. . . . . . . 1.) The sentence can be written symbolically as

$ ( \forall x)( \exists y) P(x, y) $ .

A denial is

$ \sim ( \forall x)( \exists y) P(x, y) $

is equivalent to

$ ( \exists x) \sim ( \exists y) P(x, y) $ . . . . . . (by Theorem 1.3 a.)

is equivalent to

$ ( \exists x)( \forall y) \sim P(x, y) $ . . . . . . (by Theorem 1.3 b.)

is equivalent to

There is some real number $ x $, so that for all real numbers $ y $, $ x^3 - x \ne y $ .


. . . . . . . 2.) The sentence can be written symbolically as

$ ( \forall x)( \exists y)( P(x, y) \Rightarrow ( Q(x, y) \vee R(x, y) ) ) $ .

A denial is

$ \sim ( \forall x)( \exists y)( P(x, y) \Rightarrow ( Q(x, y) \vee R(x, y) ) ) $

is equivalent to

$ ( \exists x) \sim ( \exists y)( P(x, y) \Rightarrow ( Q(x, y) \vee R(x, y) ) ) $ . . . . . . (by Theorem 1.3 a.)

is equivalent to

$ ( \exists x) ( \forall y) \sim ( P(x, y) \Rightarrow ( Q(x, y) \vee R(x, y) ) ) $ . . . . . . (by Theorem 1.3 b.)

is equivalent to

$ ( \exists x) ( \forall y) \sim ( ( Q(x, y) \vee R(x, y) ) \vee \sim P(x, y) ) $ . . . . . . (by Theorem 1.2 d.)

is equivalent to

$ ( \exists x) ( \forall y) ( \sim ( Q(x, y) \vee R(x, y) ) \wedge P(x, y) ) $ . . . . . . (by Theorem 1.2 c.)

is equivalent to

$ ( \exists x) ( \forall y) ( P(x, y) \wedge \sim ( Q(x, y) \vee R(x, y) ) ) $

is equivalent to

$ ( \exists x) ( \forall y) ( P(x, y) \wedge ( \sim Q(x, y) \wedge \sim R(x, y) ) ) $ . . . . . . (by Theorem 1.2 c.)

is equivalent to

There is some real number $ x $, so that for all real numbers $ y $, $ x^2 \le y^2 $ but $ x-y \not\le 0 $ and $ x+y \not\le 0 $ .

is equivalent to

There is some real number $ x $, so that for all real numbers $ y $, $ x^2 \le y^2 $ but $ x-y>0 $ and $ x+y>0 $ .



RETURN to problem set.



Duane Kouba 2002-06-06