+1 vote
302 views

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.

retagged | 302 views

https://photos.app.goo.gl/u5CTGAINE7vrsLen2

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).

0

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