hcju/proofwikiqa huggingface dataset:
{{begin-eqn}} {{eqn | l = a       | o = \equiv       | r = b       | rr= \pmod m       | c =  }} {{eqn | l = x       | o = \equiv       | r = y       | rr= \pmod m       | c =  }} {{eqn | ll= \leadsto       | l = a \bmod m       | r = b \bmod m       | c = Definition of Congruence }} {{eqn | l = x \bmod m       | r = y \bmod m       | c =  }} {{eqn | ll= \leadsto       | lo= \exists k_1 \in \Z:       | l = a       | r = b + k_1 z       | c =  }} {{eqn | lo= \exists k_2 \in \Z:       | l = x       | r = y + k_2 z       | c =  }} {{eqn | ll= \leadsto       | l = a x       | r = \left({b + k_1 z}\right) \left({y + k_2 z}\right)       | c = Definition of Multiplication }} {{eqn | r = b y + b k_2 z + y k_1 z + k_1 k_2 z^2       | c = Integer Multiplication Distributes over Addition }} {{eqn | r = b y + \left({b k_2 + y k_1 + k_1 k_2 z}\right) z       | c =  }} {{end-eqn}} But it is not necessarily the case that: :$b k_2 + y k_1 + k_1 k_2 z$ is an integer. In fact, $b k_2 + y k_1 + k_1 k_2 z$ can only be ''guaranteed'' to be an integer if each of $b, y, z \in \Z$. Hence $a b$ is not necessarily congruent to $x y$. {{qed}}
Let $y \in \R: y \ne 0$. Then: {{begin-eqn}} {{eqn | l = a       | o = \equiv       | r = b       | rr= \pmod z       | c = }} {{eqn | ll= \iff       | l = a \bmod z       | r = b \bmod z       | c = {{Defof|Congruence (Number Theory)|Congruence}} }} {{eqn | ll= \iff       | l = y \left({a \bmod z}\right)       | r = y \left({b \bmod z}\right)       | c = Left hand implication valid only when $y \ne 0$ }} {{eqn | ll= \iff       | l = \left({y a}\right) \bmod \left({y z}\right)       | r = \left({y b}\right) \bmod \left({y z}\right)       | c = Product Distributes over Modulo Operation }} {{eqn | l = y a       | o = \equiv       | r = y b       | rr= \pmod {y z}       | c = {{Defof|Congruence (Number Theory)|Congruence}} }} {{end-eqn}} Hence the result. Note the invalidity of the third step when $y = 0$. {{qed}}
Let $a \in x^R$. By definition of $R$-segment of $x$: :$\left({a, x}\right) \in R$ By definition of auxiliary relation: :$a \preceq x$ Thus by definition of lower closure of element: :$a \in x^\preceq$ {{qed}}
Let $r, s \in \Z$. Then: {{begin-eqn}} {{eqn | l = \map {f_c} {r s}       | r = \paren {r s}^c       | c =  }} {{eqn | r = r^c s^c       | c = Exponent Combination Laws: Product of Powers }} {{eqn | r = \map {f_c} r \map {f_c} s       | c =  }} {{end-eqn}} {{Qed}}
Let $a \in x^R$. By definition of $R$-segment of $x$: :$\left({a, x}\right) \in R$ By definition of reflexivity: :$a \preceq a$ By definition of auxiliary relation: :$\left({a, y}\right) \in R$ Thus by definition of $R$-segment of $y$: :$a \in y^R$ {{qed}}
