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I'm trying to use resolution to prove that:

$\forall x(A(x) \lor B(x)) \land \neg B(a) \models A(a)$

My attempt was to try resolve $\neg B(a)$ and $B(x)$ by forcing the substitution of x for a. This gives

$\forall aA(a) \models A(a)$ which is true.

Is this correct? I am unsure as I am still trying to understand resolution in predicate logic.

Thanks for any help!

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To do resolution, you need to negate the conclusion, put that into clauses, add that to the clauses from the premises, and then derive the empty clause.

Thus, you get the clause $\{\neg A(a) \}$ from the negation of the conclusion, and clauses $\{ A(x), B(x) \}$ and $\{ \neg B(a) \}$ from the premises.

Now you can resolve the latter two clauses by setting $x=a$, resulting in the clause $\{ A(a) \}$.

And that clause resolves with the $\{ \neg A(a) \}$ clause, resulting in the empty clause $\{ \}$

That ends the resolution, but to explain why that proves the argument to be valid: the empty represents a contradiction, and so what you have shown is that if you assume th conclusion is not true, you get a contradiction which, by Proof by Contradiction, shows that the conclusion has to be true given the turht of the premises ... meaning the argument is valid.

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