Uses Modus ponens $(A, A → B |= B)$ or resolution to show that the following set is inconsistent:
where $x$ and $y$ are universally quantified variables, $a$ is a constant and $P, Q, R$ are monadic predicates.
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).
why can't we further resolve ∼Q(a) (obtained by resolution) with Q(a) (given in question)?
@Arjun sir, i haven't got my consignment. the ...
Nice to see your plenty ...
For those who knows only about ...