Let $r, s \in \Z$ such that $r \perp s$. === Case 1: Either factor is not square-free === Let either $r \notin S$ or $s \notin S$ or both. Then either: :$\exists k \in \Z_{>1}: k^2 \divides r$ or: :$\exists k \in \Z_{>1}: k^2 \divides s$ Thus either: :$\map {\chi_S} r = 0$ or: :$\map {\chi_S} s = 0$ and so: :$\map {\chi_S} r \, \map {\chi_S} s = 0$ But then: :$\exists k \in \Z_{>1}: k^2 \divides r s$ and so: :$\map {\chi_S} {r s} = 0$ demonstrating that: :$\map {\chi_S} r \, \map {\chi_S} s = \map {\chi_S} {r s}$ {{qed|lemma}} === Case 1: Both factors are square-free === Let $r \in S$ and $s \in S$. Thus: :$\nexists k \in \Z_{>1}: k^2 \divides r$ and: :$\nexists k \in \Z_{>1}: k^2 \divides s$ Hence: :$\map {\chi_S} r = 1$ and: :$\map {\chi_S} s = 1$ and so: :$\map {\chi_S} r \, \map {\chi_S} s = 1$ Because $r \perp s$: :$\nexists k \in \Z_{>1}: k \divides r, k \divides s$ Hence there can be no $k \in \Z_{>1}$ whose multiplicity in $r s$ is greater than $1$. Thus: :$\nexists k \in \Z_{>1}: k^2 \divides {r s}$ and so: :$\map {\chi_S} {r s} = 1$ once more demonstrating that: :$\map {\chi_S} r \, \map {\chi_S} s = \map {\chi_S} {r s}$ {{qed|lemma}} In both cases: :$\map {\chi_S} r \, \map {\chi_S} s = \map {\chi_S} {r s}$ Hence the result, by definition of multiplicative function. {{Qed}}
$f$ is well-defined because by Relation Segment of Auxiliary Relation is Ideal: :$\forall x \in S: x^r$ is ideal in $L$ Let $x, y \in S$ such that :$x \preceq y$ By Preceding implies Inclusion of Segments of Auxiliary Relation: :$x^r \subseteq y^r$ Thus by definitions of $\precsim$ and $f$: :$f\left({x}\right) \precsim f\left({y}\right)$ Thus by definition: :$f$ is increasing mapping. {{qed}}
Let $r, s \in \Z$ such that $r \perp s$. Let $r$ be composed of $p$ distinct primes: :$r_1, r_2, \ldots r_p$ Thus: :$f \left({r}\right) = c^p$ Let $s$ be composed of $q$ distinct primes: :$s_1, s_2, \ldots s_q$ Thus: :$f \left({s}\right) = c^q$ As $r \perp s$, all the $r_k$ and $s_k$ are distinct. Thus $r s$ is composed of: :the $p$ distinct primes $r_1, r_2, \ldots r_p$ and: :the $q$ distinct primes $s_1, s_2, \ldots s_q$ which is a total of $p + q$ distinct primes. Thus: {{begin-eqn}} {{eqn | l = f \left({r s}\right)       | r = c^{p + q}       | c = Definition of $f$ }} {{eqn | r = c^p c^q       | c = Exponent Combination Laws: Product of Powers }} {{eqn | r = f \left({r}\right) f \left({s}\right)       | c = from above }} {{end-eqn}} Hence the result. {{Qed}}
Let $f$ and $g$ be multiplicative. Let $m \perp n$. Then: {{begin-eqn}} {{eqn | l = \map {f \times g} {m \times n}       | r = \map f {m \times n} \times \map g {m \times n}       | c = {{Defof|Pointwise Multiplication of Integer-Valued Functions}} }} {{eqn | r = \map f m \times \map f n \times \map g m \times \map g n       | c = {{Defof|Multiplicative Arithmetic Function}} }} {{eqn | r = \map f m \times \map g m \times \map f n \times \map g n       | c = Integer Multiplication is Commutative }} {{eqn | r = \paren {\map {f \times g} n} \times \paren {\map {f \times g} n}       | c = {{Defof|Pointwise Multiplication of Integer-Valued Functions}} }} {{end-eqn}} Hence the result by definition of multiplicative function. {{qed}}
