hcju/proofwikiqa huggingface dataset:
Let $m, n$ be coprime integers. Then in particular, $m, n \in \Z$. Hence, since $f$ is completely additive: :$f \left({m \times n}\right) = f \left({m}\right) + f \left({n}\right)$ and $f$ is additive, as desired. {{qed}} Category:Number Theory Category:Ring Theory okn9cbt94w16hdr1f8wcq1ox7gqmhnb
From Sum of General Logarithms: :$\log_b x + \log_b y = \log_b \left({x y}\right)$ The result follows from the definition of complete additivity. {{qed}} Category:Logarithms Category:Number Theory 8wryrdf91qouep0q87julzna5febp3i
Evaluating the equation for $\tuple {r, s - t, t, n - 1}$: {{begin-eqn}} {{eqn | l = \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {\paren {s - t} - t \paren {\paren {n - 1} - k} } {\paren {n - 1} - k} \frac r {r - t k}       | r = \binom {r + \paren {s - t} - t \paren {n - 1} } {n - 1}       | c =  }} {{eqn | ll= \leadsto       | l = \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t - t n + t + t k} {n - k - 1} \frac r {r - t k}       | r = \binom {r + s - t - t n + 1} {n - 1}       | c =  }} {{eqn | ll= \leadsto       | l = \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t \paren {n - k} } {n - k - 1} \frac r {r - t k}       | r = \binom {r + s - t n} {n - 1}       | c =  }} {{end-eqn}} Adding the equation in $\tuple {r, s, t, n}$: {{begin-eqn}} {{eqn | l = \sum_{k \mathop \ge 0} \binom {r - t k} k \paren {\binom {s - t \paren {n - k} } {n - k - 1} + \binom {s - t \paren {n - k} } {n - k} } \frac r {r - t k}       | r = \binom {r + s - t n} {n - 1} + \binom {r + s - t n} n       | c =  }} {{eqn | l = \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s + 1 - t \paren {n - k} } {n - k } \frac r {r - t k}       | r = \binom {r + s + 1 - t n} n       | c = Pascal's Rule }} {{end-eqn}} Hence the equation holds for $\tuple {r, s + 1, t, n}$ {{qed}}
From Sum over $k$ of $\left({-1}\right)^k$ by $\dbinom n k$ by $\dbinom {r - k t} n$ by $\dfrac r {r - k t}$ and renaming variables: :$\displaystyle \sum_j \left({-1}\right)^j \dbinom k j \dbinom {r - j t} k \dfrac r {r - j t} = \delta_{k 0}$ where $\delta_{k 0}$ is the Kronecker delta. Thus: :$\displaystyle \sum_{j, k} \left({-1}\right)^j \dbinom k j \dbinom {r - j t} k \dfrac r {r - j t} w^k = 1$ We have: {{begin-eqn}} {{eqn | o =        | r = \sum_{j, k} \left({-1}\right)^j \dbinom k j \dbinom {r - j t} k \dfrac r {r - j t} w^k       | c =  }} {{eqn | r = \sum_j \left({-1}\right)^j \dfrac r {r - j t} \sum_k \dbinom k j \dbinom {r - j t} k w^k       | c =  }} {{eqn | r = \sum_j \left({-1}\right)^j \dfrac r {r - j t} \sum_k \dbinom {r - j t} j \dbinom {r - j t - j} {k - j} w^k        | c = Product of $\dbinom r m$ with $\dbinom m k$ }} {{eqn | r = \sum_j \left({-1}\right)^j \dfrac r {r - j t} \dbinom {r - j t} j \sum_k \dbinom {r - j t - j} {k - j} w^k         | c =  }} {{eqn | r = \sum_j \left({-1}\right)^j A_j \left({r, t}\right) \sum_k \dbinom {r - j t - j} {k - j} w^k         | c = Definition of $A_j \left({r, t}\right)$ }} {{eqn | r = \sum_j \left({-1}\right)^j A_j \left({r, t}\right) \left({1 + w}\right)^{r - j t - j} w^j       | c = Binomial Theorem }} {{end-eqn}} Now let: :$x = \dfrac 1 {1 + w}$ {{begin-eqn}} {{eqn | l = x       | o = :=       | r = \dfrac 1 {1 + w}       | c =  }} {{eqn | ll= \leadsto       | l = z       | r = -\frac w {\left({1 + w}\right)^{1 + t} }       | c =  }} {{eqn | ll= \leadsto       | l = \left({1 + w}\right)^{r - j k - j} w^j \left({-1}\right)^j       | r = z^r z^j       | c =  }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \sum_j A_j \left({r, t}\right) z^j \left({1 + w}\right)^r       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = \sum_j A_j \left({r, t}\right) z^j       | r = \left({1 + w}\right)^{- r}       | c =  }} {{eqn | r = x^r       | c =  }} {{end-eqn}} {{finish|Establish the fact that we have convergence by ratio test and estimates for large $k$. Or use complex variable theory to establish that the function is analytic around $x {{=}} 1$.}} {{qed}}
By definition of approximating relation: :$x = \sup \left({x^{\mathcal R} }\right)$ By definition of supremum: :$y$ is upper bound for $x^{\mathcal R} \implies x \preceq y$ By definition of upper bound: :$\exists u \in S: u \in x^{\mathcal R} \land u \npreceq y$ Thus by definition of $\mathcal R$-segment: :$\exists u \in S: \left({u, x}\right) \in \mathcal R \land u \npreceq y$ {{qed}}
By definition of auxiliary relation: :$x \preceq z$ By definition of $\prec$: :$x \prec z$ By definition of antisymmetry: :$z \nprec x$ Then :$z \npreceq x$ By Not Preceding implies Approximating Relation and not Preceding: :$\exists u \in S: \left({u, z}\right) \land u \npreceq x$ Define $y = x \vee u$. Thus by Join Succeeds Operands: :$x \preceq y$ Thus by definition of auxiliary relation: :$\left({y, z}\right) \in \mathcal R$ Thus by Preceding iff Join equals Larger Operand: :$x \ne y$ {{qed}}
=== Sufficient Condition === Let :$x \preceq y$ By definition of join: :$x \vee y = \sup \left\{ {x, y}\right\}$ By definitions of upper bound and reflexivity: :$y$ is upper bound for $\left\{ {x, y}\right\}$ and :$\forall z \in S: z$ is upper bound for $\left\{ {x, y}\right\} \implies y \preceq z$ Thus by definition of supremum: :$y = \sup \left\{ {x, y}\right\} = x \vee y$ {{qed|lemma}} === Necessary Condition === Let :$x \vee y = y$ Thus by Join Succeeds Operands: :$x \preceq y$ {{qed}}
The proof proceeds by induction. For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition: :$\displaystyle \sum_k \left({-1}\right)^k \dbinom n k \dbinom {r - k t} n \dfrac r {r - k t} = \delta_{n 0}$ === Basis for the Induction === $P \left({0}\right)$ is the case: {{begin-eqn}} {{eqn | l = \sum_k \left({-1}\right)^k \dbinom 0 k \dbinom {r - k t} 0 \dfrac r {r - k t}       | r = \sum_k \left({-1}\right)^k \delta_{0 k} \dbinom {r - k t} 0 \dfrac r {r - k t}       | c = Zero Choose n }} {{eqn | r = \dbinom r 0 \dfrac r r       | c = All terms but for $k = 0$ vanish }} {{eqn | r = 1       | c = Binomial Coefficient with Zero }} {{eqn | r = \delta_{0 0}       | c =  }} {{end-eqn}} Thus $P \left({0}\right)$ is seen to hold. === Induction Hypothesis === Now it needs to be shown that, if $P \left({j}\right)$ is true, where $j \ge 0$, then it logically follows that $P \left({j + 1}\right)$ is true. So this is the induction hypothesis: :$\displaystyle \sum_k \left({-1}\right)^k \dbinom j k \dbinom {r - k t} j \dfrac r {r - k t} = \delta_{j 0}$ from which it is to be shown that: :$\displaystyle \sum_k \left({-1}\right)^k \dbinom {j + 1} k \dbinom {r - k t} {j + 1} \dfrac r {r - k t} = \delta_{\left({j + 1}\right) 0}$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | o =        | r = \sum_k \left({-1}\right)^k \dbinom {j + 1} k \dbinom {r - k t} {j + 1} \dfrac r {r - k t}       | c =  }} {{eqn | r = \sum_k \left({-1}\right)^k \left({\dbinom j k + \dbinom j {k - 1} }\right) \dbinom {r - k t} {j + 1} \dfrac r {r - k t}       | c = {{Defof|Binomial Coefficient}} }} {{end-eqn}} {{finish|brain not working tonight. Ought to be able to split binomial coeffs and reduce but it didn't do what it should. Perhaps try to solve it by reducing to factorials.