Let θ(x, y, z) be the statement “x + y = z”, and let there be two…

Concept

In a quantified statement the order of the quantifiers fixes what each variable is allowed to depend on. In the prefix ∀x ∀y ∃z the existential variable is introduced last, so the witness z may be chosen after x and y are already known, and it may be a different value for every pair. In the prefix ∃z ∀x ∀y the existential variable is fixed first, so one single constant z must work for every pair that follows. The two prefixes are therefore different claims: ∃z ∀x ∀y P(x, y, z) implies ∀x ∀y ∃z P(x, y, z), but the reverse implication does not hold in general.

Application

Here the predicate is θ(x, y, z) : x + y = z, with x, y and z ranging over the real numbers.

1. Statement (i) is ∀x ∀y ∃z (x + y = z). Fix arbitrary real numbers x and y, and choose z = x + y.

2. The real numbers are closed under addition, so this z is itself a real number and it satisfies x + y = z.

3. A witness therefore exists for every pair x, y, so statement (i) is true.

4. Statement (ii) is ∃z ∀x ∀y (x + y = z). Suppose one fixed real number z worked for every pair.

5. Putting x = 0 and y = 0 forces z = 0 + 0 = 0, while putting x = 1 and y = 0 forces z = 1 + 0 = 1.

6. That single constant would have to equal both 0 and 1, which is impossible, so no such z exists and statement (ii) is false.

Cross-check

The two prefixes side by side:

Hence statement (i) is true and statement (ii) is false. The one-way implication is a useful check: had (ii) been true, (i) would follow at once, whereas (i) being true says nothing about (ii).