=== Condition $(1)$ === By Segment of Auxiliary Relation Mapping is Increasing: :$f$ is an increasing mapping. By Segment of Auxiliary Relation is Subset of Lower Closure: :$\forall x \in S: x^r \subseteq x^\preceq$ By definition of $f$: :$\forall x \in S: f\left({x}\right) \subseteq x^\preceq$ Thus :$f \in F$ {{qed|lemma}} === Condition $(2)$ === By Lower Closure is Increasing: :$\forall x, y \in S: x \preceq y \implies x^\preceq \subseteq y^\preceq$ By definitions of $h$ and $\precsim$: :$\forall x, y \in S: x \preceq y \implies h\left({x}\right) \precsim h\left({y}\right)$ By definition: :$h$ is an increasing mapping. By definition of $h$: :$\forall x \in S: h\left({x}\right) \subseteq x^\preceq$ Thus :$h \in F$ By Segment of Auxiliary Relation is Subset of Lower Closure: :$\forall x \in S: x^r \subseteq x^\preceq$ By definitions of $f$ and $h$: :$\forall x \in S: f\left({x}\right) \subseteq h\left({x}\right)$ By definition of $\precsim$: :$\forall x \in S: f\left({x}\right) \precsim h\left({x}\right)$ Thus by definition of ordering of $M$: :$f \preccurlyeq h$ {{qed|lemma}} === Condition $(3)$ === $k$ is well-defined because by Singleton of Bottom is Ideal: :$\left\{ {\bot}\right\}$ is an ideal in $R$. By Mapping is Constant iff Increasing and Decreasing: :$k$ is an increasing mapping. By definition of smallest element: :$\forall x \in S: \bot \preceq x$ By definition of lower closure of element: :$\forall x \in S: \bot \in x^\preceq$ By definition of subset: :$\forall x \in S: \left\{ {\bot}\right\} \subseteq x^\preceq$ By definition of $k$: :$\forall x \in S: k\left({x}\right) \subseteq x^\preceq$ Thus :$k \in F$ By definition of auxiliary relation: :$\forall x \in S: \left({\bot, x}\right) \in r$ By definition of $r$-segment: :$\forall x \in S: \bot \in x^r$ By definition of subset: :$\forall x \in S: \left\{ {\bot}\right\} \subseteq x^r$ By definitions of $k$ and $f$: :$\forall x \in S: k\left({x}\right) \subseteq f\left({x}\right)$ By definition of $\precsim$: :$\forall x \in S: k\left({x}\right) \precsim f\left({x}\right)$ Thus by definition of ordering of $M$: :$k \preccurlyeq f$ {{qed}}
=== Non-empty === By definition of singleton: :$\bot \in \set \bot$ By definition: :$\set \bot$ is a non-empty set. {{qed|lemma}} === Directed === Thus by Singleton is Directed and Filtered Subset: :$\set \bot$ is directed. {{qed|lemma}} === Lower === Let $x \in \set \bot, y \in S$ such that :$y \preceq x$ By definition of singleton: :$x = \bot$ By definition of smallest element: :$\bot \preceq y$ By definition of antisymmetry: :$y = \bot$ Thus by definition of singleton: :$y \in \set \bot$ Thus by definition: :$\set \bot$ is a lower set. {{qed|lemma}} Thus by definition: :$\set \bot$ is an ideal in $\struct {S, \preceq}$ {{qed}}