}} So $P \left({r}\right) \implies P \left({r + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \Z_{\ge 0} \sum_k \left({-1}\right)^k \dbinom n k \dbinom {r - k t} n \dfrac r {r - k t} = \delta_{n 0}$
From  Sum over $k$ of $\dbinom r k$ by $\dbinom {s - k t} r$ by $\paren {-1}^k$ and renaming variables: :$\displaystyle \sum_j \paren {-1}^j \binom k j \binom {r - j t} k = t^k$ Thus: {{begin-eqn}} {{eqn | l = \sum_{j, k} \binom k j \binom {r - j t} k \paren {-1}^j       | r = \sum_{k \mathop \ge 0} t^k       | c = when $k < 0$ we have $\dbinom {r - j t} k = 0$ }} {{eqn | ll= \leadsto       | l = \sum_j \paren {-1}^j \sum_k \binom k j \binom {r - j t} k       | r = \frac 1 {1 - t}       | c = Sum of Infinite Geometric Sequence }} {{eqn | ll= \leadsto       | l = \sum_j \paren {-1}^j \sum_k \binom {r - j t} j \binom {r - j t - k} {j - k}       | r = \frac 1 {1 - t}       | c = Product of $\dbinom r m$ with $\dbinom m k$ }} {{eqn | ll= \leadsto       | l = \sum_j \paren {-1}^j \binom {r - j t} j \sum_k \binom {r - j t - k} {j - k}       | r = \frac 1 {1 - t}       | c =  }} {{end-eqn}} {{finish|... and so on}}
From Sum over $k$ of $\dbinom {r - k t} k$ by $\dfrac r {r - k t}$ by $z^k$: :$\displaystyle (1): \quad \sum_k \map {A_k} {r, t} z^k = x^r$ where: :$\map {A_n} {x, t}$ is the polynomial of degree $n$ defined as: ::$\map {A_n} {x, t} := \dbinom {x - n t} n \dfrac x {x - n t}$ :for $x \ne n t$ :$z = x^{t + 1} - x^t$. Differentiating $(1)$ {{WRT|Differentiation}} $z$: {{begin-eqn}} {{eqn | l = \sum_k \map {A_k} {r, t} k z^{k - 1}       | r = \frac \d {\d z} x^r       | c = Power Rule for Derivatives }} {{eqn | ll= \leadsto       | l = \sum_k \map {A_k} {r, t} k z^k       | r = z \frac \d {\d z} x^r       | c =  }} {{eqn | r = \paren {x^{t + 1} - x^t} r x^{r - 1} \frac {\d x} {\d z}       | c = Chain Rule for Derivatives }} {{end-eqn}} We have: {{begin-eqn}} {{eqn | l = z       | r = x^{t + 1} - x^t       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d z} {\d x}       | r = \paren {t + 1} x^t - t x^{t - 1}       | c = Power Rule for Derivatives }} {{eqn | ll= \leadsto       | l = \frac {\d x} {\d z}       | r = \frac 1 {\paren {t + 1} x^t - t x^{t - 1} }       | c = Derivative of Inverse Function }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = \sum_k k \map {A_k} {r, t} z^k       | r = \frac {\paren {x^{t + 1} - x^t} r x^{r - 1} } {\paren {t + 1} x^t - t x^{t - 1} }       | c =  }} {{eqn | r = \frac {\paren {x - 1} r x^r} {\paren {t + 1} x - t}       | c = simplifying }} {{eqn | ll= \leadsto       | l = \sum_k \map {A_k} {r, t} z^k - \frac t r \sum_k k \map {A_k} {r, t} z^k       | r = x^r - \frac t r \frac {\paren {x - 1} r x^r} {\paren {t + 1} x - t}       | c =  }} {{eqn | ll= \leadsto       | l = \sum_k \paren {1 - \frac t r k} \map {A_k} {r, t} z^k        | r = \frac {\paren {\paren {t + 1} x - t} x^r - t \paren {x - 1} x^r} {\paren {t + 1} x - t}       | c =  }} {{eqn | r = \frac {x^{r + 1} } {\paren {t + 1} x - t}       | c = simplifying }} {{eqn | ll= \leadsto       | l = \sum_k \frac {r - t k} r \dbinom {r - k t} k \dfrac r {r - k t} z^k        | r = \frac {x^{r + 1} } {\paren {t + 1} x - t}       | c = substituting for $\map {A_k} {r, t}$ }} {{eqn | ll= \leadsto       | l = \sum_k \dbinom {r - k t} k z^k        | r = \frac {x^{r + 1} } {\paren {t + 1} x - t}       | c = simplifying }} {{end-eqn}} {{qed}}
