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😃: Show that ∀xP(x) ∨∀xQ(x) and ∀x∀y(P(x)∨Q(y)),
where all quantifiers have the same nonempty domain,
are logically equivalent. (The new variable y is used to
combine the quantifications correctly.)
🤖: To show that two statements are logically equivalent, we need to prove that one can be deduced from the other using logical rules of inference. In this case, we want to show that $\forall x P(x) \lor \forall x Q(x)$ and $\forall x \forall y (\exists z (P(z) \lor Q(z)))$ are logically equivalent.

First, let's rewrite the second statement in a more convenient form:
$$\mathrm{\forall }x\mathrm{\forall }y(\mathrm{\exists }z(P(z)\vee Q(z)))\backslash forall\; x\; \backslash forall\; y\; (\backslash exists\; z\; (P(z)\; \backslash lor\; Q(z)))$$
Using the principle of universal instantiation, we can substitute $z = x$ into the formula to get:
$$\mathrm{\forall }x\mathrm{\forall }y(\mathrm{\exists }x(P(x)\vee Q(x)))\backslash forall\; x\; \backslash forall\; y\; (\backslash exists\; x\; (P(x)\; \backslash lor\; Q(x)))$$
Now, we can see that the formula reduces to:
$$\mathrm{\forall }x\mathrm{\forall }yP(x)\vee \mathrm{\forall }x\mathrm{\forall }yQ(x)\backslash forall\; x\; \backslash forall\; y\; P(x)\; \backslash lor\; \backslash forall\; x\; \backslash forall\; y\; Q(x)$$
which is identical to the first statement. Therefore, we have shown that $\forall x P(x) \lor \forall x Q(x)$ and $\forall x \forall y (\exists z (P(z) \lor Q(z)))$ are logically equivalent.