result follows by the Principle of Structural Induction. {{qed}}
Proceed by the Principle of Structural Induction on the bottom-up specification of predicate logic, applied to $\mathbf A$. If $\mathbf A = p \left({\tau_1, \ldots, \tau_n}\right)$, then: :$\mathbf A \left({x \gets \tau}\right) = p \left({\tau_1 \left({x \gets \tau}\right), \ldots, \tau_n \left({x \gets \tau}\right)}\right)$ where $\tau_i \left({x \gets \tau}\right)$ is the substitution instance of $\tau_i$. By Substitution Instance of Term is Term, each such $\tau_i \left({x \gets \tau}\right)$ is again a term. It follows that $\mathbf A \left({x \gets \tau}\right)$ is again a WFF. If $\mathbf A = \neg \mathbf B$ and the induction hypothesis applies to $\mathbf B$, then: :$\mathbf A \left({x \gets \tau}\right) = \neg \mathbf B \left({x \gets \tau}\right)$ and it follows from the induction hypothesis that $\mathbf A \left({x \gets \tau}\right)$ is a WFF of predicate logic. If $\mathbf A = \mathbf B \circ \mathbf B'$ for $\circ$ one of $\land, \lor, \implies, \iff$ and the induction hypothesis applies to $\mathbf B, \mathbf B'$: :$\mathbf A \left({x \gets \tau}\right) = \mathbf B \left({x \gets \tau}\right) \circ \mathbf B' \left({x \gets \tau}\right)$ and it follows from the induction hypothesis that $\mathbf A \left({x \gets \tau}\right)$ is a WFF of predicate logic. If $\mathbf A = \exists x: \mathbf B$, and the induction hypothesis applies to $\mathbf B$: :$\mathbf A \left({x \gets \tau}\right) = \exists x: \mathbf B \left({x \gets \tau}\right)$ and it follows from the induction hypothesis that $\mathbf A \left({x \gets \tau}\right)$ is a WFF of predicate logic. If $\mathbf A = \forall x : \mathbf B$, and the induction hypothesis applies to $\mathbf B$: :$\mathbf A \left({x \gets \tau}\right) = \forall x: \mathbf B \left({x \gets \tau}\right)$ and it follows from the induction hypothesis that $\mathbf A \left({x \gets \tau}\right)$ is a WFF of predicate logic. The result follows by the Principle of Structural Induction. {{qed}}
In the language of symbolic logic: {{begin-eqn}} {{eqn | lo= \forall x:       | l = \map P x       | o =       | c =  }} {{eqn | ll= \therefore       | l = \map P {\mathbf a}       | o =       | c =  }} {{end-eqn}} {{qed}}
Let $\AA$ be a structure on a set $A$, and let $\sigma$ be an assignment for $\forall x: \map {\mathbf A} x \implies \map {\mathbf A} \tau$. Define: :$a_\tau := \mathop{ \operatorname{val}_\AA \left({\tau}\right) } \sqbrk \sigma$ the value of $\tau$ under $\sigma$. From the definition of value under $\sigma$: :$\map {\mathrm {val}_\AA} {\forall x: \map {\mathbf A} x \implies \map {\mathbf A} \tau} \sqbrk \sigma = \map {f^\to} {\map {\mathrm {val}_\AA} {\forall x: \map {\mathbf A} x} \sqbrk \sigma, \map {\mathrm {val}_\AA} {\map {\mathbf A} \tau} \sqbrk \sigma}$ where $f^\to$ is the truth function of $\implies$. We thus need to ascertain that if: :$\map {\mathrm {val}_\AA} {\forall x: \map {\mathbf A} x} \sqbrk \sigma = T$ then also: :$\map {\mathrm {val}_\AA} {\map {\mathbf A} \tau} \sqbrk \sigma = T$ By definition of value under $\sigma$, the former amounts to: :$\map {\mathrm {val}_\AA} {\map {\mathbf A} x} \sqbrk {\sigma + \paren {x / a} } = T$ for all $a \in A$. By the Substitution Theorem for Well-Formed Formulas: :$\map {\mathrm {val}_\AA} {\map {\mathbf A} \tau} \sqbrk \sigma = \map {\mathrm {val}_\AA} {\mathbf A} \sqbrk {\sigma + \paren {x / a_\tau} }$ Since $a_\tau \in A$, the conclusion follows, and $\forall x: \map {\mathbf A} x \implies \map {\mathbf A} \tau$ is a tautology. {{qed}}
