Prove That ∃x (A(x) → B(x)) ≡ ∀x A(x) → ∃x B(x).
My Attempt -: RHS will be always true as in ∀x A(x) → ∃x B(x),∀x A(x) will be always false as not "All"number in the Universe of discourse will satisfy A(x).
coming on LHS it can be True or false ..please help me out
$$\exists x(A(x)\to B(x))\equiv\exists x(\neg A(x)\lor B(x))\equiv\exists x\neg A(x)\lor\exists xB(x)\equiv\neg\forall xA(x)\lor\exists xB(x)\equiv\forall xA(x)\to\exists xB(x)$$