Show that ∃x(P(x)∨Q(x)) and ∃xP(x)∨∃xQ(x) are logically equivalent

To show that ∃x(P(x)∨Q(x)) and ∃xP(x)∨∃xQ(x) are logically equivalent, we need to show that they imply each other.

1. Assume ∃x(P(x)∨Q(x)) is true. This means that there exists an x such that either P(x) or Q(x) is true.

2. If P(x) is true for some x, then ∃xP(x) is true, and hence ∃xP(x)∨∃xQ(x) is true.

3. If Q(x) is true for some x, then ∃xQ(x) is true, and hence ∃xP(x)∨∃xQ(x) is true.

So, we have shown that if ∃x(P(x)∨Q(x)) is true, then ∃xP(x)∨∃xQ(x) is also true.

Now, let's show the other direction.

1. Assume ∃xP(x)∨∃xQ(x) is true. This means that either there exists an x such that P(x) is true, or there exists an x such that Q(x) is true.

2. If ∃xP(x) is true, then for some x, P(x) is true. Hence, for that x, P(x)∨Q(x) is true, and hence ∃x(P(x)∨Q(x)) is true.

3. If ∃xQ(x) is true, then for some x, Q(x) is true. Hence, for that x, P(x)∨Q(x) is true, and hence ∃x(P(x)∨Q(x)) is true.

So, we have shown that if ∃xP(x)∨∃xQ(x) is true, then ∃x(P(x)∨Q(x)) is also true.

Therefore, ∃x(P(x)∨Q(x)) and ∃xP(x)∨∃xQ(x) are logically equivalent.