Let $x \in S$. Define $\mathcal R := \mathord\ll$. By definitions of way below closure and $\mathcal R$-segment: :$x^\ll = x^{\mathcal R}$ where: :$x^\ll$ denotes the way below closure of $x$ :$x^{\mathcal R}$ denotes the $\mathcal R$-segment of $x$ By definition of continuous: :$L$ satisfies axiom of approximation. Thus by axiom of approximation: :$x = \sup \left({x^\ll}\right) = \sup \left({ x^{\mathcal R} }\right)$ Hence $\ll$ is an approximating relation on $S$. {{qed}}
By Intersection of Ideals with Suprema Succeed Element equals Way Below Closure of Element: :$\displaystyle \bigcap \left\{ {I \in \mathit{Ids}: x \preceq \sup I}\right\} = x^\ll$ where $\mathit{Ids}$ denotes the set of all ideals in $L$. For all $I \in \mathit{Ids}$ define a mapping $m_I: S \to \mathit{Ids}$: :$\forall x \in S: x \preceq \sup I \implies m_I\left({x}\right) = \left\{ {x \wedge i: i \in I}\right\}$ and :$\forall x \in S: x \npreceq \sup I \implies m_I\left({x}\right) = x^\preceq$ By Intersection of Applications of Down Mappings at Element equals Way Below Closure of Element: :$\forall x \in S: \displaystyle \bigcap \left\{ {m_I\left({x}\right): I \in \mathit{Ids} }\right\} = x^\ll$ We will prove that :$\left\{ {m_I\left({x}\right): I \in \mathit{Ids} }\right\} \subseteq \left\{ {x^{\mathcal R}: \mathcal R \in \mathit{App}\left({L}\right)}\right\}$ Let $a \in \left\{ {m_I\left({x}\right): I \in \mathit{Ids} }\right\}$ Then :$\exists I \in \mathit{Ids}: a = m_I\left({x}\right)$ By Down Mapping is Generated by Approximating Relation: :$\exists \mathcal R \in \mathit{App}\left({L}\right): \forall s \in S: m_I(s) = s^{\mathcal R}$ Then :$a = x^{\mathcal R}$ Thus :$a \in \left\{ {x^{\mathcal R}: \mathcal R \in \mathit{App}\left({L}\right)}\right\}$ {{qed|lemma}} By Intersection of Family is Subset of Intersection of Subset of Family: :$\displaystyle \bigcap \left\{ {x^{\mathcal R}: \mathcal R \in \mathit{App}\left({L}\right)}\right\} \subseteq x^\ll$ We will prove that :$\left\{ {x^{\mathcal R}: \mathcal R \in \mathit{App}\left({L}\right)}\right\} \subseteq \left\{ {I \in \mathit{Ids}: x \preceq \sup I}\right\}$ Let $a \in \left\{ {x^{\mathcal R}: \mathcal R \in \mathit{App}\left({L}\right)}\right\}$ Then :$\exists \mathcal R \in \mathit{App}\left({L}\right): a = x^{\mathcal R}$ By definition of approximating relation: :$x = \sup a$ By Relation Segment of Auxiliary Relation is Ideal: :$a \in \mathit{Ids}$ Thus by definition of reflexivity: :$a \in \left\{ {I \in \mathit{Ids}: x \preceq \sup I}\right\}$ {{qed|lemma}} By Intersection of Family is Subset of Intersection of Subset of Family: :$x^\ll \subseteq \displaystyle \bigcap \left\{ {x^{\mathcal R}: \mathcal R \in \mathit{App}\left({L}\right)}\right\}$ Hence the result by definition of set equality. {{qed}}
Let $\epsilon > 0$. We have to find a value of $\delta > 0$ such that $\forall x: b - \delta < x < b: \size {\map f x - L} < \epsilon$. That is, that $L - \epsilon < \map f x < L + \epsilon$. As $L$ is an upper bound for $f$ on $\openint a b$, $\map f x < L + \epsilon$ automatically happens. Since $L - \epsilon$ is ''not'' an upper bound for $f$ on $\openint a b$, it follows that: :$\exists y \in \openint a b: \map f y > L - \epsilon$ But $f$ increases on $\openint a b$. So: :$\forall x: y < x < b: L - \epsilon < \map f y \le \map f x$ We choose $\delta = b - y$ and hence the result. {{qed}}
