Difference between revisions of "Exercícios de Dedução Natural"

Revision as of 13:26, 14 September 2020

$\varphi \land \psi \vdash \psi \land \varphi$
$(\varphi \land \psi )\land \delta \vdash \varphi \land (\psi \land \delta )$
$\varphi \vdash \varphi \land \varphi$
$\alpha \land \beta ,\gamma \land \delta \vdash \gamma \land \beta$
$\alpha \to (\beta \to \gamma )\vdash \beta \to (\alpha \to \gamma )$
$\vdash \alpha \to (\beta \to \alpha )$
$\vdash \alpha \to \alpha$
$\alpha \to \beta \vdash \alpha \to (\alpha \to \beta )$
$\alpha \to (\alpha \to \beta )\vdash \alpha \to \beta$
$\gamma \to \alpha ,\gamma \to \beta \vdash \gamma \to (\alpha \land \beta )$
$\beta \lor (\alpha \land \beta )\vdash \beta$
$(\alpha \lor \beta )\to \gamma \vdash (\alpha \to \gamma )\land (\beta \to \gamma )$
$\alpha \lor \beta \not \vdash \alpha \land \beta$
$\alpha \vdash \neg \neg \alpha$
$\beta \to \alpha ,\beta \to \neg \alpha \vdash \neg \beta$
$\alpha ,\neg \alpha \vdash \neg \beta$
$\alpha \lor \beta ,\neg \alpha \lor \gamma \vdash \beta \lor \gamma$
$\alpha \to \beta \vdash \neg \beta \to \neg \alpha$
$\neg (\alpha \lor \beta )\dashv \vdash \neg \alpha \land \neg \beta$

Terceiro Excluído / Tertium Non Datur: $\vdash \varphi \lor \neg \varphi$
Tarefa: Demonstrar a mesma fórmula, invertendo a ordem de aplicação das regras de introdução da disjunção.

@@ Line 39: / Line 39: @@
 #<math> \alpha\to\beta \vdash \neg\beta\to\neg\alpha</math><!--
 --><p>{{#ev:youtube|-Exorelokdo|||||start=741}}</p>
-#<math>\neg(\alpha \lor \beta) \dashv\vdash \neg\alpha\lor\neg\beta</math><!--
+#<math>\neg(\alpha \lor \beta) \dashv\vdash \neg\alpha\land\neg\beta</math><!--
---><p>{{#ev:youtube|-Exorelokdo|||||start=753}}</p>
+--><p>{{#ev:youtube|-Exorelokdo|||||start=817}}</p>
 ==Dedução Natural para a Lógica Proposicional Clássica==