Answer
$a) ∀xF(x, Fred)$
$b) ∀yF(Evelyn, y)$
$c) ∀x∃yF(x, y)$
$d) ¬∃x∀yF(x, y) $
$e) ∀y∃xF(x, y)$
$f) ¬∃x(F(x, Fred) ∧ F(x, Jerry))$
$g) ∃y1∃y2(F(Nancy, y1) ∧ F(Nancy, y2) ∧ y1 , = y2 ∧ ∀y(F(Nancy, y) → (y = y1 ∨ y = y2)))$
$h) ∃y(∀xF(x, y) ∧ ∀z(∀xF(x, z) → z = y)) $
$i) ¬∃xF(x, x)$
$j) ∃x∃y(x , = y ∧ F(x, y) ∧ ∀z((F(x, z) ∧ z , = x) → z = y)) $
Work Step by Step
j) We do not assume that this sentence is asserting that this person can or cannot fool her/himself.
