hcju/proofwikiqa huggingface dataset:
The proof proceeds by induction on $n$. For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition: :$\displaystyle \forall n \in \Z_{\ge 0}: \sum_{k \mathop \le n} {k \brack m} \frac {n!} {k!} = {n + 1 \brack m + 1}$ === Basis for the Induction === $\map P 0$ is the case: {{begin-eqn}} {{eqn | r = \sum_{k \mathop \le 0} {k \brack m} \frac {0!} {k!}       | o =        | c =  }} {{eqn | r = {0 \brack m} \frac 1 {0!}       | c = {{Defof|Unsigned Stirling Numbers of the First Kind}} }} {{eqn | r = \delta_{0 m}       | c = Unsigned Stirling Number of the First Kind of 0 }} {{eqn | r = \delta_{1 \paren {m + 1} }       | c =  }} {{eqn | r = {1 \brack m + 1}       | c = Unsigned Stirling Number of the First Kind of 1 }} {{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: :$\displaystyle \sum_{k \mathop \le r} {k \brack m} \frac {r!} {k!} = {r + 1 \brack m + 1}$ from which it is to be shown that: :$\displaystyle \sum_{k \mathop \le r + 1} {k \brack m} \frac {\paren {r + 1}!} {k!} = {r + 2 \brack m + 1}$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | o =        | r = \sum_{k \mathop \le r + 1} {k \brack m} \frac {\paren {r + 1}!} {k!}       | c =  }} {{eqn | r = {r + 1 \brack m} \frac {\paren {r + 1}!} {\paren {r + 1}!} + \paren {r + 1} \paren {\sum_{k \mathop \le r} {k \brack m} \frac {r!} {k!} } }} {{eqn | r = {r + 1 \brack m} + \paren {r + 1} {r + 1 \brack m + 1} }} {{eqn | r = {r + 2 \brack m + 1}       | c = {{Defof|Unsigned Stirling Numbers of the First Kind}} }} {{end-eqn}} So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall m, n \in \Z_{\ge 0}: \sum_{k \mathop \le n} {k \brack m} \frac {n!} {k!} = {n + 1 \brack m + 1}$ {{qed}}
{{begin-eqn}} {{eqn | l = \left\{ {m_I\left({x}\right): I \in \mathit{Ids} }\right\}       | r = \left\{ {m_I\left({x}\right): I \in \mathit{Ids} \land x \preceq \sup I}\right\} \cup \left\{ {m_I\left({x}\right): I \in \mathit{Ids} \land x \npreceq \sup I}\right\} }} {{eqn | r = \left\{ {\left\{ {x \wedge i: i \in I}\right\}: I \in \mathit{Ids} \land x \preceq \sup I}\right\} \cup \left\{ {x^\preceq: I \in \mathit{Ids} \land x \npreceq \sup I}\right\}       | c = definition of $m_I$ }} {{eqn | r = \left\{ {\left({x^\preceq}\right) \cap I: I \in \mathit{Ids} \land x \preceq \sup I}\right\} \cup \left\{ {x^\preceq: I \in \mathit{Ids} \land x \npreceq \sup I}\right\}       | c = Intersection of Lower Closure of Element with Ideal equals Meet of Element and Ideal }} {{end-eqn}} Thus {{begin-eqn}} {{eqn | l = \bigcap \left\{ {m_I\left({x}\right): I \in \mathit{Ids} }\right\}       | r = \bigcap \left\{ {\left({x^\preceq}\right) \cap I: I \in \mathit{Ids} \land x \preceq \sup I}\right\} \cap \bigcap \left\{ {x^\preceq: I \in \mathit{Ids} \land x \npreceq \sup I}\right\}       | c =  }} {{eqn | r = \bigcap \left\{ {\left({x^\preceq}\right) \cap