Answer / Step-by-step explanation:
Given the statement:
∃x∈D,(P(x)∧Q(x)) and (∃x∈D,P(x))∧(∃x∈D,Q(x)) ,
Then,
(a), The variable used in a ∃ statement does not matter, thus, we can change the appearance of one of the variable used in the ∃ statement.
That is:
(∃x∈D,P(x))∧(∃x∈D,Q(x)) = (∃x∈D,P(x))∧(∃y∈D,Q(y))
Where
(∃x∈D,P(x))∧(∃x∈D,Q(x)) = (∃x∈D,P(x))∧(∃y∈D,Q(y)) implies that P(x) is true for some element x in D and Q(y) is true for some element y in D. However, x and y are not necessary the same element and thus, we cannot be sure that
P(x) ∧ Q(x) or P(y) ∧ Q(y) is true.
Moreover, if P(x) is only true for x and no other element in the domain D, and if Q(y) is only true for y and no other element in the domain D, while x ≠ y,
Then, P(x) ∧ Q(x) is false and P(y) ∧ Q(y) is also false. Moreover, there is no other known element (z) such that P(z) ∧ Q(z) is true and thus the statement
∃x∈D,(P(x)∧Q(x)) is false while the statement (∃x∈D,P(x))∧(∃x∈D,Q(x)) is true.
(b)
If the statement P(x) is only true for x and no other element in the domain D, and if Q(y) is only true for y and no other element in the domain D, while x ≠ y, then Then, P(x) ∧ Q(x) is false and P(y) ∧ Q(y) is also false. Moreover, there is no other known element (z) such that P(z) ∧ Q(z) is true and thus the statement
∃x∈D,(P(x)∧Q(x)) is false while the statement (∃x∈D,P(x))∧(∃x∈D,Q(x)) is true.
So in summary, we can say for:
(a) the statement does not contain the same truth value.
(b) The statement depicts there is such a choice in the first place.
