hcju/proofwikiqa huggingface dataset:
{{begin-eqn}} {{eqn | l = \sum_{k \mathop = 1}^n \left({\dfrac {\displaystyle \prod_{\substack {1 \mathop \le r \mathop \le n \\ r \mathop \ne m} } \left({x + k - r}\right)} {\displaystyle \prod_{\substack {1 \mathop \le r \mathop \le n \\ r \mathop \ne k} } \left({k - r}\right)} }\right)       | r = \sum_{k \mathop = 1}^n \left({\dfrac {k^{n - 1} + P \left({k}\right)} {\displaystyle \prod_{\substack {1 \mathop \le r \mathop \le n \\ r \mathop \ne k} } \left({k - r}\right)} }\right)       | c = where $P \left({k}\right)$ is a polynomial of degree $n - 2$ }} {{eqn | r = \sum_{k \mathop = 1}^n \left({\dfrac {k^{n - 1} } {\displaystyle \prod_{\substack {1 \mathop \le r \mathop \le n \\ r \mathop \ne k} } \left({k - r}\right)} }\right) + \sum_{k \mathop = 1}^n \left({\dfrac {P \left({k}\right)} {\displaystyle \prod_{\substack {1 \mathop \le r \mathop \le n \\ r \mathop \ne k} } \left({k - r}\right)} }\right)       | c =  }} {{eqn | r = 1 + 0       | c = Summation of Powers over Product of Differences }} {{end-eqn}} {{qed}}
This is an example of Summation by k of Product by r of x plus k minus r over Product by r less k of k minus r: :$\displaystyle \sum_{k \mathop = 1}^n \left({\dfrac {\displaystyle \prod_{\substack {1 \mathop \le r \mathop \le n \\ r \mathop \ne m} } \left({x + k - r}\right)} {\displaystyle \prod_{\substack {1 \mathop \le r \mathop \le n \\ r \mathop \ne k} } \left({k - r}\right)} }\right) = 1$ where $n = 4$ and $m = 2$. {{qed}}
All $n^2$ elements of $C_n^{-1}$ have a term $\dfrac {-y} {x \left({x + n y}\right)}$. Further to this, the $n$ elements on the main diagonal contribute an extra $\dfrac {x + n y} {x \left({x + n y}\right)}$ to the total. Hence: {{begin-eqn}} {{eqn | l = \sum_{1 \mathop \le i, \ j \mathop \le n} b_{i j}       | r = n^2 \dfrac {-y} {x \left({x + n y}\right)} + n \dfrac {x + n y} {x \left({x + n y}\right)}       | c =  }} {{eqn | r = \dfrac {- n^2 y + n \left({x + n y}\right)} {x \left({x + n y}\right)}       | c =  }} {{eqn | r = \dfrac {- n^2 y + n x + n^2 y} {x \left({x + n y}\right)}       | c =  }} {{eqn | r = \dfrac {n x} {x \left({x + n y}\right)}       | c =  }} {{eqn | r = \dfrac n {x + n y}       | c =  }} {{end-eqn}} {{qed}}
Sum of Elements of Invertible Matrix implies the sum of elements in $V_n^{-1}$ is: :$ 1 - \det \paren { V_n^{-1} } \det \paren { V_n - J_n } $. The plan is to expand $\det \paren { V_n - J_n } $ and simplify. The method is efficiently communicated in the $3\times 3$ case: {{begin-eqn}} {{eqn   | l = \det \paren { V_3 - J_3 } =   | r = \det \paren { \begin{smallmatrix}                     x_1 - 1 & x_2 - 1 & x_3 - 1 \\                     x_1^2 - 1 & x_2^2 - 1 & x_3^2 - 1 \\                     x_1^3 - 1 & x_2^3 - 1 & x_3^3 - 1 \\                     \end{smallmatrix} }   | c = $J_3$ is the $3\times 3$ ones matrix. }} {{eqn   | l =   | r =  \paren {x_1 - 1} \paren {x_2 - 1} \paren {x_3 - 1}          \det \paren { \begin{smallmatrix}                      1 & 1 & 1 \\                      x_1+1 & x_2+1 & x_3+1 \\                      x_1^2+x+1+1 & x_2^2+x_2+1 & x_3^2+x_3+1 \\                      \end{smallmatrix} }    | c = Use $a^n - b^n$ factorization. Effect of Elementary Row Operations on Determinant collects common column factors outside the determinant. }} {{eqn   | l =   | r =\paren {x_1 - 1} \paren {x_2 - 1} \paren {x_3 - 1}        \det \paren { \begin{smallmatrix}        1 & 1 & 1 \\        x_1 & x_2 & x_3 \\        x_1^2 & x_2^2 & x_3^2 \\        \end{smallmatrix} }    | c = Effect of Elementary Row Operations on Determinant simplifies the determinant. }} {{end-eqn}} The proof is completed by '''induction''' using the lemmas below. === Lemma 1 === Let: :$\displaystyle A_n = \det \paren { \begin{smallmatrix} 1 & \cdots & 1 \\ x_1 & \cdots & x_n \\ \vdots & \cdots & \vdots \\ x_1^{n-1} & \cdots & x_n^{n-1} \\ \end{smallmatrix} }$  Then: :$\displaystyle \det \paren { V_n } =  \paren { \prod_{k \mathop = 1}^n x_k } \det \paren { A_n }$  Effect of Elementary Row Operations on Determinant === Lemma 2 === :$\displaystyle \det(V_n - J_n) = \prod_{k \mathop = 1}^n \paren { x_k - 1 }  \det \paren { A_n }$ === Lemma 3 === :$\displaystyle \det \paren { V_n^{-1} } \det(V_n - J_n) =  \dfrac { \prod_{k \mathop = 1}^n \paren { x_k - 1 }   } { \prod_{k \mathop = 1}^n { x_k }  }$  by Lemmas 1 and 2. {{qed}}
By definition of Hilbert matrix, the element $a_{i j}$ is: :$a_{i j} = \dfrac 1 {i + j - 1}$ For all $i, j \in \Z$ such that $1 \le i \le n$ and $1 \le j \le n$, let: :$x_i = i$ :$y_j = j - 1$ Then: :$a_{i j} = \dfrac 1 {x_i + y_j}$ The result follows by definition of a Cauchy matrix. {{qed}}
From Inverse of Hilbert Matrix, $H_n^{-1} = \left[{b}\right]_n$ can be specified as: :$\begin{bmatrix} b_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac {\left({-1}\right)^{i + j} \left({i + n - 1}\right)! \left({j + n - 1}\right)!} {\left({\left({i - 1}\right)!}\right)^2 \left({\left({j - 1}\right)!}\right)^2 \left({n - i}\right)! \left({n - j}\right)! \left({i + j - 1}\right)} \end{bmatrix}$ Thus: {{begin-eqn}} {{eqn | l = b_{i j}       | r = \frac {\left({-1}\right)^{i + j} \left({i + n - 1}\right)! \left({j + n - 1}\right)!} {\left({\left({i - 1}\right)!}\right)^2 \left({\left({j - 1}\right)!}\right)^2 \left({n - i}\right)! \left({n - j}\right)! \left({i + j - 1}\right)}       | c =  }} {{eqn | r = \left({\frac {\left({-1}\right)^{i + j} } {i + j - 1} }\right) \left({\frac {\left({i + n - 1}\right)!} {\left({i - 1}\right)! \, n!} }\right) \left({\frac {\left({j + n - 1}\right)!} {\left({j - 1}\right)! \, n!} }\right) \left({\frac {n! \, i} {i! \, \left({n - i}\right)!} }\right) \left({\frac {n! \, j} {j! \, \left({n - j}\right)! } }\right)       | c =  }} {{eqn | r = \frac {\left({-1}\right)^{i + j} i j} {i + j - 1} \binom {i + n - 1} n \binom {j + n - 1} n \binom n i \binom n j       | c = Definition of Binomial Coefficient }} {{eqn | r = \left({-1}\right)^{i + j} j \binom {i + n - 1} {i - i} \binom {j + n - 1} {n - 1} \binom {i + j - 2} {n - i} \binom n j       | c =  }} {{end-eqn}} All of the factors of the above expression are integers, from Binomial Coefficient is Integer. {{qed}}
From Hilbert Matrix is Cauchy Matrix, $H_n$ is a special case of a Cauchy matrix: :$\begin{bmatrix} c_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac 1 {x_i + y_j} \end{bmatrix}$ where: :$x_i = i$ :$y_j = j - 1$ Then: {{begin-eqn}} {{eqn | l = \sum_{1 \mathop \le i, \ j \mathop \le n} b_{i j}       | r = \sum_{k \mathop = 1}^n x_k + \sum_{k \mathop = 1}^n y_k       | c = Sum of Elements in Inverse of Cauchy Matrix }} {{eqn | r = \sum_{k \mathop = 1}^n k + \sum_{k \mathop = 1}^n \paren {k - 1}       | c =  }} {{eqn | r = 2 \sum_{k \mathop = 1}^n k - n       | c =  }} {{eqn | r = n \paren {n + 1} - n       | c = Closed Form for Triangular Numbers }} {{eqn | r = n^2       | c =  }} {{end-eqn}} {{qed}}