I: I \in \mathit{Ids} \land x \preceq \sup I}\right\} \cap \bigcap \left\{ {x^\preceq}\right\} }} {{eqn | r = \bigcap \left\{ {\left({x^\preceq}\right) \cap I: I \in \mathit{Ids} \land x \preceq \sup I}\right\} \cap x^\preceq       | c = Intersection of Singleton }} {{eqn | r = \left({x^\preceq}\right) \cap \bigcap \left\{ {I: I \in \mathit{Ids} \land x \preceq \sup I}\right\} \cap x^\preceq       | c = Intersection Distributes over Intersection }} {{eqn | r = \left({x^\preceq}\right) \cap \left({x^\ll}\right) \cap x^\preceq       | c = Intersection of Ideals with Suprema Succeed Element equals Way Below Closure of Element }} {{eqn | r = x^\ll       | c = Intersection with Subset is Subset as $x^\ll \subseteq x^\preceq$ by Way Below implies Preceding }} {{end-eqn}} {{qed}}
The proof proceeds by induction on $n$. For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition: :$\displaystyle \forall n \in \Z_{\ge 0}: \displaystyle \sum_{k \mathop \le n} {k \brace m} \paren {m + 1}^{n - k} = {{n + 1} \brace {m + 1}}$ === Basis for the Induction === $\map P 0$ is the case: {{begin-eqn}} {{eqn | r = \sum_{k \mathop \le 0} {k \brace m} \paren {m + 1}^{0 - k}       | o =        | c =  }} {{eqn | r = {0 \brace m} \paren {m + 1}^{0 - 0} }} {{eqn | r = \delta_{0 m}       | c = {{Defof|Stirling Numbers of the Second Kind}} }} {{eqn | r = \delta_{1 \paren {m + 1} } }} {{eqn | r = {1 \brace {m + 1} }       | c = Stirling Number of the Second Kind of 1 }} {{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: :$\displaystyle \sum_{k \mathop \le r} {k \brace m} \paren {m + 1}^{r - k} = {{r + 1} \brace {m + 1}}$ from which it is to be shown that: :$\displaystyle \sum_{k \mathop \le {r + 1} } {k \brace m} \paren {m + 1}^{r + 1 - k} = { {r + 2} \brace {m + 1} }$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | o =       | r = \sum_{k \mathop \le {r + 1} } {k \brace m} \paren {m + 1}^{r + 1 - k} }} {{eqn | r = { {r + 1} \brace m} \paren {m + 1}^{r + 1 - r - 1} + \paren {m + 1} \paren {\sum_{k \mathop \le r} {k \brace m} \paren {m + 1}^{r - k} }       | c =  }} {{eqn | r = { {r + 1} \brace m} + \paren {m + 1} { {r + 1} \brace {m + 1} } }} {{eqn | r = { {r + 2} \brace {m + 1} }       | c = {{Defof|Stirling Numbers of the Second Kind}} }} {{end-eqn}} So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \sum_{k \mathop \le n} {k \brace m} \paren {m + 1}^{n - k} = {{n + 1} \brace {m + 1}}$ {{qed}}
By Binomial Coefficient with Zero: :$\forall r \in \R: \dbinom r 0 = 1$ Hence directly: :$\dbinom 0 0 = 1$
By Zero Choose n: :$\dbinom 0 n = \delta_{0 n}$ where: :$\dbinom 0 n$ denotes a binomial coefficient :$\delta_{0 n}$ denotes the Kronecker delta. Hence directly: :$\dbinom 0 0 = 1$