From Stirling's Formula: :$n! \sim \sqrt {2 \pi n} \left({\dfrac n e}\right)^n$ from which the result can be calculated. To count the number of digits: {{begin-eqn}} {{eqn | l = \log_{10} n!       | o = \sim       | r = \log_{10} \left({\sqrt {2 \pi n} \left({\dfrac n e}\right)^n}\right)       | c = Stirling's Formula }} {{eqn | r = \log_{10} \left({\sqrt {2 \pi n} }\right) + \log_{10} \left({\left({\dfrac n e}\right)^n}\right)       | c = Sum of Logarithms }} {{eqn | r = \frac 1 2 \log_{10} 2 \pi n + n \log_{10} \left({\dfrac n e}\right)       | c = Logarithm of Power }} {{eqn | r = \frac 1 2 \left({\log_{10} 2 + \log_{10} \pi}\right) + \frac 1 2 \log_{10} n + n \left({\log_{10} n - \log_{10} e}\right)       | c = Sum of Logarithms, Difference of Logarithms }} {{end-eqn}} We have: : Common Logarithm of $2$: $\ln 2 \approx 0.30102 \ 9996$ : Common Logarithm of $\pi$: $\ln \pi \approx 0.49714 \ 9873$ : Common Logarithm of $e$: $\ln e \approx 0.43429 \ 4481$ Hence: {{begin-eqn}} {{eqn | l = \log_{10} n!       | o = \sim       | r = \frac 1 2 \left({0.30102 \ 9996 + 0.49714 \ 9873}\right) + \frac 1 2 \log_{10} n + n \left({\log_{10} n - 0.43429 \ 4481}\right)       | c =  }} {{eqn | r = \left({n + \frac 1 2}\right) \log_{10} n - 0.43429 \ 4481 \, n + 0.39908 \ 9934       | c =  }} {{end-eqn}} Hence the result from Number of Digits in Number. {{qed}} Category:Factorials boax76h7gvqf13y3w9tvz95nqxlydm3
{{begin-eqn}} {{eqn | l = x^{\overline n}       | r = \prod_{j \mathop = 0}^{n - 1} \left({x + j}\right)       | c = Definition of Rising Factorial }} {{eqn | r = x \left({x + 1}\right) \cdots \left({x + n - 1}\right)       | c = Definition of Product Notation }} {{eqn | r = \left({x + n - 1}\right) \left({x + n - 2}\right) \cdots \left({x + 1}\right) x       | c = reversing the order }} {{eqn | r = \prod_{j \mathop = 0}^{n - 1} \left({\left({x + n - 1}\right) - j}\right)       | c = Definition of Falling Factorial }} {{eqn | r = \left({x + n - 1}\right)^{\underline n}       | c = Definition of Product Notation }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \paren {-x}^{\underline k}       | r = \prod_{j \mathop = 0}^{k - 1} \paren {-x - j}       | c = {{Defof|Falling Factorial}} }} {{eqn | r = \prod_{j \mathop = 0}^{k - 1} \paren {-1} \paren {x + j}       | c =  }} {{eqn | r = \prod_{j \mathop = 0}^{k - 1} \paren {-1} \prod_{j \mathop = 0}^{k - 1} \paren {x + j}       | c = Product of Products }} {{eqn | r = \paren {-1}^k \prod_{j \mathop = 0}^{k - 1} \paren {x + j}       | c =  }} {{eqn | r = \paren {-1}^k x^{\overline k}       | c = {{Defof|Rising Factorial}} }} {{eqn | ll= \leadsto       | l = \paren {-1}^k \paren {-x}^{\underline k}       | r = \paren {-1}^{2 k} x^{\overline k}       | c = multiplying both sides by $\paren {-1}^k$ }} {{eqn | r = x^{\overline k}       | c = $-1$ to an even power is $1$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = 1^{\overline n}       | r = \dfrac {\paren {1 + n - 1}!} {\paren {1 - 1}!}       | c = Rising Factorial as Quotient of Factorials }} {{eqn | r = \dfrac {n!} {0!}       | c =  }} {{eqn | r = n!       | c = Factorial of Zero }} {{end-eqn}} {{qed}} Category:Rising Factorials Category:Factorials 4dcqx0irfqi9tnnvf7ysu9tdczqi1vu