Let $f \left({r, s, t, n}\right)$ be the function defined as: :$\displaystyle f \left({r, s, t, n}\right) := \sum_k \dbinom {r + t k} k \dbinom {s - t k} {n - k}$ We have: {{begin-eqn}} {{eqn | o =        | r = \sum_k \dbinom {r + t k} k \dbinom {s - t k} {n - k}       | c =  }} {{eqn | r = \sum_k \dbinom {r + t k} k \dbinom {s - t k} {n - k} \frac {r + t k} {r + t k}       | c =  }} {{eqn | r = \sum_k \dbinom {r + t k} k \dbinom {s - t k} {n - k} \frac r {r + t k} + \sum_k \dbinom {r + t k} k \dbinom {s - t k} {n - k} \frac {t k} {r + t k}       | c =  }} {{end-eqn}} {{proof wanted|in progress}}
By Way Below is Approximating Relation and Way Below Relation is Auxiliary Relation: :$\ll$ is an auxiliary approximating relation on $S$. Thus by Auxiliary Approximating Relation has Interpolation Property: :$\exists y \in S: x \ll y \land y \ll z \land x \ne y$ {{qed}}
Case $x \ne z$: By Way Below has Strong Interpolation Property: :$\exists y \in S: x \ll y \land y \ll z \land x \ne y$ Thus :$\exists y \in S: x \ll y \land y \ll z$ Case $x = z$: Define $y = x$ Thus :$x \ll y \land y \ll z$ {{qed}}
{{begin-eqn}} {{eqn | l = x H       | r = H       | c =  }} {{eqn | ll= \leadstoandfrom       | l = x H       | r = e H       | c = Left Coset by Identity: $e H = H$ }} {{eqn | ll= \leadstoandfrom       | l = x e^{-1}       | o = \in       | r = H       | c = Left Cosets are Equal iff Product with Inverse in Subgroup }} {{eqn | ll= \leadstoandfrom       | l = x       | o = \in       | r = H       | c = Group Properties }} {{end-eqn}} {{Qed}}
{{begin-eqn}} {{eqn | l = H x       | r = H       | c =  }} {{eqn | ll= \leadstoandfrom       | l = H x       | r = H e       | c = Right Coset by Identity: $H = H e$ }} {{eqn | ll= \leadstoandfrom       | l = x e^{-1}       | o = \in       | r = H       | c = Right Cosets are Equal iff Product with Inverse in Subgroup }} {{eqn | ll= \leadstoandfrom       | l = x       | o = \in       | r = H       | c = Group Properties }} {{end-eqn}} {{Qed}}
Let: $x H$ denote the left coset of $H$ by $x$ First we note that, from Congruence Class Modulo Subgroup is Coset, we have that the left cosets of $G$ form a partition of $G$. === Sufficient Condition === Suppose $x, y$ are in the same left coset of $H$. It follows from Congruence Class Modulo Subgroup is Coset that: : $x \in y H$ and $y \in x H$ {{iff}} $x H = y H$ From Cosets are Equal iff Product with Inverse in Subgroup, we have that: : $x H = y H \iff x^{-1} y \in H$ So if $x, y$ are in the same left coset of $H$ then $x^{-1} y \in H$. {{qed|lemma}} === Necessary Condition === Suppose that $x^{-1} y \in H$. From Left Cosets are Equal iff Product with Inverse in Subgroup, we have that: : $x H = y H \iff x^{-1} y \in H$ Again, it follows from Congruence Class Modulo Subgroup is Coset that: : $x \in y H$ and $y \in x H$ iff $x H = y H$ and so $x, y$ are in the same left coset of $H$. {{qed}}