The total number of cards in a standard deck is $52$. The number of cards in a single bridge hand is $13$. Thus $N$ is equal to the number of ways $13$ things can be chosen from $52$. Thus: {{begin-eqn}} {{eqn | l = N       | r = \dbinom {52} {23}       | c = Cardinality of Set of Subsets }} {{eqn | r = \frac {52!} {13! \left({52 - 13}\right)!}       | c = {{Defof|Binomial Coefficient}} }} {{eqn | r = \frac {52!} {13! \, 39!}       | c =  }} {{eqn | r = 635 \ 013 \ 559 \ 600       | c = after calculation }} {{end-eqn}} {{qed}}
=== First Inclusion === Let $z \in \paren {x^\preceq} \cap I$ By definition of intersection: :$z \in x^\preceq$ and $z \in I$ By definition of lower closure of element: :$z \preceq x$ By Preceding iff Meet equals Less Operand: :$x \wedge z = z$ Thus: :$z \in \set {x \wedge i: i \in I}$ {{qed|lemma}} === Second Inclusion === Let $z \in \set {x \wedge i: i \in I}$ Then :$\exists i \in I: z = x \wedge i$ By Meet Precedes Operands: :$z \preceq x$ and $z \preceq i$ By definition of ideal in ordered set: :$I$ is a lower set. By definitions of lower set and lower closure of element: :$z \in I$ and $z \in x^\preceq$ Thus by definition of intersection: :$z \in \struct {x^\preceq} \cap I$ {{qed|lemma}} Thus the result by definition of set equality. {{qed}}
From Condition for Increasing Binomial Coefficients: :$\forall k \in \Z_{> 0}: \dbinom n k > \dbinom n {k - 1} \iff k \le \dfrac n 2$ Let $n$ be an even integer, such that: :$n = 2 r$ Then: :$r = \floor {\dfrac n 2} = \ceiling {\dfrac n 2}$ and: :$\forall k < r: \dbinom n r > \dbinom n k$ From Symmetry Rule for Binomial Coefficients: :$\forall k < r: \dbinom n r > \dbinom n {n - k}$ and so: :$\forall k > r: \dbinom n r > \dbinom n k$ We also have: :$\dbinom n r = \dbinom n {n - r} = \dbinom n {n / 2}$ Thus for all $k \ne r$: :$\dbinom n k < \dbinom n {n / 2}$ and so: :$k_\max = \floor {\dfrac n 2} = \ceiling {\dfrac n 2}$ {{qed|lemma}} Let $n$ be an odd integer, such that: :$n = 2 r + 1$ Then: :$r = \floor {\dfrac n 2}$ As $r < n$: :$\forall k < r: \dbinom n r > \dbinom n k$ When $k = r$ we also have: :$\dbinom n r = \dbinom n {n - r} = \dbinom n {r + 1}$ Then from Symmetry Rule for Binomial Coefficients: :$\forall k < r: \dbinom n r > \dbinom n {n - k}$ and so: :$\forall k > r + 1: \dbinom n r > \dbinom n k$ We have that: :$r = \floor {\dfrac n 2}$ and: :$r + 1 = \ceiling {\dfrac n 2}$ Hence the result. {{qed}}
The $n$th row of Pascal's triangle consists of the finite sequence: :$\dbinom n 0, \dbinom n 1, \dbinom n 2, \ldots, \dbinom n {n - 2}, \dbinom n {n - 1}, \dbinom n n$ By the Symmetry Rule for Binomial Coefficients: :$\dbinom n m = \dbinom n {n - m}$ Hence we can write the $n$th row in reverse order: {{begin-eqn}} {{eqn | o =        | r = \dbinom n n, \dbinom n {n - 1}, \dbinom n {n - 2}, \ldots, \dbinom n 2, \dbinom n 1, \dbinom n 0       | c =  }} {{eqn | r = \dbinom n {n - n}, \dbinom n {n - \left({n - 1}\right)}, \dbinom n {n - \left({n - 2}\right)}, \ldots, \dbinom n {n - 2}, \dbinom n {n - 1}, \dbinom n {n - 0}       | c =  }} {{eqn | r = \dbinom n 0, \dbinom n 1, \dbinom n 2, \ldots, \dbinom n {n - 2}, \dbinom n {n - 1}, \dbinom n n       | c =  }} {{end-eqn}} and the sequences are seen to be the same. {{qed}}