Let $\epsilon > 0$. We have to find a value of $\delta > 0$ such that: :$\forall x: b - \delta < x < b: \size {\map f x - l} < \epsilon$ That is: :$l - \epsilon < \map f x < l + \epsilon$ As $l$ is a lower bound for $f$ on $\openint a b$: :$l - \epsilon < \map f x$ automatically happens Because $l + \epsilon$ is ''not'' a lower bound for $f$ on $\openint a b$: :$\exists y \in \openint a b: \map f y < l + \epsilon$ But $f$ decreases on $\openint a b$. So: :$\forall x: y < x < b: \map f x \le \map f y < l + \epsilon$ We choose $\delta = y - a$ and hence the result. {{qed}}
The proof proceeds by induction on $m$. For all $m \in \Z_{\ge 0}$, let $\map P n$ be the proposition: :$\displaystyle \forall n \in \Z_{\ge 0}: \sum_k \binom m k \paren {-1}^{m - k} k^n = m! {n \brace m}$ === Basis for the Induction === $\map P 0$ is the case: {{begin-eqn}} {{eqn | r = \sum_k \binom 0 k \paren {-1}^{0 - k} k^n       | o =        | c =  }} {{eqn | r = \sum_k \delta_{0 k} \paren {-1}^{- k} k^n       | c = Zero Choose n }} {{eqn | r = \paren {-1}^{- 0} 0^n       | c = all terms vanish except for $k = 0$ }} {{eqn | r = \delta_{0 n}       | c = $0^n = 0$ except when $n = 0$, as $0^0 = 1$ }} {{eqn | r = 0! {n \brace 0}       | c = {{Defof|Stirling Numbers of the Second Kind}} }} {{end-eqn}} So $\map P 0$ 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 $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true. So this is the induction hypothesis: :$\displaystyle \sum_k \binom r k \paren {-1}^{r - k} k^n = r! {n \brace r}$ from which it is to be shown that: :$\displaystyle \sum_k \binom {r + 1} k \paren {-1}^{r + 1 - k} k^n = \paren {r + 1}! {n \brace r + 1}$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | o =        | r = \sum_k \binom {r + 1} k \paren {-1}^{r + 1 - k} k^n       | c =  }} {{eqn | r = \sum_k \paren {\binom r k + \binom r {k - 1} } \paren {-1}^{r + 1 - k} k^n       | c = Pascal's Rule }} {{eqn | r = \sum_k \binom r k \paren {-1}^{r + 1 - k} k^n + \sum_k \binom r {k - 1} \paren {-1}^{r + 1 - k} k^n       | c =  }} {{eqn | r = -\sum_k \binom r k \paren {-1}^{r - k} k^n + \sum_k \binom r {k - 1} \paren {-1}^{r - \paren {k - 1} } k^n       | c =  }} {{eqn | r = -r! {n \brace r} + \sum_k \binom r {k - 1} \paren {-1}^{r - \paren {k - 1} } k^n       | c = Induction Hypothesis }} {{eqn | r = -r! {n \brace r} + \sum_k \binom r k \paren {-1}^{r - k} \paren {k + 1}^n       | c = Translation of Index Variable of Summation }} {{end-eqn}} {{finish|The next step to take is not clear.}} So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall m, n \in \Z_{\ge 0}: \sum_k \binom m k \paren {-1}^{m - k} k^n = m! {n \brace m}$ {{qed}}
The proof proceeds by induction on $m$. For all $m \in \Z_{\ge 0}$, let $\map P n$ be the proposition: :$\displaystyle \forall n \in \Z_{\ge 0}: \sum_k \binom {m - n} {m + k} \binom {m + n} {n + k} {m + k \brace k} = {n \brack n - m}$ $\map P 0$ is the case: {{begin-eqn}} {{eqn | r = \sum_k \binom {0 - n} {0 + k} \binom {0 + n} {n + k} {0 + k \brace k}       | o =        | c =  }} {{eqn | r = \sum_k \binom {- n} k \binom n {n + k}       | c = Stirling Number of the Second Kind of Number with Self }} {{eqn | r = \sum_k \binom {- n} k \delta_{0 k}       | c = as $\dbinom n {n + k} = 0$ for $k = 0$, and Binomial Coefficient with Self }} {{eqn | r = \binom {- n} 0       | c = All terms but where $k = 0$ vanish }} {{eqn | r = 1       | c = Binomial Coefficient with Zero }} {{eqn | r = {n \brack n - 0}       | c = Unsigned Stirling Number of the First Kind of Number with Self }} {{end-eqn}} So $\map P 0$ is seen to hold. === Basis for the Induction === $\map P 1$ is the case: {{finish|Needs to be established whether the case $\map P 1$ actually needs to be proved separately, or whether we can get away with using $\map P 0$ as our base case.