=== Sufficient Condition === {{:Sequential Continuity is Equivalent to Continuity in the Reals/Sufficient Condition}} === Necessary Condition === {{:Sequential Continuity is Equivalent to Continuity in the Reals/Necessary Condition}}
{{begin-eqn}} {{eqn | l = x^{\overline n}       | r = \prod_{j \mathop = 0}^{n - 1} \paren {x + j}       | c = {{Defof|Rising Factorial}} }} {{eqn | r = x \paren {x + 1} \paren {x + 2} \dotsm \paren {x + n - 1}       | c =  }} {{eqn | r = \dfrac {\paren {x + n - 1}!} {\paren {x - 1}!}       | c = {{Defof|Factorial}} }} {{eqn | r = \dfrac {\map \Gamma {x + n} } {\map \Gamma x}       | c = Gamma Function Extends Factorial }} {{end-eqn}} {{qed}}
Let $\oplus$ be a bilinear map with the property that for all $a \in A_R$, $a \oplus a = 0$. Then for all $u, v \in A_R$: {{begin-eqn}} {{eqn | l = 0       | r = \left({u + v}\right) \oplus \left({u + v}\right) }} {{eqn | r = u \oplus \left({u + v}\right) + v \oplus \left({u + v}\right) }} {{eqn | r = u \oplus v + v \oplus u = 0  }} {{end-eqn}} {{qed|lemma}} Let $\oplus$ be a bilinear map with the property that for all $a, b \in A_R$, $a \oplus b + b \oplus a = 0$. Let $u \in A_R$. Then: {{begin-eqn}} {{eqn | l = 0       | r = u \oplus u + u \oplus u }}  {{eqn | r = 2 \left({u \oplus u}\right) }} {{end-eqn}} Since the characteristic of $R$ is not $2$, thus: :$u \oplus u = 0$ {{qed}} Category:Algebra kvu4p5d5p00w9uwep4sjnbm6jkg7k3j
Let $D$ be a directed subset of $\tau$ such that :$y \preceq \sup D$ By proof of Topology forms Complete Lattice: :$y \subseteq \bigcup D$ By Subset Relation is Transitive: :$H \subseteq \bigcup D$ By definition: :$D$ is open cover of $H$ By definition of compact: :$H$ has finite subcover $\mathcal G$ of $D$ By Directed iff Finite Subsets have Upper Bounds: :$\exists d \in D: d$ is upper bound for $\mathcal G$ By definitions of upper bound and of $\preceq$: :$\forall z \in \mathcal G: z \subseteq d$ By Union is Smallest Superset/Set of Sets: :$\bigcup \mathcal G \subseteq d$ By definition of cover: :$H \subseteq \bigcup \mathcal G$ By Subset Relation is Transitive: :$x \subseteq d$ Thus by definition of $\preceq$: :$x \preceq d$ Thus by definition of way below relation: :$x \ll y$ {{qed}}
=== Sufficient Condition === Let $T$ be locally compact. Let $U \in \tau, x \in U$. Since $T$ is locally compact there exists a neighborhood basis $\mathcal B$ of $x$ consisting of compact sets.  From Set is Open iff Neighborhood of all its Points, $U$ is a neighborhood of $x$. By the definition of a neighborhood basis: :$\exists K \in \mathcal B : K \subseteq U$. By the definition of a neighborhood of $x$: :$\exists V \in \tau : x \in V \subseteq K$ By the definition of the interior of a subset then: :$x \in V \subseteq K^\circ$ Thus :$\exists K \subseteq S: x \in K^\circ \land K \subseteq U \land K$ is compact {{qed|lemma}} === Necessary Condition === Let :$\forall U \in \tau, x \in U: \exists K \subseteq S: x \in K^\circ \land K \subseteq U \land K$ is compact Let $\mathcal B$ be the set of all compact neighborhoods of $x$. It is shown that $\mathcal B$ is a neighborhood basis of $x$. Let $N$ be a neighborhood of $x$. By the definition of a neighborhood of $x$: :$\exists U \in \tau : x \in U \subseteq N$ By assumption: :$\exists K \subseteq S: x \in K^\circ \land K \subseteq U \land K$ is compact By the definition of the interior of a subset $K^\circ$ is open. By the definition of a neighborhood then $K$ is a neighborhood of $x$. Hence $K \in \mathcal B$ and $K \subseteq U \subseteq N$. As $N$ was arbitrary, then $B$ is a neighborhood basis of $x$ consisting of compact sets.
