SINCE THE FIRST TWO QUANTIFIERS ARE UNIVERSAL AND THE LAST QUANTIFIER IS EXISTENTIAL, FARLEY CHOOSESX AND Y, AFTER WHICH, YOU CHOOSE Z

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.