Proving ∀x(A(x) ∨ B) → ∀xA(x) ∨ B, with x is not in B, by natural deduction

Question

how can prove ∀x(A(x) ∨ B) → ∀xA(x) ∨ B where x is not in B using natural deduction.

i am not sure how should use for all introduction rule here.

any help wpuld be highly appreciate.

Cheers

Answer 1

If you are using a Fitch style proof system (i.e. One with Introduction and Elimination rules), I would set this up as a proof by contradiction: Assume $\neg (\forall x \ A(x) \lor B)$, use your rules to derive $\neg \forall x \ A(x))$ and $\neg B$, and use those in combination with your $\forall x (A(x) \lor B)$ to derive a contradiction (this should not be hard: from $\neg \forall x \ A(x)$ you should be able to derive $\exists x \neg A(x)$ (this may require its own proof by contradiction), then witness that existential with an $a$ to get $\neg A(a)$, instantiate the universal to get $(A(a) \lor B)$, and combine those two and the $\neg B$ to get the contradiction)