67. Since the first two quantifiers are universal and the last quantifier is existential, Farley chooses
x and y, after which, you choose z. If Farley chooses values such that x ≥ y, the proposition
(x < y) → ((z > x) ∧ (z < y))
is true by default (i.e., it is true regardless of what value you choose for z). If Farley chooses
values such that x < y, you can choose z = (x + y)/2 and again the proposition
is true. Since you can always win the game, the quantified propositional function is true.
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