{{begin-eqn}} {{eqn | l = x^{\overline 1}       | r = \dfrac {\Gamma \left({x + 1}\right)}{\Gamma \left({x}\right)}       | c = Rising Factorial as Quotient of Factorials }} {{eqn | r = x       | c = Gamma Difference Equation }} {{end-eqn}} {{qed}} Category:Rising Factorials aeu50xkg5nt93tq2vghn7q97cbnuqvj
{{begin-eqn}} {{eqn | l = n^{\underline n}       | r = \dfrac {n!} {\paren {n - n}!}       | c = Falling Factorial as Quotient of Factorials }} {{eqn | r = \dfrac {n!} {0!}       | c =  }} {{eqn | r = n!       | c = Factorial of Zero }} {{end-eqn}} {{qed}} Category:Falling Factorials opk3w0uaoprxmpgvd9rzvqako48cfx9
{{begin-eqn}} {{eqn | l = x^{\underline 1}       | r = \dfrac {\Gamma \left({x + 1}\right)} {\Gamma \left({x - 1 + 1}\right)}       | c = Falling Factorial as Quotient of Factorials }} {{eqn | r = \dfrac {\Gamma \left({x + 1}\right)} {\Gamma \left({x}\right)}       | c =  }} {{eqn | r = x       | c = Gamma Difference Equation }} {{end-eqn}} {{qed}} Category:Falling Factorials jjmyfhhwutxv5dyo2xhk3s6plb6w4nx
Define relation $\mathcal R$ on $S$: :$\forall x, y \in S: \left({x, y}\right) \in \mathcal R \iff x \in f\left({y}\right)$ We will prove that :$(i): \quad \forall x, y \in S: \left({x, y}\right) \in \mathcal R \implies x \preceq y$ Let $x, y \in S$ such that : $\left({x, y}\right) \in \mathcal R$ By definitions of $\mathcal R$ and $F$: :$x \in f\left({y}\right) \subseteq y^\preceq$ By definition of subset; :$x \in y^\preceq$ Thus be definition of lower closure of element: :$x \preceq y$ {{qed|lemma}} We will prove that :$(ii): \quad \forall x, y, z, u \in S: x \preceq y \land \left({y, z}\right) \in \mathcal R \land z \preceq u \implies \left({x, u}\right) \in \mathcal R$ Let $x, y, z, u \in S$ such that : $x \preceq y \land \left({y, z}\right) \in \mathcal R \land z \preceq u$ By definition of $\mathcal R$: :$y \in f\left({z}\right)$ By definition of increasing mapping: :$f\left({z}\right) \precsim f\left({u}\right)$ By definition of $\precsim$: :$f\left({z}\right) \subseteq f\left({u}\right)$ By definition of subset: :$y \in f\left({u}\right)$ By definition of $F$: :$f\left({u}\right)$ is an ideal in $R$. By definition of ideal: :$f\left({u}\right)$ is a lower set. By definition of lower set: :$x \in f\left({u}\right)$ Thus by definition of $\mathcal R$: :$\left({x, u}\right) \in \mathcal R$ {{qed|lemma}} We will prove that :$(iii): \quad \forall x, y, z \in S: \left({x, z}\right) \in \mathcal R \land \left({y, z}\right) \in \mathcal R \implies \left({x \vee y, z}\right) \in \mathcal R$ Let $x, y, z \in S$ such that : $\left({x, z}\right) \in \mathcal R \land \left({y, z}\right) \in \mathcal R$ By definition of $\mathcal R$: :$x \in f\left({z}\right)$ and $y \in f\left({z}\right)$ By definition of $F$: :$f\left({z}\right)$ is an ideal in $R$. By definition of ideal: :$f\left({z}\right)$ is a directed lower set. By definition of directed subset: :$\exists d \in f\left({z}\right): x \preceq d \land y \preceq d$ By definition of supremum: :$x \vee y \preceq d$ By definition of lower set: :$x \vee y \in f\left({z}\right)$ Thus by definition of $\mathcal R$: :$\left({x \vee y, z}\right) \in \mathcal