Let $\tuple {k_1, k_2, \ldots, k_m}$ be an ordered $m$-tuple of integers. Let $\map \epsilon {k_1, k_2, \ldots, k_m}$ denote the sign of $\tuple {k_1, k_2, \ldots, k_m}$. Let $\tuple {l_1, l_2, \ldots, l_m}$ be the same as $\tuple {k_1, k_2, \ldots, k_m}$ except for $k_i$ and $k_j$ having been transposed. Then from Transposition is of Odd Parity: :$\map \epsilon {l_1, l_2, \ldots, l_m} = -\map \epsilon {k_1, k_2, \ldots, k_m}$ Let $\tuple {j_1, j_2, \ldots, j_m}$ be the same as $\tuple {k_1, k_2, \ldots, k_m}$ by arranged into non-decreasing order. That is: :$j_1 \le j_2 \le \cdots \le j_m$ Then it follows that: :$\map \det {\mathbf B_{k_1 \cdots k_m} } = \map \epsilon {k_1, k_2, \ldots, k_m} \map \det {\mathbf B_{j_1 \cdots j_m} }$ Hence: {{begin-eqn}} {{eqn | l = \map \det {\mathbf A \mathbf B}       | r = \sum_{1 \mathop \le l_1, \mathop \ldots \mathop , l_m \mathop \le m} \map \epsilon {l_1, \ldots, l_m} \paren {\sum_{k \mathop = 1}^n a_{1 k} b_{k l_1} } \cdots \paren {\sum_{k \mathop = 1}^n a_{m k} b_{k l_m} }       | c = {{Defof|Determinant of Matrix}} }} {{eqn | r = \sum_{1 \mathop \le k_1, \mathop \ldots \mathop , k_m \mathop \le n} a_{1 k_1} \cdots a_{m k_m} \sum_{1 \mathop \le l_1, \mathop \ldots \mathop , l_m \mathop \le m} \map \epsilon {l_1, \ldots, l_m} b_{k_1 l_1} \cdots b_{k_m l_m}        | c =  }} {{eqn | r = \sum_{1 \mathop \le k_1, \mathop \ldots \mathop , k_m \mathop \le n} a_{1 k_1} \cdots a_{m k_m} \map \det {\mathbf B_{k_1 \cdots k_m} }       | c =  }} {{eqn | r = \sum_{1 \mathop \le k_1, \mathop \ldots \mathop , k_m \mathop \le n} a_{1 k_1} \map \epsilon {k_1, \ldots, k_m} \cdots a_{m k_m} \map \det {\mathbf B_{j_1 \cdots j_m} }       | c =  }} {{eqn | r = \sum_{1 \mathop \le j_1 \mathop \le j_2 \mathop \le \cdots \mathop \le j_m \le n} \map \det {\mathbf A_{j_1 \cdots j_m} } \map \det {\mathbf B_{j_1 \cdots j_m} }       | c =  }} {{end-eqn}} If two $j$s are equal: :$\map \det {\mathbf A_{j_1 \cdots j_m} } = 0$ {{qed}}
Let $x, y, z, u \in S$ such that :$\left({x, z}\right) \in \mathcal R \land \left({y, u}\right) \in \mathcal R$ By definition of reflexivity: :$x \preceq x$ and $y \preceq y$ By Join Succeeds Operands: :$z \preceq z \vee u$ and $u \preceq z \vee u$ By condition $(ii)$ of auxiliary relation: :$\left({x, z \vee u}\right) \in \mathcal R$ and $\left({y, z \vee u}\right) \in \mathcal R$ Thus by condition $(iii)$ of auxiliary relation: :$\left({x \vee y, z \vee u}\right) \in \mathcal R$ {{qed}}
Let $x, y, z \in S$ such that :$\left({x, y}\right) \in \mathcal R \land \left({y, z}\right) \in \mathcal R$ By definition of reflexivity: :$z \preceq z$  By condition $(i)$ of auxiliary relation: :$x \preceq y$ Thus by condition $(ii)$ of auxiliary relation: :$\left({x, z}\right) \in \mathcal R$ Thus by definition :$\mathcal R$ is a transitive relation. {{qed}}
