Question

"some people like discrete math" which one would be correct? 1) ∃ x likediscretemaths(x) 2) ∃ x(people(x) -> likes(x,discretemaths))

Answer 1

Suppose vv maps variable xx to some object that we denote by cc, and suppose we have an II that sets ∀xA(x)→∃y B(c,y)∀xA(x)→∃y B(c,y) to true.

We'll show by a proof by contradiction that this means that II will set ∃x(A(x)→∃y B(x,y))∃x(A(x)→∃y B(x,y)) to true. Thus, suppose ¬∃x(A(x)→∃y B(x,y))¬∃x(A(x)→∃y B(x,y)). The latter can be rewritten as:

¬∃x(A(x)→∃y B(x,y))⇔¬∃x(A(x)→∃y B(x,y))⇔

∀x¬(A(x)→∃y B(x,y))⇔∀x¬(A(x)→∃y B(x,y))⇔

∀x(A(x)∧¬∃y B(x,y))⇔∀x(A(x)∧¬∃y B(x,y))⇔

∀xA(x)∧∀x¬∃y B(x,y)∀xA(x)∧∀x¬∃y B(x,y)

So, we have ∀xA(x)∀xA(x), and thus by the first Assumption we have ∃y B(c,y)∃y B(c,y), but we also have ∀x¬∃y B(x,y)∀x¬∃y B(x,y), and therefore ¬∃y B(c,y)¬∃y B(c,y), and thus we obtain a contradiction as desired.

And so yes, this is indeed a valid formula


hope it's help you