Discrete Mathematics > The Logic of Quantified Statements > Predicates and Quantified Statements > Negation of A Universal Conditional Statement

Question

There are various questions in this topic, but none were covering my particular question. Can you please help me with the following: I have a Universal Conditional Statement (Universal Implication). Please note that it is NOT a simple straight forward Conditional (Implication) Statement. It has a UNIVERSAL QUANTIFIER:

$\forall , , \in \Bbb Z$, $(−)$ $(−)$ ℎ $(−)$ .

Now I am looking to find the negation of the "Universal Conditional Statement. If I rewrite the question:

Originally, we have:

$P(x) = (−)$ $(−)$ = $R(x) \wedge S(x)$

$R(x) = (−)$

$S(x) = (−)$

$Q(x) = (−)$

which is

$\forall , , \in \Bbb Z$, $P(x)$ ℎ $Q(x)$

or in another word:

$\forall , , \in \Bbb Z$, $R(x) \wedge S(x)$ ℎ $Q(x)$

Will the negation of the above 'Universal Conditional Statement' be:

$\exists , , \in \Bbb Z$, $P(x)$ ℎ $Q(x)$

$\exists , , \in \Bbb Z$, $(−)$ $(−)$ ℎ $(−)$ not e.

This is as per page $111$ of Discrete Mathematics from Susan Epp $4$th edition (or page $125$ of the same book $5$th edition)

or will it be:

$\exists , , \in \Bbb Z, P(x)$ and $\neg Q(x)$?

$\exists , , \in \Bbb Z, (−)$ $(−)$ and $(−)$ not e.

Answer 1

The negation of $\forall m(T(m))$ is $\exists m(\neg T(m))$. Here $m$ represents the ordered triplet $(a,b,c)\in\Bbb Z^3$ and $T(m)$ is the statement $P(a,b)\implies Q(a,b)$. The negation of $T(m)$ is $P(a,b)\wedge\neg Q(a,b)$ (read this post!) so the required negation is$$\exists m(\neg T(m))\iff \exists a,b,c\in\Bbb Z,(a-b)\text{ is even and }(b-c)\text{ is even and }(a-c)\text{ is not even.}$$i.e. your second statement.