hcju/proofwikiqa huggingface dataset:
{{begin-eqn}} {{eqn | l = \sum_k \binom r k \binom s k k       | r = \sum_k \binom r k \binom {s - 1} {k - 1} s       | c = Factors of Binomial Coefficient }} {{eqn | r = s \sum_k \binom r k \binom {s - 1} {k - 1}       | c = as $s$ is constant }} {{eqn | r = s \binom {r + s - 1} {r - 1}       | c = Sum over $k$ of $\dbinom r {m + k} \dbinom s {n + k}$: $s \gets s - 1$, $m \gets 0$, $n \gets -1$ }} {{end-eqn}} {{qed}}
Let $y \in x^\RR$. By definition of $\RR$-segment: :$\tuple {y, x} \in \RR$ By definition of subset: :$\tuple {y, x} \in \QQ$ Thus by definition of $\QQ$-segment: :$y \in x^\QQ$ {{qed}}
{{begin-eqn}} {{eqn | r = \sum_k \binom {n + k} {2 k} \binom {2 k} k \frac {\left({-1}\right)^k} {k + 1}       | o =        | c =  }} {{eqn | r = \sum_k \binom {n + k} k \binom n k \frac {\left({-1}\right)} {k + 1}       | c = Product of $\dbinom r m$ with $\dbinom m k$ }} {{eqn | r = \sum_k \binom {n + k} k \binom {n + 1} {k + 1} \frac {\left({-1}\right)^k} {n + 1}       | c = Factors of Binomial Coefficient }} {{eqn | r = \frac 1 {n + 1} \sum_{k \mathop \ge 0} \binom {n + k} n \binom {n + 1} {k + 1} \left({-1}\right)^k       | c = Symmetry Rule for Binomial Coefficients and rearranging }} {{eqn | r = -\frac 1 {n + 1} \sum_{k \mathop \ge 1} \binom {n - 1 + k} n \binom {n + 1} k \left({-1}\right)^k       | c = Translation of Index Variable of Summation }} {{eqn | r = -\frac 1 {n + 1} \sum_{k \mathop \ge 0} \binom {n - 1 + k} n \binom {n + 1} k \left({-1}\right)^k + \frac 1 {n + 1} \binom {n - 1} n       | c = separating out the case where $k = 0$ }} {{eqn | r = -\frac 1 {n + 1} \left({-1}\right)^{n + 1} \binom {n - 1} {-1} + \frac 1 {n + 1} \binom {n - 1} n       | c = Sum over $k$ of $\dbinom r k \dbinom {s + k} n (-1)^{r - k}$ }} {{eqn | r = \frac 1 {n + 1} \binom {n - 1} n       | c = as $\dbinom {n - 1} {-1} = 0$ }} {{eqn | r = \left[{n = 0}\right]       | c = as $\dbinom {n - 1} n = 1 \iff n = 0$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | r = \sum_k \binom {n + k} {2 k} \binom {2 k} k \frac {\left({-1}\right)^k} {k + 1}       | o =        | c =  }} {{eqn | r = \sum_k \binom {n + k} k \binom n k \frac {\left({-1}\right)} {k + 1}       | c = Product of $\dbinom r m$ with $\dbinom m k$ }} {{eqn | r = \sum_k \binom {n + k} k \binom {n + 1} {k + 1} \frac {\left({-1}\right)^k} {n + 1}       | c = Factors of Binomial Coefficient }} {{eqn | r = \frac 1 {n + 1} \sum_k \binom {- \left({n + 1}\right)} k \binom {n + 1} {k + 1}       | c = Negated Upper Index of Binomial Coefficient }} {{eqn | r = \frac 1 {n + 1} \binom {n + 1 - \left({n + 1}\right)} {n + 1 - 1 + 0}       | c = Sum over $k$ of $\dbinom r {m + k} \dbinom s {n + k}$ }} {{eqn | r = \frac 1 {n + 1} \binom 0 n       | c =  }} {{eqn | r = \left[{n = 0}\right]       | c = {{Defof|Binomial Coefficient}} }} {{end-eqn}} {{qed}}
