Respuesta :
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.
In this exercise we have to use the knowledge of sets to identify which of the statements is true and false, thus we can state that:
A) the statement does not contain the same truth value.
B) The statement depicts there is such a choice in the first place.
Then, the first statement says that:
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:
[tex](\exists \ x \in D,P(x)) \wedge ( \exists \ x\in D,Q(x)) = (\exists \ x \in D,P(x)) \wedge (\exists \ y \in D,Q(y))[/tex]
Where the equation above 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 is true.
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. Then, [tex]P(x) \wedge Q(x)[/tex] is false and [tex]P(y) \wedge Q(y)[/tex] is also false.
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. Then, [tex]P(x) \wedge Q(x) \ or \ P(y) \wedge Q(y)[/tex] is also false.
Moreover, there is no other known element (z) such that is true and thus the statement.
See more about sets at : brainly.com/question/8053622
