3 votes 3 votes Use rules of inference to show that if $\forall x (P(x) \vee Q(x))$ and $\forall x ((\sim P(x) \wedge Q(x)) \rightarrow R(x))$ are true, then $\forall x (\sim R(x) \rightarrow P(x))$ is also true, where the domains of all quantifiers are the same. Mathematical Logic kenneth-rosen discrete-mathematics mathematical-logic propositional-logic + – Pooja Khatri asked Mar 19, 2019 Pooja Khatri 408 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 6 votes 6 votes $ \begin{flalign} \text{Proof} \\ && &1.\ \forall x (P(x) \lor Q(x)) \tag*{Premise} \\ && &2.\ P(a) \lor Q(a) \tag*{Universal instantiation from (1)} \\ && &3.\ P(a) \lor Q(a) \lor R(a) \tag*{Addition from (2)} \\ && &4.\ \forall x ((\neg P(x) \land Q(x)) \to R(x)) \tag*{Premise} \\ && &5.\ \forall x (P(x) \lor \neg Q(x) \lor R(x)) \tag*{From (4)} \\ && &6.\ P(a) \lor \neg Q(a) \lor R(a) \tag*{Universal instantiation from (5)} \\ && &7.\ P(a) \lor R(a) \tag*{Resolution from (3) & (6)} \\ && &8.\ \forall x (\neg R(x) \to P(x)) \tag*{Universal generalization from (7)} \\ \\ &&&& \blacksquare \end{flalign} $ thehitchh1ker answered May 7, 2023 edited May 7, 2023 by thehitchh1ker thehitchh1ker comment Share Follow See all 0 reply Please log in or register to add a comment.