Let $H x$ denote the right coset of $H$ by $x$. First we note that, from Congruence Class Modulo Subgroup is Coset, we have that the right cosets of $G$ form a partition of $G$. === Sufficient Condition === Suppose $x, y$ are in the same left coset of $H$. It follows from Congruence Class Modulo Subgroup is Coset that: : if $x \in H y$ and $y \in H x$ iff $H x = H y$ From Cosets are Equal iff Product with Inverse in Subgroup, we have that: : $H x = H y \iff x y^{-1} \in H$ So if $x, y$ are in the same right coset of $H$ then $x y^{-1} \in H$. {{qed|lemma}} === Necessary Condition === Now suppose that $x y^{-1} \in H$. From Right Cosets are Equal iff Product with Inverse in Subgroup, we have that: : $H x = H y \iff x y^{-1} \in H$ It follows from Congruence Class Modulo Subgroup is Coset that: : $x \in H y$ and $y \in H x$ iff $H x = H y$ and so: : $x, y$ are in the same left coset of $H$ Hence $x, y$ are in the same right coset of $H$ if $x y^{-1} \in H$. {{qed}}
{{begin-eqn}} {{eqn | o =        | r = \sum_k \binom {m - r + s} k \binom {n + r - s} {n - k} \binom {r + k} {m - n}       | c =  }} {{eqn | r = \sum_k \binom {m - r + s} k \binom {n + r - s} {n - k} \left({\sum_j \binom r {m + n - j} \binom k j}\right)       | c = Chu-Vandermonde Identity }} {{eqn | r = \sum_k \sum_j \binom {m - r + s} j \binom {m - r + s - j} {k - j} \binom {n + r - s} {n - k} \binom r {m + n - j}       | c = Product of $\dbinom r m$ with $\dbinom m k$ }} {{eqn | r = \sum_j \binom {m - r + s} j \binom r {m + n - j} \left({\sum_k \binom {m - r + s - j} {k - j} \binom {n + r - s} {n - k} }\right)       | c =  }} {{eqn | r = \sum_j \binom {m - r + s} j \binom r {m + n - j} \binom {m + n - j} {n - j}       | c = Chu-Vandermonde Identity }} {{eqn | r = \sum_j \binom {m - r + s} j \binom r {m + n - j} \binom {m + n - j} m       | c = Symmetry Rule for Binomial Coefficients }} {{eqn | r = \sum_j \binom {m - r + s} j \binom r m \binom {r - m} {n - j}       | c = Product of $\dbinom r m$ with $\dbinom m k$ }} {{eqn | r = \binom r m \left({\sum_j \binom {m - r + s} j \binom {r - m} {n - j} }\right)       | c =  }} {{eqn | r = \binom r m \binom s n       | c = Chu-Vandermonde Identity }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \sum_k \left({-1}\right)^{n - k} \left[{n \atop k}\right] x^k       | r = x^{\underline n}       | c = {{Defof|Unsigned Stirling Numbers of the First Kind}} }} {{eqn | ll= \leadsto       | l = \sum_k \left({-1}\right)^{n - k} \left[{n \atop k}\right] \left({-x}\right)^k       | r = \left({-x}\right)^{\underline n}       | c = putting $-x$ for $x$ }} {{eqn | ll= \leadsto       | l = \left({-1}\right)^n \sum_k \left[{n \atop k}\right] x^k       | r = \left({-x}\right)^{\underline n}       | c =  }} {{eqn | ll= \leadsto       | l = \sum_k \left[{n \atop k}\right] x^k       | r = \left({-1}\right)^n \left({-x}\right)^{\underline n}       | c =  }} {{eqn | r = x^{\overline n}       | c = Rising Factorial in terms of Falling Factorial of Negative }} {{end-eqn}} {{qed}}
