Question

Identify the error in this argument that supposedly shows that if ∃xp(x) ∧ ∃xq(x) is true then ∃x(p(x) ∧ q(x)) is true

Answer 1

The problem is that ∃xp(x) is true, and ∃xq(x) is true means that for some x=x1, p(x) is true, and for some x=x2, q(x) is true.  There is no guarantee that x1=x2, although it is not impossible.

However, ∃x(p(x) ∧ q(x)) means that for the same x=x0, p(x0) is true, and q(x0) is true.  This cannot be implied from the first proposition, therefore

∃xp(x) ∧ ∃xq(x)  =>  ∃x(p(x) ∧ q(x))    is a false statement.