R$ {{qed|lemma}} We will prove that :$(iv): \quad \forall x \in S: \left({\bot, x}\right) \in \mathcal R$ where $\bot$ denotes the smallest element in $R$. Let $x \in S$. By definition of $F$: :$f\left({x}\right)$ is an ideal in $R$. By definition of ideal: :$f\left({z}\right)$ is a non-empty lower set. By definition of non-empty set: :$\exists z: z \in f\left({x}\right)$ By definition of smallest element: :$\bot \preceq z$ By definition of lower set: :$\bot \in f\left({x}\right)$ Thus by definition of $\mathcal R$: :$\left({\bot, x}\right) \in \mathcal R$ {{qed|lemma}} By definition: :$\mathcal R$ is an auxiliary relation on $S$. Thus by definitions of $\mathcal R$ and $\mathcal R$-segment: :$\forall x \in S: f\left({x}\right) = x^{\mathcal R}$ {{qed}}
From the definition of binomial coefficient: :$\dbinom n 2 = \dfrac {n!} {2! \paren {n - 2}!}$ The result follows directly from the definition of the factorial: :$2! = 1 \times 2$ {{qed}}
By Segment of Auxiliary Relation Mapping is Element of Increasing Mappings Satisfying Inclusion in Lower Closure define $G:\mathit{Aux}\left({R}\right) \to F$ :$\forall \mathcal R \in \mathit{Aux}\left({R}\right): G\left({\mathcal R}\right) = \left({S \ni x \mapsto x^{\mathcal R} }\right)$ We will prove by Order Isomorphism is Surjective Order Embedding that :$G$ is an order isomorphism. === Surjection === By Element of Increasing Mappings Satisfying Inclusion in Lower Closure is Generated by Auxiliary Relation: :$\forall f \in F: \exists \mathcal R \in \mathit{Aux}\left({R}\right): f = \left({S \ni x \mapsto x^{\mathcal R} }\right)$ By definition of image of mapping: :$\forall f \in F: f \in \operatorname{Im}\left({G}\right)$ Thus by definition: :$G$ is a surjection. {{qed|lemma}} === Order Embedding === Let $\mathcal R, \mathcal Q \in \mathit{Aux}\left({R}\right)$ We will prove that :$\mathcal R \precsim' \mathcal Q \implies G\left({\mathcal R}\right) \preccurlyeq G\left({\mathcal Q}\right)$ Let :$\mathcal R \precsim' \mathcal Q$ By definition of $\precsim'$: :$\mathcal R \subseteq \mathcal Q$ Let $x \in S$. By Relation Segment is Increasing :$x^{\mathcal R} \subseteq x^{\mathcal Q}$ By Relation Segment of Auxiliary Relation is Ideal: :$x^{\mathcal R} \in \mathit{Ids}\left({R}\right)$ and $x^{\mathcal Q} \in \mathit{Ids}\left({R}\right)$ By definition of $\precsim$: :$x^{\mathcal R} \precsim x^{\mathcal Q}$ By definition of ordering on mappings: :$\left({S \ni x \mapsto x^{\mathcal R} }\right) \precsim \left({S \ni x \mapsto x^{\mathcal Q} }\right)$ Thus by definitions of $G$ and $\preccurlyeq$: :$G\left({\mathcal R}\right) \preccurlyeq G\left({\mathcal Q}\right)$ {{qed|lemma}} We will prove that :$G\left({\mathcal R}\right) \preccurlyeq G\left({\mathcal Q}\right) \implies \mathcal R \precsim' \mathcal Q$ Let :$G\left({\mathcal R}\right) \preccurlyeq G\left({\mathcal Q}\right)$ By definition of $G$ and $\preccurlyeq$: :$\left({S \ni x \mapsto x^{\mathcal R} }\right) \precsim \left({S \ni x \mapsto x^{\mathcal Q} }\right)$ Let $x, y \in