The Cauchy-Binet Formula gives: :$\displaystyle \det \left({\mathbf A \mathbf B}\right) = \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop < \cdots \mathop < j_m \le n} \det \left({\mathbf A_{j_1 j_2 \ldots j_m} }\right) \det \left({\mathbf B_{j_1 j_2 \ldots j_m}}\right)$ where: :$\mathbf A$ is an $m \times n$ matrix :$\mathbf B$ is an $n \times m$ matrix. :For $1 \le j_1, j_2, \ldots, j_m \le n$: ::$\mathbf A_{j_1 j_2 \ldots j_m}$ denotes the $m \times m$ matrix consisting of columns $j_1, j_2, \ldots, j_m$ of $\mathbf A$. ::$\mathbf B_{j_1 j_2 \ldots j_m}$ denotes the $m \times m$ matrix consisting of rows $j_1, j_2, \ldots, j_m$ of $\mathbf B$. When $m = 1$, the relation: :$1 \le j_1 < j_2 < \cdots < j_m \le n$ degenerates to: :$1 \le j \le n$ By definition of order $1$ determinant: :$\det \left({\left[{a_j}\right]}\right) = a_j$ :$\det \left({\left[{b_j}\right]}\right) = b_j$ Hence the result. {{qed}}
:$\forall x, y \in S: x \preceq y \implies x \preceq y$ Then condition $(i)$ of auxiliary relation is satisfied. By definition of transitivity: :$\forall x, y, z, u \in S: x \preceq y \preceq z \preceq u \implies x \preceq u$ Then the condition $(ii)$ of auxiliary relation is satisfied. By  definition of supremum: :$\forall x, y, z \in S: x \preceq z \land y \preceq z \implies x \vee y \preceq z$ Then the condition $(iii)$ of auxiliary relation is satisfied. By definition of smallest element: :$\forall x \in S: \bot \preceq x$ Then the condition $(iv)$ of auxiliary relation is satisfied. Thus the result by definition auxiliary relation. {{qed}}
The Cauchy-Binet Formula gives: :$\displaystyle \det \left({\mathbf A \mathbf B}\right) = \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop < \cdots \mathop < j_m \le n} \det \left({\mathbf A_{j_1 j_2 \ldots j_m}}\right) \det \left({\mathbf B_{j_1 j_2 \ldots j_m}}\right)$ where: :$\mathbf A$ is an $m \times n$ matrix :$\mathbf B$ is an $n \times m$ matrix. :For $1 \le j_1, j_2, \ldots, j_m \le n$: ::$\mathbf A_{j_1 j_2 \ldots j_m}$ denotes the $m \times m$ matrix consisting of columns $j_1, j_2, \ldots, j_m$ of $\mathbf A$. ::$\mathbf B_{j_1 j_2 \ldots j_m}$ denotes the $m \times m$ matrix consisting of rows $j_1, j_2, \ldots, j_m$ of $\mathbf B$. From the definition of transpose $\mathbf A^\intercal$ is an $n \times m$ matrix. Hence the Cauchy-Binet Formula can be applied directly: :$\displaystyle \det \left({\mathbf A \mathbf A^\intercal}\right) = \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop < \cdots \mathop < j_m \le n} \det \left({\mathbf A_{j_1 j_2 \ldots j_m} }\right) \det \left({\mathbf A^\intercal_{j_1 j_2 \ldots j_m} }\right)$ Note that by construction: :$\mathbf A_{j_1 j_2 \ldots j_m}$ is a square matrix Also, by definition of transpose: :$\mathbf A^\intercal_{j_1 j_2 \ldots j_m} = \left({\mathbf A_{j_1 j_2 \ldots j_m} }\right)^\intercal$ The result follows from Determinant of Transpose: :$\det \left({\mathbf A}\right) = \det \left({\mathbf A^\intercal}\right)$ {{qed}}
By Preceding is Auxiliary Relation: :$\preceq \mathop \in {\it Aux}\left({L}\right)$ By definition :$\preceq$ is lower bound for $\varnothing$ in $P$ We will prove that :$\forall R \in {\it Aux}\left({L}\right): R$ is lower bound for $\varnothing \implies R \mathop \precsim \preceq$ Let $R \in {\it Aux}\left({L}\right)$ By condition $(i)$ of definition of auxiliary relation: :$R \mathop \subseteq \preceq$ Thus by definition of $\precsim$: :$R \mathop \precsim \preceq$ {{qed|lemma}} By definition of infimum: :$\preceq \mathop = \inf_P \varnothing$ Thus by Infimum of Empty Set is Greatest Element: :$\preceq \mathop = \top_P$ {{qed}}