The proof proceeds by induction on $n$. For all $n \in \Z_{> 0}$, let $P \left({n}\right)$ be the proposition: :$\displaystyle \left({x + y}\right)^{\overline n} = \sum_k \binom n k x^{\overline k} y^{\overline {n - k} }$ === Basis for the Induction === $P \left({1}\right)$ is the case: {{begin-eqn}} {{eqn | l = \left({x + y}\right)^{\overline 1}       | r = x^{\overline 1} + y^{\overline 1}       | c = Number to Power of One Rising is Itself }} {{eqn | r = x^{\overline 0} y^{\overline 1} + x^{\overline 1} y^{\overline 0}       | c = Number to Power of Zero Rising is One }} {{eqn | r = \binom 1 0 x^{\overline 0} y^{\overline 1} + \binom 1 1 x^{\overline 1} y^{\overline 0}       | c = One Choose n }} {{eqn | r = \sum_k \binom 1 k x^{\overline k} y^{\overline {1 - k} }       | c =  }} {{end-eqn}} Thus $P \left({1}\right)$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({m}\right)$ is true, where $m \ge 1$, then it logically follows that $P \left({m + 1}\right)$ is true. So this is the induction hypothesis: :$\displaystyle \left({x + y}\right)^{\overline m} = \sum_k \binom m k x^{\overline k} y^{\overline {m - k} }$ from which it is to be shown that: :$\displaystyle \left({x + y}\right)^{\overline {m + 1} } = \sum_k \binom {m + 1} k x^{\overline k} y^{\overline {m + 1 - k} }$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \left({x + y}\right)^{\overline {m + 1} }       | r = \left({x + y}\right)^{\overline m} \left({x + y + m}\right)       | c = Definition of Rising Factorial }} {{eqn | r = \left({\sum_k \binom m k x^{\overline k} y^{\overline {m - k} } }\right) \left({x + y + m}\right)       | c = Induction Hypothesis }} {{eqn | r = \left({\sum_k \binom m k x^{\overline k} y^{\overline {m - k} } }\right) \left({x + k + y + m - k}\right)       | c =  }} {{eqn | r = \sum_k \binom m k x^{\overline k} \left({x + k}\right) y^{\overline {m - k} } + \sum_k \binom m k x^{\overline k} y^{\overline {m - k} } \left({y + m - k}\right)       | c =  }} {{eqn | r = \sum_k \binom m k x^{\overline {k + 1} } y^{\overline {m - k} } + \sum_k \binom m k x^{\overline k} y^{\overline {m - k + 1} }       | c = Definition of Rising Factorial }} {{eqn | r = \sum_k \binom m k x^{\overline {k + 1} } y^{\overline {m - k} } + \sum_k \binom m {k + 1} x^{\overline {k + 1} } y^{\overline {m - k} }       | c = Translation of Index Variable of Summation }} {{eqn | r = \sum_k \binom {m + 1} k x^{\overline {k + 1} } y^{\overline {m - k} }       | c = Pascal's Rule }} {{end-eqn}} So $P \left({m}\right) \implies P \left({m + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \Z_{> 0}: \left({x + y}\right)^{\overline n} = \sum_k \binom n k x^{\overline k} y^{\overline {n - k} }$
From Rising Factorial as Factorial by Binomial Coefficient: :$\left({x + y}\right)^{\overline n} = n! \dbinom {x + y + n - 1} n$ Recall Sum over $k$ of $\dbinom {r - t k} k$ by $\dbinom {s - t \left({n - k}\right)} {n - k}$ by $\dfrac r {r - t k}$: :$\displaystyle \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t \left({n - k}\right)} {n - k} \frac r {r - t k} = \binom {r + s - t n} n$ Let the following substitutions be made: :$r \gets x$ :$t \gets -\left({1 - z}\right)$ :$s \gets y - 1 + n z$ and so to obtain: :$\displaystyle \dbinom {x + y + n - 1} n = \sum_k \dbinom {x + \left({1 - z}\right) k} k \dbinom {y - 1 + n z + \left({n - k}\right) \left({1 - z}\right)} {n - k} \dfrac x {x + \left({1 - z}\right) k}$ Then: {{begin-eqn}} {{eqn | l = \dbinom {x + \left({1 - z}\right) k} k       | r = \dfrac {\left({x - k z + 1}\right)^{\overline k} } {k!}       | c = Rising Factorial as Factorial by Binomial Coefficient }} {{eqn | ll= \leadsto       | l = \dfrac x {x + \left({1 - z}\right) k} \dbinom {x + \left({1 - z}\right) k} k       | r = \dfrac {x \left({x - k z + 1}\right)^{\overline {k - 1} } } {k!