離散數學 truth value

(a) Consider x = -20, y = -10x < y but x < exp(-y) is wrongSo,∀x∀y (x < y -> x < exp(-y)) is wrong(b) Consider x = 1, there is no suitable y such that x < y andx < exp(-y). So, the statement So,∀x∃y (x < y -> x < exp(-y)) is is also wrong(c) Consider x = 0, then for all y > 0, exp(-y) > 0So, ∃x∀y (x < y -> x < exp(-y)) is true(d) As (c) is true, (d) should also be true