By definition of smallest element: :$\forall x \in S: \bot \preceq x$ Thus by definition of $B$: :$\forall x, y \in S: \left({x, y}\right) \in B \implies x \preceq y$ We will prove that :$\forall x, y, z, u \in S: x \preceq y \land \left({y, z}\right) \in B \land z \preceq u \implies \left({x, u}\right) \in B$ Let $x, y, z, u \in S$ such that :$x \preceq y \land \left({y, z}\right) \in B \land z \preceq u$ By definition of $B$: :$y = \bot$ By definition of smallest element: :$\bot \preceq x$ By definition of antisymmetry: :$x = \bot$ Thus by definition of $B$: :$\left({x, u}\right) \in B$ {{qed|lemma}} We will prove that :$\forall x, y, z \in S: \left({x, z}\right) \in B \land \left({y, z}\right) \in B \implies \left({x \vee y, z}\right) \in B$ Let $x, y, z \in S$ such that : $\left({x, z}\right) \in B \land \left({y, z}\right) \in B$ By definition of $B$: :$x = y = \bot$ By Join is Idempotent: :$x \vee y = \bot$ Thus by definition of $B$: :$\left({x \vee y, z}\right) \in B$ {{qed|lemma}} Thus by definition of $B$: :$\forall x \in S: \left({\bot, x}\right) \in B$ Thus by definition: :$B$ is auxiliary relation. {{qed}}
By Bottom Relation is Auxiliary Relation: :$B \mathop \in {\it Aux}\left({L}\right)$ By definition :$B$ is upper bound for $\varnothing$ in $P$ We will prove that :$\forall R \in {\it Aux}\left({L}\right): R$ is upper bound for $\varnothing \implies B \mathop \precsim R$ Let $R \in {\it Aux}\left({L}\right)$ By condition $(iv)$ of definition of auxiliary relation: :$B \mathop \subseteq R$ Thus by definition of $\precsim$: :$B \mathop \precsim R$ {{qed|lemma}} By definition of supremum: :$B \mathop = \sup_P \varnothing$ Thus by Supremum of Empty Set is Smallest Element: :$B \mathop = \bot_P$ {{qed}}
The inverse of $p \implies q$ is: :$\neg p \implies \neg q$ The contrapositive of $p \implies q$ is: :$\neg q \implies \neg p$ The converse of $\neg q \implies \neg p$ is: :$\neg p \implies \neg q$ The two are seen to be equal. {{qed}} Category:Implication 3evau7fkxx96gwhtt8hqi0rpefw4w1z
The inverse of $p \implies q$ is: :$\neg p \implies \neg q$ The converse of $p \implies q$ is: :$q \implies p$ The contrapositive of $q \implies p$ is: :$\neg p \implies \neg q$ The two are seen to be equal. {{qed}} Category:Implication 9b8kfzy0f04jach47afrtwkv29o2wg3
The converse of $p \implies q$ is: :$q \implies p$ The inverse of $p \implies q$ is: :$\neg p \implies \neg q$ The contrapositive of $\neg p \implies \neg q$ is: :$\neg\neg q \implies \neg\neg p$ By Double Negation, the two are seen to be equal. {{qed}} Category:Implication plj00q6y1603076aolfow75v0tbe9y3
The converse of $p \implies q$ is: :$q \implies p$ The contrapositive of $p \implies q$ is: :$\neg q \implies \neg p$ The inverse of $\neg q \implies \neg p$ is: :$\neg\neg q \implies \neg\neg p$ By Double Negation, the two are seen to be equal. {{qed}} Category:Implication 3mmpmjh5n6hqtx9aqgmij19sqj146rm
