Let P(x) and Q(x) be predicates and suppose D is the domain of x. In 55–58, for the statement forms in each pair, determine whether (a) they have the same truth value for every choice of P(x), Q(x), and D, or (b) there is a choice of P(x), Q(x), and D for which they have opposite truth values.
∃x ∈ D, (P(x) ∧ Q(x)), and (∃x ∈ D, P(x)) ∧ (∃x ∈ D, Q(x))
Let \(P(x)\) and \(Q(x)\) be predicates and suppose \(D\) is the domain of \(x\).
Let the statement is
\(\exists x \in D,(P(x) \wedge Q(x))\), and \((\exists x \in D, P(x)) \wedge(\exists x \in D, Q(x))\)
\((a)\)
No matter what the domain \(D\) or the predicates \(P(x)\) and \(Q(x)\) are, the given statements have the same truth value.
If the statement \(" \exists x \in D,(P(x) \wedge Q(x)) '\) is true, then \(P(x) \wedge Q(x)\) is true for some \(x\) in \(D\). Which implies that either \(P(x)\) is true for some \(x\) in \(D\) and \(Q(x)\) is true for some \(x\) in \(D\). But then \(P(x)\) is true for some \(x\) in \(D\) and \(Q(x)\) is true for some \(x\) in \(D .\) So, the statement \("(\exists x \in D, P(x)) \wedge(\exists x \in D, Q(x))^{\prime}\) is true.
\((b)\)
Conversely, if the statement \("(\exists x \in D, P(x)) \wedge(\exists x \in D, Q(x))^{\prime}\) is true, then either \(P(x)\) is true for some \(x\) in \(D\) and \(Q(x)\) is true for some \(x\) in \(D\).
This implies that either \(P(x)\) and \(Q(x)\) is true for some \(x\) in \(D .\) And so \(P(x) \wedge Q(x)\) is true for some \(x\) in \(D\)
Hence the statement " \(\exists x \in D,(P(x) \wedge Q(x))\) ' is true.