A) ∀x [P(x)] ---> ∃x [P(x)]
It Convert to digital logic expression
P1.P2 -------> P1 + P2
You can not find 1--------->0
I mean If P1=1, P2=1 then P1.P2 =1 (true) otherwise 0 (false),So P1 + P2 =1 +1 =1
So Option (A) is True always
But
B) ∃x [P(x)] ---> ∀x [P(x)]
It Convert to digital logic expression
P1 + P2 ------> P1.P2
You can easily find 1------------>0
I mean If P1=1 ,P2=0 or P1=0 ,P2=1 then P1 + P2 =1(true) otherwise 0(false) So P1.P2 = 1.0 = 0.1 = 0
So this is Not true always