Express the negations of each of these statements so that
all negation symbols immediately precede predicates.
∃x∃y(Q(x, y) ↔ Q(y, x))
By steps: 1)∀ x∀ y¬(Q(x,y)↔ Q(y,x))2)∀ x∀ y(Q(x,y) ⊕ Q(y,x))where ⊕ is exclusive and exhaustivehence answer is ∀x∀y(Q(x,y)⊕Q(y,x))By \space steps: \space \\ 1)\forall \space x\forall \space y\neg(Q(x,y)\leftrightarrow \space Q(y,x))\\ 2)\forall \space x\forall \space y(Q(x,y) \space \oplus \space Q(y,x)) \\ where \space \oplus \space is \space exclusive \space and \space exhaustive\\ hence \space answer \space is \space ∀x∀y(Q(x,y)⊕Q(y,x))By steps: 1)∀ x∀ y¬(Q(x,y)↔ Q(y,x))2)∀ x∀ y(Q(x,y) ⊕ Q(y,x))where ⊕ is exclusive and exhaustivehence answer is ∀x∀y(Q(x,y)⊕Q(y,x))