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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?

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

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

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

QED.

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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 :).

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You are correct, $\lnot P(c)\lor R(c)$ is not the result of resolution of those lines.   Fortunately it is also not what you want to obtain from the resolution.

$\{\{p, q\},\{p,\lnot q,r\}\}$ resolves to $\{\{p,r\}\}$ ...which is also $\{\{p,\lnot\lnot r\}\}$

Thusly, line 7 should be $P(c)\lor R(c)$, which is equivalent to $P(c)\lor\lnot\lnot R(c)$, and thus $\lnot R(c)\to P(c)$, as you require.

  1. $\forall x~(P(x) \lor Q(x))$ (Premise I)
  2. $\forall x~((\lnot P(x) \land Q(x)) \to R(x)$ (Premise II)
  3. $\quad P(c) \lor Q(c)$ (Universal Instantiation from 1 to arbitrary $c$)
  4. $\quad (\lnot P(c) \land Q(c)) \to R(c)$ (Universal Instantiation from 2)
  5. $\quad \lnot(\lnot P(c) \land Q(c)) \lor R(c)$ (Conditional Logical Equivalence)
  6. $\quad P(c) \lor \lnot Q(c) \lor R(c)$ (DeMorgan's Law)
  7. $\quad P(c) \lor R(c)$ (Resolution 3 and 6)
  8. $\quad \lnot R(c)\to P(c)$ (Conditional Logical Equivalence from 7)
  9. $\forall x~(\lnot R(x)\to P(x))$ (Universal Generalisation, from 8 and arbitrary $c$)
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