From Sum over $k$ of $\dbinom {r - k} m \dbinom {s + k} n$: :$\displaystyle \sum_{k \mathop = 0}^r \binom {r - k} m \binom {s + k} n = \binom {r + s + 1} {m + n + 1}$ Setting $r = n + k - 1$, $m = 2 k$, $s = 0$, $n = m - 1$ and using $j$ as the index variable, this gives us: :$\displaystyle \sum_{j \mathop = 0}^{n + k - 1} \binom {n + k - 1 - j} {2 k} \binom j {m - 1} = \dbinom {n + k} {m + 2 k}$ This leads to: :$\displaystyle \sum_k \binom {n + k} {m + 2 k} \binom {2 k} k \frac {\left({-1}\right)^k} {k + 1} = \sum_k \sum_{j \mathop = 0}^{n + k - 1} \binom {n + k - 1 - j} {2 k} \binom {2 k} k \binom j {m - 1} \frac {\left({-1}\right)} {k + 1}$ Note that the terms of the above are $0$ when $n \le j < n + k - 1$. This means that $k \ge 1$. So: :$0 \le n + k - 1 - j \le k - i < 2 k$ Thus the first binomial coefficient in the above will vanish. The condition on the second summation can then be replaced by $0 \le j < n$. After exchange of summation and application of Sum over $k$ of $\dbinom {n + k} {2 k} \dbinom {2 k} k \dfrac {-1^k} {k + 1}$ converts it into: :$\displaystyle \sum{_0 \mathop \le j < n} \binom j {m - 1} \delta_{\left({n \mathop - 1 \mathop j}\right) 0}$ where $\delta_{\left({n \mathop - 1 \mathop j}\right) 0}$ denotes the Kronecker delta. All terms are therefore $0$ except where $j = n - 1$. Hence the result. {{qed}}
Let: :$\displaystyle S := \sum_k \binom {n + k} {m + 2 k} \binom {2 k} k \frac {\paren {-1}^k} {k + 1}$ Then: {{begin-eqn}} {{eqn | l = \binom {2 k + 1} {k + 1}       | r = \binom {2 k} k \frac {2 k + 1} {k + 1}       | c = Factors of Binomial Coefficient }} {{eqn | ll= \leadsto       | l = \binom {2 k} k       | r = \binom {2 k + 1} {k + 1} \frac {k + 1} {2 k + 1}       | c =  }} {{eqn | r = \binom {2 k + 1} k \frac {k + 1} {2 k + 1}       | c = Symmetry Rule for Binomial Coefficients }} {{end-eqn}} Also: {{begin-eqn}} {{eqn | l = \binom {n + k} {m + 2 k}       | r = \paren {-1}^{n + k - m - 2 k} \binom {-\paren {m + 2 k + 1} } {n + k - m - 2 k}       | c = Moving Top Index to Bottom in Binomial Coefficient }} {{eqn | r = \paren {-1}^{n - m - k} \binom {- m - 2 k - 1} {n - m - k}       | c =  }} {{end-eqn}} Reassembling $S$: :$\displaystyle S = \sum_k \binom {1 + 2 k} k \binom {-m - 2 k - 1} {n - m - k} \frac {\paren {-1}^{n - m} } {1 + 2 k}$ Recall Sum over $k$ of $\dbinom {r - t k} k$ by $\dbinom {s - t \paren {n - k} } {n - k}$ by $\dfrac r {r - t k}$: :$\displaystyle \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t \paren {n - k} } {n - k} \frac r {r - t k} = \binom {r + s - t n} n$ Making the following substitutions: :$r \gets 1$ :$t \gets -2$ :$n \gets n - m$ we obtain: {{begin-eqn}} {{eqn | l = s + 2 \paren {n - m - k}       | r = - m - 2 k - 1 }} {{eqn | ll= \leadsto       | l = s       | r = m - 2 n - 1 }} {{eqn | ll= \leadsto       | l = \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t \paren {n - k} } {n - k} \frac r {r - t k}       | r = \sum_{k \mathop \ge 0} \binom {1 + 2 k} k \binom {- m - 2 k - 1} {n - m - k} \frac 1 {1 + 2 k} }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = S       | r = \paren {-1}^{n - m} \sum_k \binom {1 + 2 k} k \binom {- m - 2 k - 1} {n - m - k} \frac 1 {1 + 2 k}       | c =  }} {{eqn | r = \paren {-1}^{n - m} \binom {1 + \paren {m - 2 n - 1} + 2 \paren {n - m} } {n - m}       | c =  }} {{eqn | r = \paren {-1}^{n - m} \binom {- m} {n - m}       | c =  }} {{eqn | r = \binom {n - 1} {n - m}       | c = Negated Upper Index of Binomial Coefficient }} {{eqn | r = \binom {n - 1} {m - 1}       | c = Symmetry Rule for Binomial Coefficients }} {{end-eqn}} {{qed}}
By Supremum of Lower Closure of Element: :$\sup \left({x^\preceq}\right) = x$ By Lower Closure of Element is Ideal: :$x^\preceq \in \mathit{Ids}$ Then by definition of reflexivity: :$x^\preceq \in \left\{ {I \in \mathit{Ids}: x \preceq \sup I}\right\}$ We will prove that :$\displaystyle \bigcap \left\{ {I \in \mathit{Ids}: x \preceq \sup I}\right\} \subseteq x^\ll$ Let $z \in \displaystyle \bigcap \left\{ {I \in \mathit{Ids}: x \preceq \sup I}\right\}$ By definition of intersection: :$\forall I \in \mathit{Ids}: x \preceq \sup I \implies z \in I$ By Way Below iff Second Operand Preceding Supremum of Ideal implies First Operand is Element of Ideal: :$z \ll x$ Thus by definition of way below closure: :$z \in x^\ll$ {{qed|lemma}} It by definition of set equality remains to prove opposite inclusion. Let $z \in x^\ll$ By definition of way below closure: :$z \ll x$ We will prove that :$\forall Y \in \left\{ {I \in \mathit{Ids}: x \preceq \sup I}\right\}: z \in Y$ Let $Y \in \left\{ {I \in \mathit{Ids}: x \preceq \sup I}\right\}$ Then :$Y \in \mathit{Ids}$ and $x \preceq \sup Y$ Thus Way Below iff Second Operand Preceding Supremum of Ideal implies First Operand is Element of Ideal: :$z \in Y$ {{qed|lemma}} Thus by definition of intersection: :$z \in \displaystyle \bigcap \left\{ {I \in \mathit{Ids}: x \preceq \sup I}\right\}$ {{qed}}
By Way Below Closure is Directed in Bounded Below Join Semilattice: :$x^\ll$ is a non-empty directed set. Let $y \in x^\ll, z \in S$ such that :$z \preceq y$ By definition of way below closure: :$y \ll x$ By definition of reflexivity: :$x \preceq x$ By Preceding and Way Below implies Way Below: :$z \ll x$ Thus by definition of way below closure: :$z \in x^\ll$ Thus by definition :$x^\ll$ is a lower set. Thus by definition :$x^\ll$ is an ideal in $L$. {{qed}}
Let: :$\displaystyle S = \sum_k A_k \left({r, t}\right) A_{n - k} \left({s, t}\right)$ Both sides of the statement of the theorem are polynomials in $r$, $s$ and $t$. Therefore it can be assumed that $r \ne k t \ne s$ for $0 \le k \le n$ or something will become undefined. By replacing the polynomials $A_n$ with their binomial coefficient definitions, the theorem can be expressed as: :$\displaystyle S = \sum_k \dbinom {r - k t} k  \dbinom {s - \left({n - k}\right) t} {n - k} \dfrac r {r - k t} \dfrac s {s - \left({n - k}\right) t}$ Using the technique of partial fractions: :$\dfrac 1 {r - k t} \dfrac 1 {s - \left({n - k}\right) t} = \dfrac 1 {r + s - n t} \left({\dfrac 1 {r - k t} + \dfrac 1 {s - \left({n - k}\right) t} }\right)$ Thus: :$\displaystyle S = \frac s {r + s - n t} \sum_k \dbinom {r - k t} k \dbinom {s - \left({n - k}\right) t} {n - k} \dfrac r {r - k t} + \frac r {r + s - n t} \sum_k \dbinom {r - k t} k \dbinom {s - \left({n - k}\right) t} {n - k} \dfrac s {s - \left({n - k}\right) t}$ From Sum over $k$ of $\dbinom {r - t k} k \dbinom {s - t \left({n - k}\right)} {n - k} \dfrac r {r - t k}$: :$\displaystyle \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t \left({n - k}\right)} {n - k} \frac r {r - t k} = \binom {r + s - t n} n$ for $r, s, t \in \R, n \in \Z$. Thus we have: :$S = \dfrac s {r + s - n t} \dbinom {r + s - t n} n + \dfrac r {r + s - n t} \dbinom {s + r - t n} n$ after changing $k$ to $n - k$ in the second term. That is: :$S = \dbinom {r + s - n t} n \dfrac {r + s} {r + s - n t}$ which is $A_n \left({r + s, t}\right)$. {{qed}}
From Sum over $k$ of $\dbinom {r - k t} k$ by $\dfrac r {r - k t}$ by $z^k$: :$\displaystyle \sum_k A_k \left({r, t}\right) z^k = x^r$ and: :$\displaystyle \sum_k A_k \left({s, t}\right) z^k = x^s$ Hence: {{begin-eqn}} {{eqn | l = x^{r + s}       | r = \sum_k A_k \left({r, t}\right) z^k \sum_k A_k \left({s, t}\right) z^k       | c =  }} {{eqn | r = \sum_k A_k \left({r + s, t}\right) z^k       | c =  }} {{end-eqn}} Taking the $z^n$ coefficient: {{begin-eqn}} {{eqn | l = \sum_k A_k \left({r, t}\right) z^k \sum_k A_{n - k} \left({s, t}\right) z^{n - k}       | r = A_n \left({r + s, t}\right) z^n       | c =  }} {{eqn | ll= \leadsto       | l = \sum_k A_k \left({r, t}\right) \sum_k A_{n - k} \left({s, t}\right)       | r = A_n \left({r + s, t}\right) }} {{end-eqn}} {{qed}}
Let $n$ be a non-negative integer. We assume a solution of the form: :$(1): \quad n! = a_0 + a_1 n + a_2 n \left({n - 1}\right) + a_3 n \left({n - 1}\right) \left({n - 2}\right) + \cdots$ We can express $(1)$ using binomial coefficients: :$(2): \quad n! = \displaystyle \sum_k \dbinom n k k! a_k$ Then: {{begin-eqn}} {{eqn | l = \sum_n n! \binom m n \left({-1}\right)^{n - k}       | r = \sum_n \left({\sum_k \binom n k k! \, a_k}\right) \binom m n \left({-1}\right)^{n - k}       | c = substituting for $n!$ from $(2)$ }} {{eqn | r = \sum_k k! \, a_k \sum_n \binom n k \binom m n \left({-1}\right)^{m - n}       | c = changing order of summation }} {{eqn | r = \sum_k k! \, a_k \delta_{k m}       | c = Corollary to Sum over $k$ of $\dbinom r k \dbinom {s + k} n \left({-1}\right)^{r - k}$ }} {{eqn | r = m! \, a_m       | c = as the $m$th term is the only one left standing }} {{eqn | ll= \leadsto       | l = a_m       | r = \sum_{n \mathop \ge 0} \left({-1}\right)^{m - n} \dfrac {n!} {m!} \binom m n       | c =  }} {{eqn | r = \sum_{n \mathop = 0}^m \left({-1}\right)^{m - n} \dfrac {n!} {m!} \dfrac {m!} {n! \, \left({m - n}\right)!}       | c = {{Defof|Binomial Coefficient}} }} {{eqn | r = \sum_{n \mathop = 0}^m \frac {\left({-1}\right)^{m - n} } {\left({m - n}\right)!}       | c = simplifying }} {{eqn | r = \sum_{n \mathop = 0}^m \frac {\left({-1}\right)^n} {n!}       | c = Permutation of Indices of Summation }} {{end-eqn}} {{qed}}
