User:Ben Spinozoan/Wronskian&Independence

Proof

 * In the language of first-order logic, the set of functions $$\{f_1,f_2,...,f_n\}\,$$ is linearly independent, over the interval $$\Omega\,$$ in $$\R$$, iff:
 * $$ \forall \boldsymbol{\alpha}.\,\Big[\boldsymbol{\alpha}\in\R^n\land\, \boldsymbol{\alpha}\ne0\,\to\,\exists x\in\Omega\,.^\neg P(\boldsymbol{\alpha},x)\Big] $$, where $$ P(\boldsymbol{\alpha},x) \equiv \Bigg[\sum_{i=1}^n\alpha_i\,f_i(x)=0\Bigg] $$.
 * Expressed in disjunctive normal form (DNF) the above definition reads:
 * $$ LI(\Omega)\,\equiv\,\forall \boldsymbol{\alpha}.\,\Big[\,\boldsymbol{\alpha}\notin\R^n\,\lor\,\boldsymbol{\alpha}=0\,\lor\,\exists x\in\Omega\,.^\neg P(\boldsymbol{\alpha},x)\,\Big] $$,
 * where $$LI(\Omega)$$ is shorthand for the statement occurring immediately before the iff (note that negation of $$LI(\Omega)$$ gives the correct statement for linear dependence).


 * The text of our theorem "If the Wronskian is non-zero at some point in an interval, then the functions are linearly independent on the interval", now translates as
 * $$\exists y\in\Omega.\;W(y)\ne0\,\to\,LI(\Omega)$$,


 * or, in DNF,
 * $$ (1)\quad \forall \boldsymbol{\alpha}.\,\bigg[\,\forall y.\,\Big[\,y\notin\Omega\,\lor\,W(y)=0\,\Big]\lor\,\boldsymbol{\alpha}\notin\R^n\,\lor\,\boldsymbol{\alpha}=0\,\lor\,\exists x\in\Omega\,. ^\neg P(\boldsymbol{\alpha},x)\,\bigg]$$,
 * where $$W(y)$$ is the value of the Wronskian at the point $$y$$.


 * The following statement summarizes the situation when Cramer's rule is applied to the linear system associated with the Wronskian:
 * $$ \forall \boldsymbol{\alpha}.\,\Bigg[\,\boldsymbol{\alpha}\in\R^n\to\bigg[\,\exists x \in \Omega.\, \Big[\,P(\boldsymbol{\alpha},x)\,\land\, W(x)\ne0\,\Big] \to\, \boldsymbol{\alpha}=0\; \bigg]\,\Bigg]$$,
 * or,
 * $$ (2)\quad \forall \boldsymbol{\alpha}.\,\bigg[\,\forall x. \, \Big[\, x\notin\Omega \,\lor\;W(x)=0 \;\lor\, ^\neg\!P(\boldsymbol{\alpha},x)\,\Big] \lor\,\boldsymbol{\alpha}\notin\R^n\,\lor\, \boldsymbol{\alpha}=0\;\bigg] $$.


 * In first-order logic, the statement $$ \forall x. \,\big[ A(x) \lor B(x)\big] $$, entails the statement $$ \forall x.A(x) \lor \exists y.B(y) $$. Consequently, statement (2) entails statement (1), and the theorem is proved.