Question

Give an example to show that we cannot distribute an existential quantifier over a conjunction.

Answer 1

Answer:

Assume a set S={2,4,3,9}

let p(x) = "x divisible by 2 "

   q(x) = "x divisible by 3 "

LHS :: ∃x(P(x) ∧ Q(x))   is false always...as there exist no number which divisible by both 2 & 3..in above set

RHS ::∃xP(x) ∧∃xQ(x) is true always...as "there exist a number which divisible by 2 " and "there exist a number which divisible by 3 "

∃x(P(x) ∧ Q(x)) = ∃xP(x) ∧ ∃xQ(x)

      F                =     T

as LHS & RHS are not equal                 

so we cannot distribute an existential quantifier over a conjunction.