From Pascal's Rule we have: :$\forall n, k \in \Z: \dbinom n {k - 1} = \dbinom {n + 1} k - \dbinom n k$ Thus it is sufficient to prove that: :$\forall n \in \Z: \dbinom n {-1} = 0$ So: {{begin-eqn}} {{eqn | l = \dbinom n {-1}       | r = \dbinom {n + 1} 0 - \dbinom n 0       | c = Pascal's Rule }} {{eqn | r = 1 - 1       | c = Binomial Coefficient with Zero }} {{eqn | r = 0       | c =  }} {{end-eqn}} Hence the result. {{qed}} Category:Binomial Coefficients jilju9t3idp64zejhlfrap6ma23dwna
{{ProofWanted|We need to establish the classical algorithms and build up the knowledge of exactly what a carry is before tackling this delightful little number.}} {{Namedfor|Ernst Eduard Kummer|cat = Kummer}}
The statement: :$\dbinom n k$ is odd is equivalent to: :$\dbinom n k \equiv 1 \pmod p$ The corollary to Lucas' Theorem gives: :$\displaystyle \dbinom n k \equiv \prod_{j \mathop = 0}^r \dbinom {a_j} {b_j} \pmod p$ where: : $n, k \in \Z_{\ge 0}$ and $p$ is prime : the representations of $n$ and $k$ to the base $p$ are given by: ::$n = a_r p^r + \cdots + a_1 p + a_0$ ::$k = b_r p^r + \cdots + b_1 p + b_0$ When $p = 2$, the digits $a_j, b_j$ are either $0$ or $1$. In order for $\dbinom n k \equiv 1 \pmod p$, it is necessary and sufficient that $\dbinom {a_j} {b_j} \equiv 1 \pmod p$ for all $j \in \left\{ {0, 1, 2, \ldots, r}\right\}$. In order for this to happen, $\dbinom {a_j} {b_j} = \dbinom 0 0$ or $\dbinom {a_j} {b_j} = \dbinom 1 0$ or $\dbinom {a_j} {b_j} = \dbinom 1 1$. Suppose $a_i = 0$ for some $i \in \left\{ {0, 1, 2, \ldots, r}\right\}$. Then if $b_i = 1$: :$\dbinom {a_j} {b_j} = \dbinom 0 1 = 0$ and so: : $\dbinom n k \equiv 0 \pmod p$ for whichever $k$ (and there will be at least one) has digit $b_i = 1$. So the only way it can be assured that all $\dbinom {a_j} {b_j} \equiv 1 \pmod p$ for all $k \in \left\{ {0, 1, 2, \ldots, n}\right\}$ is for $a_j = 0$ for all $j \in \left\{ {0, 1, 2, \ldots, r}\right\}$. That is, for $n = 2^m - 1$ for some $m \in \Z_{\ge 0}$. {{qed}}
From Sum of Powers of Positive Integers: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 1}^n i^p       | r = 1^p + 2^p + \cdots + n^p       | c =  }} {{eqn | r = \frac {n^{p + 1} } {p + 1} + \sum_{k \mathop = 1}^p \frac {B_k \, p^{\underline {k - 1} } \, n^{p - k + 1} } {k!}       | c =  }} {{end-eqn}} where $B_k$ are the Bernoulli numbers. Setting $p = 4$: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 1}^n i^4       | r = \frac {n^{4 + 1} } {4 + 1} + \sum_{k \mathop = 1}^4 \frac {B_k \, 4^{\underline {k - 1} } \, n^{4 - k + 1} } {k!}       | c =  }} {{eqn | r = \frac {n^5} 5 + \frac {B_1 \, 4^{\underline 0} \, n^4} {1!