From the corollary to Sum over $k$ of $\dbinom r k \dbinom {s + k} n \paren {-1}^{r - k}$: :$\displaystyle \sum_k \binom r k \binom k n \paren {-1}^{r - k} = \delta_{n r}$ where $\delta_{n r}$ denotes the Kronecker delta. Thus when $n \ne r$: :$\displaystyle \sum_k \binom r k \binom k n \paren {-1}^{r - k} = 0$ and so: :$\displaystyle \sum_k \binom r k \paren {-1}^{r - k} \paren {c_0 \binom k 0 + c_1 \binom k 1 + \cdots + c_m \binom k m} = c_r$ as the only term left standing is the $r$th one. Choosing the coefficients $c_i$ as appropriate, a polynomial in $k$ can be expressed as a summation of binomial coefficients in the form: :$c_0 \dbinom k 0 + c_1 \dbinom k 1 + \cdots + c_m \dbinom k m$ Thus we can rewrite such a polynomial in $k$ as: :$b_0 + b_1 k + \cdots + b_r k^r$ Since each $c_m \dbinom k m$ is a polynomial of degree $m$, it follows that the only one with a non-zero degree $r$ term is $c_r \dbinom k r$. The coefficient of $k^r$ in $c_r \dbinom k r$ must be equal to $b_r$, that is: :$b_r = \dfrac {c_r}{r!}$ Hence the result. {{qed}}
From Summation of Powers over Product of Differences: :$\displaystyle \sum_{j \mathop = 1}^n \begin{pmatrix} {\dfrac { {x_j}^r} {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \left({x_j - x_k}\right)} } \end{pmatrix} = \begin{cases} 0 & : 0 \le r < n - 1 \\ 1 & : r = n - 1 \\ \displaystyle \sum_{j \mathop = 1}^n x_j & : r = n \end{cases}$ Now we have: {{begin-eqn}} {{eqn | l = \frac {\left({-1}\right)^k \dbinom n k} {n!}       | r = \frac {\left({-1}\right)^k } {k! \left({n - k}\right)!}       | c =  }} {{eqn | r = \frac {\left({-1}\right)^k } {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \left({k - j}\right)}       | c =  }} {{end-eqn}} {{proof wanted|in progress}}
From Sum over $k$ of $\dbinom r k \left({-1}\right)^k$ by Polynomial: :$\displaystyle \sum_k \binom r k \left({-1}\right)^{r - k} P_r \left({k}\right) = r! \, b_r$ where: :$P_r \left({k}\right) = b_0 + b_1 k + \cdots + b_r k^r$ is a polynomial in $k$ of degree $r$. {{proof wanted}}
{{begin-eqn}} {{eqn | l = \sum_{k \mathop = 0}^n \binom {-2} k       | r = \sum_{k \mathop = 0}^n \left({-1}\right) \binom {k - \left({-2}\right) - 1} k       | c = Negated Upper Index of Binomial Coefficient }} {{eqn | r = \sum_{k \mathop = 0}^n \left({-1}\right) \binom {k + 1} k       | c =  }} {{eqn | r = \sum_{k \mathop = 0}^n \left({-1}\right) \left({k + 1}\right)       | c = Binomial Coefficient with Self minus One }} {{eqn | r = 1 - 2 + 3 - 4 + \cdots \pm \left({n + 1}\right)       | c =  }} {{eqn | r = \left({1 + 2 + 3 + 4 + \cdots + \left({n + 1}\right)}\right) - 2 \times \left({2 + 4 + 6 + 8 + \cdots + m}\right)       | c = where $m = n$ or $m = n + 1$ according to whether $n$ is odd or even }} {{end-eqn}} When $n$ is even, we have: {{begin-eqn}} {{eqn | r = \left({1 + 2 + 3 + 4 + \cdots + \left({n + 1}\right)}\right) - 2 \times \left({2 + 4 + 6 + 8 + \cdots + n}\right)       | o =        | c =  }} {{eqn | r = \left({1 + 2 + 3 + 4 + \cdots + \left({n + 1}\right)}\right) - 4 \left({1 + 2 + 3 + 4 + \cdots + \frac n 2}\right)       | c =  }} {{eqn | r = \sum_{k \mathop = 2}^{n + 1} k - 4 \sum_{k \mathop = 1}^{\frac n 2} k       | c =  }} {{eqn | r = \frac {\left({n + 1}\right) \left({n + 2}\right)} 2 - 4 \frac {\frac n 2 \left({\frac n 2 + 1}\right)} 2       | c = Closed Form for Triangular Numbers }} {{eqn | r = \frac {\left({n + 1}\right) \left({n + 2}\right)} 2 - \frac {n \left({n + 2}\right)} 2       | c = simplifying }} {{eqn | r = \frac {\left({n + 2}\right) \left({n + 1 - n}\right)} 2       | c = simplifying }} {{eqn | r = \frac {n + 2} 2       | c = simplifying }} {{end-eqn}} As $n$ is even, $n + 1$ is odd, and so: :$\dfrac {n + 2} 2 = \left({-1}\right)^n \left\lceil {\dfrac {n + 1} 2}\right\rceil$ {{qed|lemma}} When $n$ is odd, we have: {{begin-eqn}} {{eqn | r = \left({1 + 2 + 3 + 4 + \cdots + \left({n + 1}\right)}\right) - 2 \times \left({2 + 4 + 6 + 8 + \cdots + n + 1}\right)       | o =        | c =  }} {{eqn | r = \left({1 + 2 + 3 + 4 + \cdots + \left({n + 1}\right)}\right) - 4 \left({1 + 2 + 3 + 4 + \cdots + \frac {n + 1} 2}\right)       | c =  }} {{eqn | r = \sum_{k \mathop = 2}^{n + 1} k - 4 \sum_{k \mathop = 1}^{\frac {n + 1} 2} k       | c =  }} {{eqn | r = \frac {\left({n + 1}\right) \left({n + 2}\right)} 2 - 4 \frac {\frac {n + 1} 2 \left({\frac {n + 1} 2 + 1}\right)} 2       | c = Closed Form for Triangular Numbers }} {{eqn | r = \frac {\left({n + 1}\right) \left({n + 2}\right)} 2 - \frac {\left({n + 1}\right) \left({n + 3}\right)} 2       | c = simplifying }} {{eqn | r = \frac {\left({n + 1}\right) \left({\left({n + 2}\right) - \left({n + 3}\right)}\right)} 2       | c = simplifying }} {{eqn | r = \left({-1}\right) \frac {n + 1} 2       | c = simplifying }} {{end-eqn}} As $n$ is odd, $n + 1$ is even, and so $\dfrac {n + 1} 2$ is an integer. Thus from Real Number is Integer iff equals Ceiling: :$\left({-1}\right) \dfrac {n + 1} 2 = \left({-1}\right)^n \left\lceil {\dfrac {n + 1} 2}\right\rceil$ {{qed|lemma}} Thus: :$\displaystyle \sum_{k \mathop = 0}^n \binom {-2} k = \left({-1}\right)^n \left\lceil {\dfrac {n + 1} 2}\right\rceil$ whether $n$ is odd or even. Hence the result. {{qed}}
{{begin-eqn}} {{eqn | l = \sum_{k \mathop = 0}^n \binom m k \left({k - \dfrac m 2}\right)       | r = \sum_{k \mathop = 0}^n k \binom m k - \dfrac m 2 \sum_{k \mathop = 0}^n \binom m k       | c =  }} {{eqn | r = \sum_{k \mathop = 0}^n m \binom {m - 1} {k - 1} - \dfrac m 2 \sum_{k \mathop = 0}^n \binom m k       | c = Factors of Binomial Coefficient }} {{eqn | r = \dfrac m 2 \sum_{k \mathop = 0}^n \left({2 \binom {m - 1} {k - 1} - \binom m k}\right)       | c =  }} {{eqn | r = \dfrac m 2 \sum_{k \mathop = 0}^n \left({2 \binom {m - 1} {k - 1} - \left({\binom {m - 1} {k - 1} + \binom {m - 1} k}\right)}\right)       | c = Pascal's Rule }} {{eqn | r = \dfrac m 2 \sum_{k \mathop = 0}^n \left({\binom {m - 1} {k - 1} - \binom {m - 1} k}\right)       | c = simplifying }} {{eqn | r = \dfrac m 2 \left({\binom {m - 1} {-1} - \binom {m - 1} n}\right)       | c = as the series telescopes }} {{eqn | r = -\dfrac m 2 \binom {m - 1} n       | c = {{Defof|Binomial Coefficient}} for negative lower index }} {{end-eqn}} {{qed}}
