1. The statement of Example 1.6.6 is
∀x∃y(x + y = 0).
As was pointed out in Example 1.6.6, this statement is true. Now
∀x∀y(x + y = 0)
is false; a counterexample is x = y = 1. Also
∃x∀y(x + y = 0)
is false since, given any x, if y = 1 − x, then x + y = 0.
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