} + \frac {B_2 \, 4^{\underline 1} \, n^3} {2!} + \frac {B_3 \, 4^{\underline 2} \, n^2} {3!} + \frac {B_4 \, 4^{\underline 3} \, n} {4!}       | c =  }} {{eqn | r = \frac {n^5} 5 + \frac 1 2 \frac {n^4} {1!} + \frac 1 6 \frac {4 n^3} {2!} + 0 \frac {4 \times 3 n^2} {3!} - \frac 1 {30} \frac {4 \times 3 \times 2 n} {4!}       | c = {{Defof|Bernoulli Numbers}} and {{Defof|Falling Factorial}} }} {{eqn | r = \frac {n^5} 5 + \frac {n^4} 2 + \frac {n^3} 3 - \frac n {30}       | c = simplifying }} {{eqn | r = \frac {6 n^5 + 15 n^4 + 10 n^3 - n} {30}       | c =  }} {{eqn | r = \frac {n \paren {n + 1} \paren {2 n + 1} \paren {3 n^2 + 3 n - 1} } {30}       | c =  }} {{end-eqn}} Hence the result. {{qed}}
By definition of binomial coefficient: {{begin-eqn}} {{eqn | l = \binom k 4       | r = \frac {k \paren {k - 1} \paren {k - 2} \paren {k - 3} } {4 !}       | c =  }} {{eqn | l = \binom k 3       | r = \frac {k \paren {k - 1} \paren {k - 2} } {3 !}       | c =  }} {{eqn | l = \binom k 2       | r = \frac {k \paren {k - 1} } {2 !}       | c =  }} {{eqn | l = \binom k 1       | r = k        | c =  }} {{eqn | ll= \leadsto       | l = 24 \binom k 4       | r = k^4 - 6 k^3 + 11 k^2 - 6 k       | c =  }} {{eqn | l = 36 \binom k 3       | r = 6 k^3 - 18 k^2 + 12 k       | c =  }} {{eqn | l = 14 \binom k 2       | r = 7 k^2 - 7 k       | c =  }} {{eqn | ll= \leadsto       | l = k^4       | r = 24 \binom k 4 + 36 \binom k 3 + 14 \binom k 2 + \binom k 1       | c =  }} {{end-eqn}} So: {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 0}^n k^4       | r = \sum_{k \mathop = 0}^n \paren {24 \binom k 4 + 36 \binom k 3 + 14 \binom k 2 + \binom k 1}       | c =  }} {{eqn | r = 24 \binom {n + 1} 5 + 36 \binom {n + 1} 4 + 14 \binom {n + 1} 3 + \binom {n + 1} 2       | c = Sum of Binomial Coefficients over Upper Index }} {{eqn | r = 24 \frac {\paren {n + 1} n \paren {n - 1} \paren {n - 2} \paren {n - 3} } {5 !}       | c = Definition of Binomial Coefficient }} {{eqn | o =        | ro= +       | r = 36 \frac {\paren {n + 1} n \paren {n - 1} \paren {n - 2} } {4 !}       | c =  }} {{eqn | o =        | ro= +       | r = 14 \frac {\paren {n + 1} n \paren {n - 1} } {3 !}       | c =  }} {{eqn | o =        | ro= +       | r = \frac {\paren {n + 1} n } {2 !}       | c =  }} {{eqn | r = \paren {n + 1} n \paren {\frac {\paren {n - 1} \paren {n - 2} \paren {n - 3} } 5 + \frac {3 \paren {n - 1} \paren {n - 2} } 2 + \frac {7 \paren {n - 1} } 3 + \frac 1 2}       | c =  }} {{eqn | r = \frac {\paren {n + 1} n} {30} \paren {6 \paren {n^3 - 6 n^2 + 11 n - 6 } + 45 \paren {n^2 - 3 n + 2} + 70 \paren {n - 1} + 15}       | c =  }} {{eqn | r = \frac {\paren {n + 1} n} {30} \paren {6 n^3 - 36 n^2 + 66 n - 36 + 45 n^2 - 135 n + 90 + 70 n - 70 + 15}       | c =  }} {{eqn | r = \frac {\paren {n + 1} n} {30} \paren {6 n^3 + 9 n^2 + n - 1}       | c =  }} {{eqn | r = \frac {\paren {n + 1} n} {30} \paren {2 n + 1} \paren {3 n^2 + 3 n - 1}       | c =  }} {{eqn | r = \frac {\paren {n + 1} n \paren {n + \frac 1 2} \paren {3 n^2 + 3 n - 1} } {15}       | c =  }} {{end-eqn}} {{qed}}
