Question

2. Let’s consider how the operations (∧) and (∨) relate to the quantifier (∃).

a) Give and explain an explicit subset S of real numbers and
predicates P(x) and Q(x) for which the statement below is false:43
∃x ∈ S, P(x) ∧ Q(x) ↔ [∃x ∈ S, P(x)] ∧ [∃x ∈ S, Q(x)].

Answer 1

Solution :

2. a)

Observe that if x S, P(x) Q(x), then [ x S, P(x)] [ x S, Q(x)] .

Let us prove that the backward implication (converse) is false.

Let S = [1,) .

For any x in S,

let P(x) : x-1=0

let Q(x) : x-1 > 0

Observe that x S , P(x) (namely, x=1). Also, x S, Q(x) (namely, x=100).

Thus, [ x S, P(x)] [ x S, Q(x)] is true.

However, x S, P(x) Q(x) (namely because x=1 and x>1 cannot be simultaneously true for a given x in S).

Thus, the converse is not true.

Hence, the equivalence is not true.