a may be false because ∃x(p(x)∧q(x)) here the vale of x is for both p(x)^q(x)
for
∃x(p(x)∧q(x))
at x=1 => p(1) and q(1)
at x=9 =>P(9) and q(9)
.
.
.
.
at x=n => p(n) and q(n)
now for
(∃xp(x)∧∃xq(x))
vale of x is distinct for both
p(1) and q(6)
p(6) and q (45)
.
.
.
p(n) and q(m)
so by this way it is FALSE...
correct me if i am incorrect