By definition of non-empty set: :$\exists \mathcal R: \mathcal R \in \mathcal F$ We will prove that :$\forall x, y \in S: \left({x, y}\right) \in \bigcap \mathcal F \implies x \preceq y$ Let $x, y \in S$ such that :$\left({x, y}\right) \in \bigcap \mathcal F$ By definition of intersection: :$\left({x, y}\right) \in \mathcal R$ Thus by definition of auxiliary relation: :$x \preceq y$ {{qed|lemma}} We will prove that :$\forall x, y, z, u \in S: x \preceq y \land \left({y, z}\right) \in \bigcap \mathcal F \land z \preceq u \implies \left({x, u}\right) \in \bigcap \mathcal F$ Let $x, y, z, u \in S$ such that :$x \preceq y \land \left({y, z}\right) \in \bigcap \mathcal F \land z \preceq u$ By definition of intersection: :$\forall r \in \mathcal F: \left({y, z}\right) \in r$ By definition of auxiliary relation: :$\forall r \in \mathcal F: \left({x, u}\right) \in r$ Thus by definition of intersection: :$\left({x, u}\right) \in \bigcap \mathcal F$ {{qed|lemma}} We will prove that :$\forall x, y, z \in S: \left({x, z}\right) \in \bigcap \mathcal F \land \left({y, z}\right) \in \bigcap \mathcal F \implies \left({x \vee y, z}\right) \in \bigcap \mathcal F$ Let $x, y, z \in S$ such that :$\left({x, z}\right) \in \bigcap \mathcal F \land \left({y, z}\right) \in \bigcap \mathcal F$ By definition of intersection: :$\forall r \in \mathcal F: \left({x, z}\right) \in r \land \left({y, z}\right) \in r$ By definition of auxiliary relation: :$\forall r \in \mathcal F: \left({x \vee y, z}\right) \in r$ Thus by definition of intersection: :$\left({x \vee y, z}\right) \in \bigcap \mathcal F$ {{qed|lemma}} By definition of auxiliary relation: :$\forall x \in S: \forall r \in \mathcal F: \left({\bot, x}\right) \in r$ Thus by definition of intersection: :$\forall x \in S: \left({\bot, x}\right) \in \bigcap \mathcal F$ Thus by definition: :$\bigcap \mathcal F$ is auxiliary relation. {{qed}}
Let $X \subseteq \operatorname {Aux} \left({L}\right)$ In the case when $X \ne \O$: By Intersection of Auxiliary Relations is Auxiliary Relation: :$\bigcap X \in \operatorname {Aux} \left({L}\right)$ By Intersection is Largest Subset, $\bigcap X$ is the infimum of $X$. In case when $X = \O$: By proof of Preceding is Top in Ordered Set of Auxiliary Relations: :$\O$ admits an infimum in $P$ Then: :$X$ (empty or non-empty) admits an infimum in $P$ By duality of Lattice is Complete iff it Admits All Suprema: :$P$ is a complete lattice. {{qed}}
The Cauchy-Binet Formula gives: :$(1): \quad \displaystyle \det \left({\mathbf A \mathbf B}\right) = \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop < \cdots \mathop < j_m \le n} \det \left({\mathbf A_{j_1 j_2 \ldots j_m}}\right) \det \left({\mathbf B_{j_1 j_2 \ldots j_m}}\right)$ where: :$\mathbf A$ is an $m \times n$ matrix :$\mathbf B$ is an $n \times m$ matrix. :For $1 \le j_1, j_2, \ldots, j_m \le n$: ::$\mathbf A_{j_1 j_2 \ldots j_m}$ denotes the $m \times m$ matrix consisting of columns $j_1, j_2, \ldots, j_m$ of $\mathbf A$. ::$\mathbf B_{j_1 j_2 \ldots j_m}$ denotes the $m \times m$ matrix consisting of rows $j_1, j_2, \ldots, j_m$ of $\mathbf B$. But here $m > n$. Therefore the set $\left\{ {j_1, j_2, \ldots, j_m}\right\}$ such that: :$1 \mathop \le j_1 \mathop < j_2 \mathop < \cdots \mathop < j_m \le n$ is the empty set. Thus the {{RHS}} of $(1)$ is a vacuous summation. Hence the result. {{qed}}