Damon Stokes

2022-06-24

Find the negation of the following quantified sentence $\left(\forall x\right)\left(p\left(x\right)\vee q\left(x\right)\to ¬q\left(x\right)\right)$
My attempt:
To negate a quantified sentence, I just need to change the quantifier and connectives.
So: $\left(\forall x\right)\left(p\left(x\right)\vee q\left(x\right)\to ¬q\left(x\right)\right)$
Denying:
$¬\left(\forall \right)=\exists$
$¬\left(\vee \right)=\wedge$
$¬\left(\to \right)=\wedge$
$¬\left(¬q\left(x\right)\right)=q\left(x\right)$
I'm pretty sure that my solution is correct, the problem is that the denial of the conditional connective is causing me doubts.

Angelo Murray

Expert

Step 1
Notes: $¬\left(\vee \right)=\wedge$
Not quite: the negation of $A\vee B$ is $\mathrm{¬}A\wedge \mathrm{¬}B.$.
Remember, you are negating a sentence, not merely its main logical connective.
$\mathrm{¬}\left(\to \right)$
Not quite: the negation of $A\to B$ is $A\wedge \mathrm{¬}B.$
$\mathrm{¬}\left(\forall \right)$
Not quite: the negation of $\mathrm{¬}\forall xA\left(x\right)$ is $\exists x\mathrm{¬}A\left(x\right).$.
$\mathrm{¬}\left(¬q\left(x\right)\right)$
Correct.
Step 2
So: $\left(\forall x\right)\left(p\left(x\right)\vee q\left(x\right)\to ¬q\left(x\right)\right)$
Liberally inserting parentheses helps disambiguate sentences. In any case, the precedence of logical connectives dictates that this sentence ought to be read as $\forall x\left(\left(p\left(x\right)\vee q\left(x\right)\right)\to ¬q\left(x\right)\right)$
(I removed the unnecessary parentheses around the quantifier).
Denying:
This is incorrect, as pointed out in the comments and other answers. My above corrections make clear why.

Lovellss

Expert

Step 1
Using the definition of $P\to Q$, which is $\left(\mathrm{¬}P\right)\vee Q$.
Step 2
So, since $\mathrm{\forall }x\phantom{\rule{thickmathspace}{0ex}}\left(p\left(x\right)\vee q\left(x\right)\to \mathrm{¬}q\left(x\right)$ is the same as $\mathrm{\forall }x\phantom{\rule{thickmathspace}{0ex}}\mathrm{¬}\left(p\left(x\right)\vee q\left(x\right)\right)\vee \mathrm{¬}q\left(x\right)$, its negation is $\mathrm{\exists }x\phantom{\rule{mediummathspace}{0ex}}\mathrm{¬}\mathrm{¬}\left(p\left(x\right)\vee q\left(x\right)\right)\wedge \mathrm{¬}\mathrm{¬}q\left(x\right),\phantom{\rule{1em}{0ex}}\text{i.e.}\phantom{\rule{1em}{0ex}}\mathrm{\exists }x\phantom{\rule{mediummathspace}{0ex}}\left(p\left(x\right)\vee q\left(x\right)\right)\wedge q\left(x\right).$

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