Question

32. Express the negations of each of these statements so thatall negation symbols immediately precede predicates.
a) \(\exists \user1{z}\forall \user1{y}\forall \user1{xT}\left( {\user1{x, y, z}} \right)\)
b)\(\exists \user1{x}\exists \user1{yP}\left( {\user1{x,y}} \right) \wedge \forall \user1{x}\forall \user1{yQ}\left( {\user1{x, y}} \right)\)
c)\(\exists \user1{x}\exists \user1{y}\left( {\user1{Q}\left( {\user1{x,y}} \right) \leftrightarrow \user1{Q}\left( {\user1{y, x}} \right)} \right)\)
d) \(\forall \user1{y}\exists \user1{x}\exists \user1{z}\left( {\user1{T}\left( {\user1{x,y,z}} \right) \vee \user1{Q}\left( {\user1{x, y}} \right)} \right)\)

Short answer

The negations of the statements are as follows,

1. \(\exists \user2{z}\forall \user2{y}\forall \user2{xT}\left( {\user2{x, y, z}} \right)\) is \(\forall \user1{z}\exists \user1{y}\exists \user1{x} \sim \user1{T}\left( {\user1{x, y, z}} \right)\)
2. \(\exists \user2{x}\exists \user2{yP}\left( {\user2{x,y}} \right) \wedge \forall \user2{x}\forall \user2{yQ}\left( {\user2{x, y}} \right)\)is\(\forall \user1{x}\forall \user1{y} \sim \user1{P}\left( {\user1{x,y}} \right) \vee \exists \user1{x}\exists \user1{y} \sim \user1{Q}\left( {\user1{x, y}} \right)\)
3. \(\exists \user2{x}\exists \user2{y}\left( {\user2{Q}\left( {\user2{x,y}} \right) \leftrightarrow \user2{Q}\left( {\user2{x, y}} \right)} \right)\)is\(\forall \user1{x}\forall \user1{y} \sim \left( {\user1{Q}\left( {\user1{x,y}} \right) \leftrightarrow \user1{Q}\left( {\user1{x, y}} \right)} \right)\)
4. \(\forall y\exists \user2{x}\exists \user2{z}\left( {\user2{T}\left( {\user2{x,y,z}} \right) \vee \user2{Q}\left( {\user2{x, y}} \right)} \right)\)is \(\forall \user1{y}\exists \user1{x}\exists \user1{z}\left( { \sim \user1{T}\left( {\user1{x,y,z}} \right) \wedge \sim \user1{Q}\left( {\user1{x, y}} \right)} \right)\)

Step-by-step solution

Introduction to the Concept

When the negation is applied to the quantifiers,\(\forall \)quantifier becomes\(\exists \)quantifier and vice versa, i.e., quantifier\(\exists \)becomes\(\forall \)quantifier.