By definition of complete lattice: :$\varnothing$ admits a supremum and an infimum. By Infimum of Empty Set is Greatest Element: :$\forall x \in S: x \preceq \inf \varnothing$ Thus by definition: :$L$ is bounded above. By Supremum of Empty Set is Smallest Element: :$\forall x \in S: \sup \varnothing \preceq x$ Thus by definition: :$L$ is bounded below. Hence $L$ is bounded. {{qed}}
For $k \ge 1$, let $f_c^k$ be the constant function of $k$ variables with value $c$. We know from Constant Function is Primitive Recursive that $f_c^1$ is primitive recursive. Now: :$\map {f_c^k} {n_1, n_2, \ldots, n_k} = \map {f_c^1} {n_1} = \map {f_c^1} {\map {\pr_1^k} {n_1, n_2, \ldots, n_k} }$ where $\pr_1^k$ is a projection function which is basic primitive recursive. So $f_c^k$ is obtained from the primitive recursive function $f_c^1$ and the basic primitive recursive function $\pr_1^k$ by substitution. Hence by definition, $f_c^k$ is primitive recursive. {{qed}} Category:Primitive Recursive Functions Category:Constant Mappings p7103pmh89wjmd8e0mfrg551rg2r1jz
We have: :$n = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}$ We also have: :$\forall i, j \in \closedint 1 n: i \ne j \implies p_i^{k_i} \perp p_j^{k_j}$ So: :$\map f {p_i^{k_i} p_j^{k_j} } = \map f {p_i^{k_i} } \, \map f {p_j^{k_j} }$ It is a simple inductive process to show that $\map f n = \map f {p_1^{k_1} } \, \map f {p_2^{k_2} } \dotsm \map f {p_r^{k_r} }$. {{handwaving}} {{qed}} Category:Multiplicative Functions 2wrphk1b2zl7ohc8fzbupknu3hs00xb
{{begin-eqn}} {{eqn | l = \paren {-1}^n \dbinom {-n} {k - 1}       | r = \paren {-1}^n \paren {\paren {-1}^{-n - \paren {k - 1} } \binom {- \paren {k - 1 + 1} } {- n - \paren {k - 1} } }       | c = Moving Top Index to Bottom in Binomial Coefficient }} {{eqn | r = \paren {-1}^{- \paren {k - 1} } \binom {- k} {- \paren {n - 1} - k}       | c =  }} {{eqn | r = \paren {-1}^k \binom {- k} {- k - \paren {n - 1} }       | c =  }} {{eqn | r = \paren {-1}^k \binom {- k} {n - 1}       | c = Symmetry Rule for Binomial Coefficients }} {{end-eqn}} {{questionable|Can someone explain the disappearance of a factor of $-1$? Seems to have to do with a negative bottom index}}
By definition of binomial coefficient: :$\dbinom m n = \begin{cases} \dfrac {m!} {n! \left({m - n}\right)!} & : 0 \le n \le m \\ & \\ 0 & : \text { otherwise } \end{cases}$ Thus when $n > 1$: :$\dbinom 1 n = 0$ and when $n < 0$: :$\dbinom 1 n = 0$ Then: {{begin-eqn}} {{eqn | l = \dbinom 1 0       | r = \dfrac {1!} {0! \left({0 - 1}\right)!}       | c = {{Defof|Binomial Coefficient}} }} {{eqn | r = 1       | c = {{Defof|Factorial}} }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \dbinom 1 1       | r = \dbinom 1 {1 - 1}       | c = Symmetry Rule for Binomial Coefficients }} {{eqn | r = \dbinom 1 0       | c =  }} {{eqn | r = 1       | c = from above }} {{end-eqn}} {{qed}} Category:Examples of Binomial Coefficients rq46fe27b1uakci157m84qzw75tysil
