1. **Problem:** Express the negations of the given statements so that all negation symbols immediately precede predicates. 2. **Recall:** The key rules for negating quantifiers and logical connectives are: - Negation of an existential quantifier: $\neg \exists x P(x) \equiv \forall x \neg P(x)$ - Negation of a universal quantifier: $\neg \forall x P(x) \equiv \exists x \neg P(x)$ - Negation of conjunction: $\neg (A \wedge B) \equiv \neg A \vee \neg B$ - Negation of disjunction: $\neg (A \vee B) \equiv \neg A \wedge \neg B$ - Negation of biconditional: $\neg (A \leftrightarrow B) \equiv (A \wedge \neg B) \vee (\neg A \wedge B)$ 3. **Step-by-step negations:** **a)** $\exists z \forall y \forall x T(x,y,z)$ Negation: $$ \neg \exists z \forall y \forall x T(x,y,z) \equiv \forall z \neg \forall y \forall x T(x,y,z) \equiv \forall z \exists y \neg \forall x T(x,y,z) \equiv \forall z \exists y \exists x \neg T(x,y,z) $$ **b)** $\exists x \exists y P(x,y) \wedge \forall x \forall y Q(x,y)$ Negation: $$ \neg (\exists x \exists y P(x,y) \wedge \forall x \forall y Q(x,y)) \equiv \neg \exists x \exists y P(x,y) \vee \neg \forall x \forall y Q(x,y) \equiv \forall x \forall y \neg P(x,y) \vee \exists x \exists y \neg Q(x,y) $$ **c)** $\exists x \exists y (Q(x,y) \leftrightarrow Q(y,x))$ Negation: $$ \neg \exists x \exists y (Q(x,y) \leftrightarrow Q(y,x)) \equiv \forall x \forall y \neg (Q(x,y) \leftrightarrow Q(y,x)) $$ Using biconditional negation: $$ \forall x \forall y ((Q(x,y) \wedge \neg Q(y,x)) \vee (\neg Q(x,y) \wedge Q(y,x))) $$ **d)** $\forall y \exists x \exists z (T(x,y,z) \vee Q(x,y))$ Negation: $$ \neg \forall y \exists x \exists z (T(x,y,z) \vee Q(x,y)) \equiv \exists y \neg \exists x \exists z (T(x,y,z) \vee Q(x,y)) \equiv \exists y \forall x \forall z \neg (T(x,y,z) \vee Q(x,y)) $$ By De Morgan: $$ \exists y \forall x \forall z (\neg T(x,y,z) \wedge \neg Q(x,y)) $$ 4. **Summary of negations with negation symbols immediately preceding predicates:** - a) $\forall z \exists y \exists x \neg T(x,y,z)$ - b) $\forall x \forall y \neg P(x,y) \vee \exists x \exists y \neg Q(x,y)$ - c) $\forall x \forall y ((Q(x,y) \wedge \neg Q(y,x)) \vee (\neg Q(x,y) \wedge Q(y,x)))$ - d) $\exists y \forall x \forall z (\neg T(x,y,z) \wedge \neg Q(x,y))$ These forms ensure all negation symbols are directly in front of predicates.