2
$\begingroup$

I have a logical statement that looks like the following for the sentence "If every cat feels wet, then every dog is happy"

$$ [\forall x \ \ C(x) \implies W(x)] \implies [\forall y \ \ D(y) \implies H(y)] $$

I want to put the NEGATED version of this statement in CNF form, so first I remove all the implications: $$ \lnot([\forall x \ \ \lnot C(x) \lor W(x)] \implies [\forall y \ \ \lnot D(y) \lor H(y)]) \\ \lnot([\lnot\forall x \ \ \lnot C(x) \lor W(x)] \lor [\forall y \ \ \lnot D(y) \lor H(y)]) \\ \lnot([\exists x \ \ C(x) \land \lnot W(x)] \lor [\forall y \ \ \lnot D(y) \lor H(y)]) \ \ \ \text{used Demorgan's Law here}\\ $$

Assuming the above steps are current, I am confused on how to distribute the outermost negation inside when there are quantifiers and logical statements.

I understand the following conversions:

$$ \lnot \forall x \ P(x) = \exists x \ \lnot P(x) \\ \lnot \exists x \ P(x) = \forall x \ \lnot P(x) \\ \lnot (a \lor b \lor...c) = \lnot (\lnot a \land \lnot b \land...\lnot c) \\ \lnot (a \land b \land...c) = \lnot (\lnot a \land \lnot b \lor...\lnot c) $$

but it's not clear to me how I can distribute the outermost negation because now it involves quantifiers. Any hints?

Edit 1

I think I may have gotten it:

$$ (\lnot[\exists x \ \ C(x) \land \lnot W(x)] \land \lnot[\forall y \ \ \lnot D(y) \lor H(y)]) \\ ([\forall x \ \ \lnot C(x) \lor W(x)] \land [\exists y \ \ D(y) \land \lnot H(y)]) \\ $$

$\endgroup$
0
$\begingroup$

One extra substitution rule to remember is Implication Negation Equivalence: $$\neg(\phi\to\psi) ~\equiv~ (\phi\wedge\neg\psi)$$

This can be derived using:

$$\begin{align}\neg(\phi\to\psi)&\quad&\\\neg(\neg\phi\vee\psi)&&&\text{Implication Equivalence}\\\neg\neg\phi\wedge\neg\psi&&&\text{de Morgan's Rule}\\\phi\wedge\neg\psi&&&\text{Double Negation Equivalence}\end{align}$$

Also vice versa.

Thus your statement's negation begins:

$$\begin{align}&\neg\Big(\big(\forall x~(Cx\to Wx)\big)\to\big(\forall y~(Dy\to Hy)\big)\Big) \\&\quad\big(\forall x~(Cx\to Wx)\big)\wedge\neg\big(\forall y~(Dy\to Hy)\big)&&\text{Implication Negation Equivalence} \\&\quad\vdots\end{align}$$

And it should be clear how to continue.


Also recall that the distribution rules for quantifiers in non-empty domains includes: $$(\forall x~P(x))\wedge (\exists y~Q(y))~~\equiv~~ \forall x~\exists y~(P(x)\wedge Q(y))$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.