Let $T$ be separable. By Space is Separable iff Density not greater than Aleph Zero: :$\map d T \le \aleph_0$ where: :$\map d T$ denotes the density of $T$ :$\aleph$ denotes the aleph mapping. By definition of density: :$\exists A \subseteq S: A$ is dense $\land \map d T = \card A$ where $\card A$ denotes the cardinality of $A$. By definition of dense set: :$A^- = S$ where $A^-$ denotes the closure of $A$. By Set in Discrete Topology is Clopen: :$A$ is closed Then by Set is Closed iff Equals Topological Closure: :$A^- = A$ Thus by Countable iff Cardinality not greater than Aleph Zero: :$S$ is countable {{qed}}
Let $T$ be separable. {{AimForCont}} $S$ is uncountable. Then by Uncountable Discrete Space is not Separable, $T$ is not separable. Hence the result by Proof by Contradiction. {{qed}}
Substituting $n - 1 - r + n t$ for $s$ in the {{RHS}}: {{begin-eqn}} {{eqn | l = \binom {r + s - t n} n       | r = \binom {r + \paren {n - 1 - r + n t} - t n} n       | c =  }} {{eqn | r = \binom {r + n - 1 - r + n t - t n} n       | c =  }} {{eqn | r = \binom {n - 1} n       | c =  }} {{end-eqn}} Substituting $n - 1 - r + n t$ for $s$ in the {{LHS}}: {{begin-eqn}} {{eqn | o =       | r = \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {n - 1 - r + t k} {n - k} \frac r {r - t k}       | c =  }} {{eqn | r = \sum_{k \mathop \ge 0} \dfrac {\paren {r - t k}! \paren {n - 1 - r + t k}! \, r} {k! \paren {r - t k - k}! \paren {n - k}! \paren {k - 1 - r + t k}! \paren {r - t k} }       | c =  }} {{eqn | r = \sum_{k \mathop \ge 0} \frac r {n!} \binom n k \dfrac {\paren {r - t k - 1}! \paren {n - 1 - r + t k}!} {\paren {r - t k - k}! \paren {k - 1 - r + t k}!}       | c =  }} {{eqn | r = \sum_{k \mathop \ge 0} \frac r {n!} \binom n k \prod_{0 \mathop < j \mathop < k} \paren {r - t k - j} \prod_{0 \mathop < j \mathop < n \mathop - k} \paren {n - 1 - r + t k - j}       | c =  }} {{eqn | r = \sum_{k \mathop \ge 0} \frac r {n!} \binom n k \paren {-1}^{k - 1} \prod_{0 \mathop < j \mathop < k} \paren {-r + t k + j} \prod_{k \mathop \le j \mathop < n} \paren {- r + t k + j}       | c =  }} {{end-eqn}} The two products give a polynomial of degree $n - 1$ in $k$. Hence the sum for all $k$ is $0$. Thus we have: {{begin-eqn}} {{eqn | l = \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {n - 1 - r + t k} {n - k} \frac r {r - t k}       | r = 0       | c =  }} {{eqn | r = \binom {n - 1} n       | c = {{Defof|Binomial Coefficient}} }} {{end-eqn}} Thus the equation indeed holds for the special case where $s = n - 1 - r + n t$.
