-1
$\begingroup$

Does $P \implies \sim R$ entail $P \implies \sim ( R $ or $ S)$?

Can anyone please help me ? Is the other way correct ?

Can I prove or disprove it using Truth Table ?

$\endgroup$
5
  • $\begingroup$ What does $\sim R$ mean? $\endgroup$ Oct 1, 2021 at 8:07
  • $\begingroup$ @Mathphysmeister $\sim$ is another symbol for $\neg$, the negation. $\endgroup$
    – user562983
    Oct 1, 2021 at 8:11
  • $\begingroup$ Suppose $P \implies \sim R$ is true, then either $P,\sim R$ both true or $P$ false. So if $P$ is false then clearly $P \implies \sim (R \ \mathrm{or} S)$. Now suppose that $P,\sim R$ are both true. Then $R$ is false. $S$ can be true or false. Suppose $S$ is true, then $\sim(R \ \mathrm{or} S)$ is false. And if $S$ is false then $\sim (R \ \mathrm{or} S)$ is true. Thus we conclude that it cannot be correct. It depends on how rigorous your course is whether you can use this type of truth table reasoning or whether you really need to use deduction rules! $\endgroup$ Oct 1, 2021 at 8:17
  • $\begingroup$ I think the simplest way is to use the definition of $\implies$: $P\implies Q$ is $(\sim P)\vee Q$. $\endgroup$
    – Bernard
    Oct 1, 2021 at 8:26
  • $\begingroup$ Yes, you can disprove it with Truth Table. $\endgroup$ Oct 1, 2021 at 14:05

1 Answer 1

1
$\begingroup$

$P \implies \sim(R \lor S)$ can be written as $P \implies \sim R \land \sim S$.

This is De Morgan's law at work and can be shown using truth tables. Alternatively, we can reason intuitively: $\sim(R \lor S)$ means neither $R$ nor $S$ is true, or, in other words, both $R$ and $S$ must be false.

From $P \implies \sim R$, if $P$ is true we know only $\sim R$. For all we know about $S$, $S$ could be true which would render $\sim R \land \sim S$ false. Thus, $P \implies \sim(R \lor S)$ is not a logical consequence of $P \implies \sim R$.

A more concise answer:

For $A$ to entail $B$, $B$ must be true under every truth assignment such that $A$ is true. Let us examine the truth assignment $v(P) = 1$, $v(R) = 0$ and $v(S) = 1$. Under this truth assignment, $A$ sure is true, but $B$ is false because $P$ is true but $\sim R \land \sim S$ is false. So we have found a truth assignment under which $A$ is true but $B$ is false, implying that $A$ does not entail $B$.

$\endgroup$

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.