We have by definition of Gaussian binomial coefficient: :$\displaystyle \dbinom r m_q = \prod_{k \mathop = 0}^{m - 1} \dfrac {1 - q^{r - k} } {1 - q^{k + 1} }$ Consider a typical factor of this product: {{begin-eqn}} {{eqn | l = \dfrac {1 - q^{r - k} } {1 - q^{k + 1} }       | r = \dfrac {\paren {1 - q^{r - k} } / \paren {1 - q} } {\paren {1 - q^{k + 1} } / \paren {1 - q} }       | c = multiplying top and bottom by $1 - q$ }} {{eqn | r = \dfrac {\sum_{j \mathop = 0}^{r - k - 1} q^j} {\sum_{j \mathop = 0}^k q^j}       | c = Sum of Geometric Sequence }} {{eqn | ll= \leadsto       | l = \lim_{q \mathop \to 1^-} \dfrac {1 - q^{r - k} } {1 - q^{k + 1} }       | r = \lim_{q \mathop \to 1^-} \dfrac {\sum_{j \mathop = 0}^{r - k - 1} q^j} {\sum_{j \mathop = 0}^k q^j}       | c =  }} {{eqn | r = \dfrac {\sum_{j \mathop = 0}^{r - k - 1} 1} {\sum_{j \mathop = 0}^k 1}       | c =  }} {{eqn | r = \dfrac {r - k} {k + 1}       | c =  }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \lim_{q \mathop \to 1^-} \dbinom r k_q       | r = \prod_{k \mathop = 0}^{m - 1} \dfrac {r - k} {k + 1}       | c =  }} {{eqn | r = \prod_{k \mathop = 1}^m \dfrac {r - k - 1} k       | c = Translation of Index Variable of Product }} {{eqn | r = \dbinom r m       | c = {{Defof|Binomial Coefficient}} }} {{end-eqn}} {{qed}}
Let $\mathcal A$ be a structure for predicate logic. Let $\sigma$ be an assignment for $\mathbf A \iff \mathbf B$ in $\mathcal A$. Then the value of $\mathbf A \iff \mathbf B$ under $\sigma$ is given by: :$f^\leftrightarrow \left({ \mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A}\right) } \left[{\sigma}\right], \mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf B}\right) } \left[{\sigma}\right] }\right)$ and from the definition of $f^\leftrightarrow$ we see that $\mathcal A, \sigma \models_{\mathrm{PL_A}} \mathbf A \iff \mathbf B$ iff: :$\mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A}\right) } \left[{\sigma}\right] = \mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf B}\right) } \left[{\sigma}\right]$ Because the possible values are just $T$ and $F$, this is equivalent to: :$\mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A}\right) } \left[{\sigma}\right] = T$ iff $\mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf B}\right) } \left[{\sigma}\right] = T$ which by definition of $\mathrm{PL_A}$-model amounts to: :$\mathcal A, \sigma \models_{\mathrm{PL_A}} \mathbf A$ iff $\mathcal A, \sigma \models_{\mathrm{PL_A}} \mathbf B$ Because $\mathcal A$ and $\sigma$ were arbitrary, the above equivalence holds for all such $\mathcal A$ and $\sigma$. The result follows by definition of tautology. {{qed}} Category:Predicate Logic c1t9ipnxtcw46n3dqnc3gnf56vucb8l
