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+3 votes

Uses Modus ponens $(A, A → B |= B)$ or resolution to show that the following set is inconsistent:

  1. $Q(x) → P (x) \vee \sim R (a)$
  2. $R (a) \wedge \sim Q(a)$
  3. $Q(a)$
  4. $\sim P (y)$

where $x$ and $y$ are universally quantified variables, $a$ is a constant and $P, Q, R$ are monadic predicates.

asked in Mathematical Logic by Veteran (59.9k points)
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1 Answer

+1 vote

Answered it using rules of inference, please see below image:

After applying appropriate rules of inference, at last we get ~Q(a), which is inconsistent (as it will give True if Q(a) is false and vice versa. Therefore, always true/always false is not possible).

answered by Active (2k points)

why can't we further resolve Q(a) (obtained by resolution) with Q(a) (given in question)?

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