Use the rules of inference to show that if $\forall x(P(x) \lor Q(x))$ and $\forall x(\lnot P(x) \land Q(x) \to R(x))$ are true.

Question

then $\forall x (\lnot R(x) \to P(x))$ is true where the domains of all quantifiers is the same.

I'm getting caught up on step 7 below.

1. $\forall x (P(x) \lor Q(x)$ (Premise)
2. $P(c) \lor Q(c)$ (Universal Instantiation from 1)
3. $\forall x (\lnot P(x) \land Q(x) \to R(x)$ (Premise)
4. $\lnot P(c) \land Q(c) \to R(c)$ (Universal Instantiation from 2)
5. $\lnot(\lnot P(c) \land Q(c)) \lor R(c)$ (Conditional Logical Equivalence)
6. $P(c) \lor \lnot Q(c) \lor R(c)$ (DeMorgan's Law)
7. $\lnot P(c) \lor R(c)$ (Resolution 2 and 6)

^^^ How can 7 be the results of the resolution of 2 and 6? When I try and work it out I'm left with the following:

2-a. $Q(c) \lor P(c)$ (Commutative Law)

6-a. $\lnot Q(c) \lor P(c) \lor R(c)$ (Commutative Law)

7-a. $P(c) \lor P(c) \lor R(c)$ (Resolution 2 and 6)

— Continuing on can I apply the Idempotent law to 7-a?

8. $R(c) \lor P(c)$ (Idempotent and Commutative Laws)

9. $\lnot R(c) \to P(c)$ (Conditional Logical Equivalence)

10. $\forall x (\lnot R(x) \to P(x))$ Universal Generalization

QED.

Answer 1

Your reasoning in 2-a through 10 is correct.

Line 7 does not follow from 2 and 6. To see this, you can use a model theoretic approach to show that $\neg P(c) \vee R(c)$ is not a logical implication of $P(c) \vee Q(c)$ and $P(c) \vee \neg Q(c) \vee R(c)$.

For example, let \begin{equation} P(c) = T\\ Q(c) = T\\ R(c) = F. \end{equation} Then we have \begin{equation} (P(c) \vee Q(c)) = T\\ (P(c) \vee \neg Q(c) \vee R(c)) = T\\ \neg P(c) \vee R(c) = F. \end{equation} Thus, the conditional $[(P(c) \vee Q(c)) \wedge (P(c) \vee \neg Q(c) \vee R(c))] \rightarrow (\neg P(c) \vee R(c))$ is false in this model, which means that you cannot infer $\neg P(c) \vee R(c)$ from the other two propositions in a proof theoretic setting. Of course, this all presupposes that the logical system you are using is sound :).