Let $\mathcal A$ be a structure for predicate logic. Let $\mathbf B$ be any universal closure of $\mathbf A$. Then $\mathbf B$ is a sentence of the form: :$\forall x_1: \cdots \forall x_n: \mathbf A$ By definition of the models relation: :$\mathcal A \models_{\mathrm{PL}} \mathbf B$ iff $\operatorname{val}_{\mathcal A} \left({\mathbf B}\right) = T$  Hence, recursively applying the [[Definition:Value of Formula under Assignment|definition of $\mathop{ \operatorname{val}_{\mathcal A} \left({\cdot}\right) } \left[{\sigma}\right]$]], we see: :$\mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf B}\right) } \left[{\varnothing}\right] = T$ iff $\forall a_1, \ldots, a_n \in A: \mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A}\right) } \left[{\frac{x_1} {a_1} + \ldots + \frac{x_n} {a_n} }\right] = T$ where $\frac{x_1} {a_1} + \ldots + \frac{x_n} {a_n}$ denotes the iterated extension of an assignment. By Value of Formula under Assignment Determined by Free Variables $\mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A}\right) } \left[{\frac{x_1} {a_1} + \ldots + \frac{x_n} {a_n} }\right]$ only depends on the $a_i$ for the free variables $x_i$ in $\mathbf A$. Because we check all possible $a_i \in A$ and all free variables $x_i$ in $\mathbf A$ are quantified over in $\mathbf B$, it follows that: :$\forall a_1, \ldots, a_n \in A: \mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A}\right) } \left[{\frac{x_1} {a_1} + \ldots + \frac{x_n} {a_n} }\right] = T$ iff: :$\forall a_1, \ldots, a_k \in A: \mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A}\right) } \left[{\frac{x_1} {a_1} + \ldots + \frac{x_k} {a_k} }\right] = T$ where $x_1, \ldots, x_k$ are the free variables of $\mathbf A$. But this last condition does not depend on $\mathbf B$ beyond that it be a universal closure. Hence, for any two universal closures $\mathbf B, \mathbf B'$ of $\mathbf A$: :$\mathcal A \models_{\mathrm{PL}} \mathbf B$ iff $\mathcal A \models_{\mathrm{PL}} \mathbf B$ The result follows by definition of semantic equivalence. {{qed}}
=== Sufficient Condition === Let $L$ be meet-continuous. Let $x$ be an element of $S$, $D$ be a directed subset of $S$ such that :$x \preceq \sup D$ Thus {{begin-eqn}} {{eqn | l = x       | r = x \wedge \sup D       | c = Preceding iff Meet equals Less Operand }} {{eqn | r = \sup \set {x \wedge d: d \in D}       | c = {{Defof|Meet-Continuous Lattice}} }} {{end-eqn}} {{qed|lemma}} === Necessary Condition === Let: :$\forall x \in S$, directed subset $D$ of $S: x \preceq \sup D \implies x = \sup \set {x \wedge d: d \in D}$ By definition of reflexivity: :$\forall x \in S$, directed subset $D$ of $S: x \preceq \sup D \implies x \preceq \sup \set {x \wedge d: d \in D}$ By Meet is Directed Suprema Preserving: :$\wedge$ preserves directed suprema as a mapping from Cartesian product $\struct {S \times S, \precsim}$ of $L$ and $L$ into $L$. Thus by Meet-Continuous iff Meet Preserves Directed Suprema :$L$ is meet-continuous. {{qed}}