S$ such that :$\left({x, y}\right) \in \mathcal R$ By definition of $\mathcal R$-segment: :$x \in y^{\mathcal R}$ By definition of ordering on mappings: :$y^{\mathcal R} \precsim y^{\mathcal Q}$ By definition of $\precsim$: :$y^{\mathcal R} \subseteq y^{\mathcal Q}$ By definition of subset: :$x \in y^{\mathcal Q}$ Thus by definition of $\mathcal Q$-segment: :$\left({x, y}\right) \in \mathcal Q$ By definition of subset: :$\mathcal R \subseteq \mathcal Q$ Thus by definition of $\precsim'$: :$\mathcal R \precsim' \mathcal Q$ {{qed}}
{{begin-eqn}} {{eqn | l = \sum_k \binom r {m + k} \binom s {n + k}       | r = \sum_k \binom r {r - m - k} \binom s {s - n - k}       | c = Symmetry Rule for Binomial Coefficients }} {{eqn | r = \sum_k \binom r k \binom s {s - n - \left({r - m - k}\right)}       | c = Change of Index Variable of Summation }} {{eqn | r = \sum_k \binom r k \binom s {r - m - k + n}       | c = Symmetry Rule for Binomial Coefficients }} {{eqn | r = \sum_k \binom r k \binom s {\left({r - m + n}\right) - k}       | c =  }} {{eqn | r = \binom {r + s} {r - m + n}       | c = Chu-Vandermonde Identity }} {{end-eqn}} {{qed}}
The proof proceeds by induction on $r$. For all $r \in \Z_{>0}$, let $P \left({r}\right)$ be the proposition: :$\displaystyle \sum_k \binom r k \binom {s + k} n \left({-1}\right)^{r - k} = \binom s {n - r}$ === Basis for the Induction === $P \left({0}\right)$ is the case: {{begin-eqn}} {{eqn | l = \sum_k \binom 0 k \binom {s + k} n \left({-1}\right)^{0 - k}       | r = \delta_{0 k} \binom {s + k} n \left({-1}\right)^{0 - k}       | c = Zero Choose n }} {{eqn | r = \binom s n       | c = All terms vanish but for $k = 0$ }} {{eqn | r = \binom s {n - 0}       | c =  }} {{end-eqn}} Thus $P \left({0}\right)$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({m}\right)$ is true, where $m \ge 0$, then it logically follows that $P \left({m + 1}\right)$ is true. This is the induction hypothesis: :$\displaystyle \sum_k \binom m k \binom {s + k} n \left({-1}\right)^{m - k} = \binom s {n - m}$ from which it is to be shown that: :$\displaystyle \sum_k \binom {m + 1} k \binom {s + k} n \left({-1}\right)^{m + 1 - k} = \binom s {n - \left({m + 1}\right)}$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | o =        | r = \sum_k \binom {m + 1} k \binom {s + k} n \left({-1}\right)^{m + 1 - k}       | c =  }} {{eqn | r = \sum_k \left({\binom m k + \binom m {k - 1} }\right) \binom {s + k} n \left({-1}\right)^{m + 1 - k}       | c = Pascal's Rule }} {{eqn | r = \sum_k \binom m k \binom {s + k} n \left({-1}\right)^{m + 1 - k} + \sum_k \binom m {k - 1} \binom {s + k} n \left({-1}\right)^{m + 1 - k}       | c =  }} {{eqn | r = -\sum_k \binom m k \binom {s + k} n \left({-1}\right)^{m - k} + \sum_k \binom m {k - 1} \binom {s + k} n \left({-1}\right)^{m + 1 - k}       | c =  }} {{eqn | r = -\binom s {n - m} + \sum_k \binom m {k - 1} \binom {s + k} n \left({-1}\right)^{m + 1 - k}        | c = Induction Hypothesis }} {{eqn | r = -\binom s {n - m} + \sum_k \binom m {k - 