}       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = \dbinom {y - 1 + n z + \left({n - k}\right) \left({1 - z}\right)} {n - k}       | r = \dbinom {y - 1 + n z + n - k - n z + k z} {n - k}       | c =  }} {{eqn | r = \dbinom {y + k z \left({n - k}\right) - 1} {n - k}       | c =  }} {{eqn | r = \frac 1 {\left({n - 1}\right)!} \left({y + k z}\right)^{\overline {n - k} }       | c = Rising Factorial as Factorial by Binomial Coefficient }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = n! \dbinom {x + y + n - 1} n       | r = \sum_k \frac {n!} {k! \left({n - k}\right)!} x \left({x - k z + 1}\right)^{\overline {k - 1} } \left({y + k z}\right)^{\overline {n - k} }       | c =  }} {{eqn | ll= \leadsto       | l = \left({x + y}\right)^{\overline n}       | r = \sum_k \binom n k x \left({x - k z + 1}\right)^{\overline {k - 1} } \left({y + k z}\right)^{\overline {n - k} }       | c =  }} {{end-eqn}} {{qed}} {{Namedfor|Ruggiero Torelli|cat = Torelli}}
=== Sufficient Condition === Let $x \ll y$. Let $D$ be a directed subset of $S$ such that :$y \preceq \sup D$ By Way Below has Interpolation Property: :$\exists x' \in S: x \ll x' \land x' \ll y$ By definition of way below relation: :$\exists d \in D: x' \preceq d$ Thus by Preceding and Way Below implies Way Below and definition of reflexivity: :$x \ll d$ {{qed|lemma}} === Necessary Condition === Assume that :for every directed subset $D$ of $S$ such that $y \preceq \sup D$ ::there exists an element $d$ of $D$: $x \ll d$ By Way Below implies Preceding: :for every directed subset $D$ of $S$ such that $y \preceq \sup D$ ::there exists an element $d$ of $D$: $x \preceq d$ Thus by definition of way below relation: $x \ll y$ {{qed}}
=== Sufficient Condition === Let $L$ be continuous. Let $x \in S$. By definition of continuous: :$x^\ll$ is directed. By Way Below Closure is Lower Set: :$x^\ll$ is a lower set. Thus by definition: :$x^\ll$ is an ideal in $L$. By definition of continuous: :$L$ satisfies axiom of approximation. By definition of axiom of approximation: :$x = \sup \left({x^\ll}\right)$ Thus by definition of reflexivity: :$x \preceq \sup \left({x^\ll}\right)$ Let $I$ be an ideal in $L$ such that :$x \preceq \sup I$ We will prove that :$x^\ll \subseteq I$ Let $y \in x^\ll$ By definition of way below closure: :$y \ll x$ Thus by Way Below iff Second Operand Preceding Supremum of Ideal implies First Operand is Element of Ideal: :$y \in I$ {{qed|lemma}} === Necessary Condition === Assume that :$\forall x \in S: x^\ll$ is an ideal in $L$ and $x \preceq \sup \left({x^\ll}\right)$ and ::for every ideal $I$ in $L$: $x \preceq \sup I \implies x^\ll \subseteq I$ Let $x \in S$. By assumption: :$x \preceq \sup \left({x^\ll}\right)$ By Way Below implies Preceding: :$x$ is upper bound for $x^\ll$ By definition of supremum: :$\sup \left({x^\ll}\right) \preceq x$ Thus by definition of antisymmetry: :$x = \sup \left({x^\ll}\right)$ Thus by definition: :$L$ satisfies axiom of approximation. Thus by assumption and definition of ideal: :$\forall x \in S: x^\ll$ is directed. Hence $L$ is continuous. {{qed}}
Recall that a closed set is by definition the relative complement of an open set. Hence Sigma-Algebra Generated by Complements of Generators applies to yield: :$\sigma \left({\mathcal O^n}\right) = \sigma \left({\mathcal C^n}\right)$ {{qed}}