=== Sufficient Condition === Let $L$ be continuous. Thus by Continuous Lattice is Meet-Continuous: :$L$ is meet-continuous. Thus by Way Below is Approximating Relation and Way Below Relation is Auxiliary Relation: :$\ll$ is auxiliary approximating relation on $S$. Thus by Continuous Lattice iff Auxiliary Approximating Relation is Superset of Way Below Relation: :for every auxiliary approximating relation $\mathcal Q$ on $S$: $\ll \subseteq \mathcal Q$ {{qed|lemma}} === Necessary Condition === Let $L$ be meet-continuous. Assume that :there exists an auxiliary approximating relation $\mathcal R$ on $S$ ::for every auxiliary approximating relation $\mathcal Q$ on $S$: $\mathcal R \subseteq \mathcal Q$ Let $x \in S$. By Intersection of Relation Segments of Approximating Relations equals Way Below Closure: : $\displaystyle \bigcap \set {x^{\mathcal Q}: \mathcal Q \in \mathit{App}\left({L}\right)} = x^\ll$ where :$x^{\mathcal Q}$ denotes the $\mathcal Q$-segment of $x$,  :$\mathit{App}\left({L}\right)$ denotes the set of all auxiliary approximating relations on $S$. By Intersection is Subset/General Result: :$x^\ll \subseteq x^{\mathcal R}$ By definition of approximating relation: :$x = \sup \left({x^{\mathcal R} }\right)$ We will prove that :$\forall a \in \set {x^{\mathcal Q}: \mathcal Q \in \mathit{App}\left({L}\right)}: x^{\mathcal R} \subseteq a$ Let $a \in \left\{ {x^{\mathcal Q}: \mathcal Q \in \mathit{App}\left({L}\right)}\right\}$ Then :$\exists \mathcal Q \in \mathit{App}\left({L}\right): a = x^{\mathcal Q}$ By assumption: :$\mathcal R \subseteq \mathcal Q$ Thus by Relation Segment is Increasing: :$x^{\mathcal R} \subseteq a$ {{qed|lemma}} By Intersection is Largest Subset/General Result: :$x^{\mathcal R} \subseteq x^\ll$ By definition of set equality: :$x^{\mathcal R} = x^\ll$ Thus :$x = \sup \left({x^\ll}\right)$ By definition: :$L$ satisfies axiom of approximation. By Way Below Closure is Directed in Bounded Below Join Semilattice: :$\forall x \in S: x^\ll$ is directed By definition of complete lattice: :$L$ is up-complete. Hence $L$ is continuous. {{qed}}
=== Sufficient Condition === Let $L$ be continuous. Let $\mathcal R \in \mathit{App}\left({L}\right)$ Let $\left({a, b}\right) \in S \times S$ such that :$a \ll b$ By definition of way below closure: :$a \in b^\ll$ where $b^\ll$ denotes the way below closure of $b$. By Complete Lattice is Bounded and Continuous Lattice is Meet-Continuous: $L$ is a bounded below meet-continuous lattice. By Intersection of Relation Segments of Approximating Relations equals Way Below Closure: : $\displaystyle \bigcap \left\{ {b^{\mathcal R}: \mathcal R \in \mathit{App}\left({L}\right)}\right\} = b^\ll$ where $b^{\mathcal R}$ denotes the $\mathcal R$-segment of $b$. By Intersection is Subset/General Result: :$b^\ll \subseteq b^{\mathcal R}$ By definition of subset: :$a \in b^{\mathcal R}$ By definition of $\mathcal R$-segment: :$\left({a, b}\right) \in \mathcal R$ Thus by definition of subset: :$\ll \subseteq \mathcal R$ Thus by Way Below is Approximating Relation: :$\ll$ is an approximating relation. {{qed|lemma}} === Necessary Condition === Let :$\forall \mathcal R \in \mathit{App}\left({L}\right): \ll \subseteq \mathcal R$ and $\ll$ is an approximating relation Define $\mathcal R := \mathord \ll$. By definition of approximating relation: :$\forall x \in S: x = \sup \left({x^{\mathcal R} }\right)$ By definitions of way below closure and $\mathcal R$-segment: :$\forall x \in S: x = \sup \left({x^\ll}\right)$ By definition: :$L$ satisfies axiom of approximation. By Way Below Closure is Directed in Bounded Below Join Semilattice: :$\forall x \in S: x^\ll$ is directed By definition of complete lattice: :$L$ is up-complete. Hence $L$ is continuous. {{qed}}