}} Thus $\map P 1$ is seen to hold for all $m$. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $\map P r$ is true, where $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true. So this is the induction hypothesis: :$\displaystyle \sum_k \binom {r - n} {r + k} \binom {r + n} {n + k} {r + k \brace k} = {n \brack n - r}$ from which it is to be shown that: :$\displaystyle \sum_k \binom {r + 1 - n} {r + 1 + k} \binom {r + 1 + n} {n + k} {r + 1 + k \brace k} = {n \brack n - r + 1}$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | o =        | r = \sum_k \binom {r + 1 - n} {r + 1 + k} \binom {r + 1 + n} {n + k} {r + 1 + k \brace k} = {n \brack n - r + 1}       | c =  }} {{eqn | r = \sum_k \paren {\binom {r - n} {r + 1 + k} + \binom {r - n} {r + k} } \binom {r + 1 + n} {n + k} {r + 1 + k \brace k} = {n \brack n - r + 1}       | c = Pascal's Rule }} {{end-eqn}} {{finish|Lots to do, not sure if this is a workable approach.}} So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall m, n \in \Z_{\ge 0}: \sum_k \binom {m - n} {m + k} \binom {m + n} {n + k} {m + k \brace k} = {n \brack n - m}$ {{qed}}
The proof proceeds by induction on $m$. For all $m \in \Z_{\ge 0}$, let $\map P n$ be the proposition: :$\forall n \in \Z_{\ge 0}: \displaystyle \sum_k \binom {m - n} {m + k} \binom {m + n} {n + k} {m + k \brack k} = {n \brace n - m}$ === Basis for the Induction === $\map P 0$ is the case: {{begin-eqn}} {{eqn | o =        | r = \sum_k \binom {0 - n} {0 + k} \binom {0 + n} {n + k} {0 + k \brack k}       | c =  }} {{eqn | r = \sum_k \binom {- n} k \binom n {n + k}       | c = Unsigned Stirling Number of the First Kind of Number with Self }} {{eqn | r = \sum_k \binom {- n} k \delta_{0 k}       | c = as $\dbinom n {n + k} = 0$ for $k = 0$, and Binomial Coefficient with Self }} {{eqn | r = \binom {- n} 0       | c = All terms but where $k = 0$ vanish }} {{eqn | r = 1       | c = Binomial Coefficient with Zero }} {{eqn | r = {n \brace n - 0}       | c = Stirling Number of the Second Kind of Number with Self }} {{end-eqn}} So $\map P 0$ 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 $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true. So this is the induction hypothesis: :$\displaystyle \sum_k \binom {r - n} {r + k} \binom {r + n} {n + k} {r + k \brack k} = {n \brace n - r}$ from which it is to be shown that: :$\displaystyle \sum_k \binom {r + 1 - n} {r + 1 + k} \binom {r + 1 + n} {n + k} {r + 1 + k \brack k} = {n \brace n - r + 1}$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | o =        | r = \sum_k \binom {r + 1 - n} {r + 1 + k} \binom {r + 1 + n} {n + k} {r + 1 + k \brack k} = {n \brace n - r + 1}       | c =  }} {{eqn | r = \sum_k \paren {\binom {r - n} {r + 1 + k} + \binom {r - n} {r + k} } \binom {r + 1 + n} {n + k} {r + 1 + k \brack k} = {n \brace n - r + 1}       | c = Pascal's Rule }} {{end-eqn}} {{finish|Lots to do, not sure if this is a workable approach.}} So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall m, n \in \Z_{\ge 0}: \displaystyle \sum_k \binom {m - n} {m + k} \binom {m + n} {n + k} {m + k \brack k} = {n \brace n - m}$ {{qed}}