1} \left({\binom {s + k - 1} n + \binom {s + k - 1} {n - 1} }\right) \left({-1}\right)^{m + 1 - k}       | c = Pascal's Rule }} {{eqn | r = -\binom s {n - m} + \sum_k \binom m {k - 1} \binom {s + k - 1} n \left({-1}\right)^{m - \left({k - 1}\right)}       | c =  }} {{eqn | o =        | r = + \sum_k \binom m {k - 1} \binom {s + k - 1} {n - 1} \left({-1}\right)^{m - \left({k - 1}\right)}       | c =  }} {{eqn | r = -\binom s {n - m} + \sum_k \binom m k \binom {s + k} n \left({-1}\right)^{m - k} + \sum_k \binom m k \binom {s + k} {n - 1} \left({-1}\right)^{m - k}       | c = Translation of Index Variable of Summation }} {{eqn | r = -\binom s {n - m} + \binom s {n - m} + \binom s {n - 1 - m}       | c = Induction Hypothesis }} {{eqn | r = \binom s {n - \left({m + 1}\right)}       | c = simplifying }} {{end-eqn}} So $P \left({m}\right) \implies P \left({m + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \sum_k \binom r k \binom {s + k} n \left({-1}\right)^{r - k} = \binom s {n - r}$ for all $s \in \R, r \in \Z_{\ge 0}, n \in \Z$. {{qed}}
{{begin-eqn}} {{eqn | o =        | r = \sum_{k \mathop = 0}^r \binom {r - k} m \binom {s + k} n       | c =  }} {{eqn | r = \sum_{k \mathop = 0}^r \binom {-\left({m + 1}\right)} {r - k - m} \binom {-\left({n + 1}\right)} {s + k - m} \left({-1}\right)^{r - m + s - m}       | c = Moving Top Index to Bottom in Binomial Coefficient }} {{eqn | r = \binom {-\left({m + 1}\right) - \left({n + 1}\right)} {r - m + s - m} \left({-1}\right)^{r - m + s - m}       | c = Chu-Vandermonde Identity }} {{eqn | r = \binom {r + s + 1} {m + n + 1}       | c = Moving Top Index to Bottom in Binomial Coefficient }} {{end-eqn}} {{qed}}
From Sum over $k$ of $\dbinom {r - k t} k$ by $\dfrac r {r - k t}$ by $z^k$: :$\displaystyle \sum_k A_k \left({r, t}\right) z^k = x^r$ where: : $A_n \left({x, t}\right)$ is the polynomial of degree $n$ defined as: ::$A_n \left({x, t}\right) := \dbinom {x - n t} n \dfrac x {x - n t}$ :for $x \ne n t$ :$z = x^{t + 1} - x^t$. From Sum over $k$ of $\dbinom {r - k t} k$ by $z^k$: :$\displaystyle \sum_k \dbinom {r - t k} k z^k = \frac {x^{r + 1} } {\left({t + 1}\right)x - t}$ Thus: {{begin-eqn}} {{eqn | l = \sum_k A_k \left({r, t}\right) z^k \sum_k \dbinom {s - t k} k z^k       | r = \frac {x^r x^{s + 1} } {\left({t + 1}\right)x - t}       | c =  }} {{eqn | r = \frac {x^{\left({r + s}\right) + 1} } {\left({t + 1}\right)x - t}       | c =  }} {{eqn | r = \sum_k \dbinom {r + s - t k} k z^k       | c =  }} {{end-eqn}} Equating coefficients of $z$: {{begin-eqn}} {{eqn | l = \sum_k A_k \left({r, t}\right) z^k \sum_k \dbinom {s - \left({n - k}\right) t} k z^{n - k}       | r = \binom {r + s - n t} n z^n       | c =  }} {{eqn | l = \sum_k \dbinom {r - k t} k \dfrac r {r - k t} z^k \dbinom {s - \left({n - k}\right) t} k z^{n - k}       | r = \binom {r + s - n t} n z^n       | c =  }} {{eqn | l = \sum_k \dbinom {r - k t} k \dbinom {s - \left({n - k}\right) t} k \dfrac r {r - k t}       | r = \binom {r + s - t n} n       | c =  }} {{end-eqn}} {{qed}}