0
$\begingroup$

Trying to prove if this is logically equivalent

$p \leftrightarrow (q \rightarrow r)$ and $(p \leftrightarrow q) \rightarrow r$

I got that they wouldn't be equal just wanted to confirm if I was right. My final answer was $((p\lor r)\land (\neg q\lor r))\land (\neg r\lor (\neg p\lor q))$ after simplifying the right hand side.

|cite|improve this question
$\endgroup$
  • 1
    $\begingroup$ To prove that they are not equal in the clearest way, exhibit a triple of truth values for $(p,q,r)$ such that, for that triple, the two expressions don't have the same truth value. An 8-line truth table will show all such a triples, but you can stop after the first mismatch. $\endgroup$ – quasi Mar 22 '17 at 7:44
  • $\begingroup$ So its pointless trying to use the laws to figure out a way to simplify ? $\endgroup$ – rprogramr Mar 22 '17 at 7:46
  • $\begingroup$ Not pointless, but a single triple yielding a mismatch disproves equivalence. $\endgroup$ – quasi Mar 22 '17 at 7:47
  • $\begingroup$ They are not equivalent; to show it, it ios enough to consider a truth assignment $v$ such that : $v(p)=$ F and $v(r)=$ t. $\endgroup$ – Mauro ALLEGRANZA Mar 22 '17 at 11:11
  • $\begingroup$ But how can you tell they aren't equivalent did you simplify first ? $\endgroup$ – rprogramr Mar 22 '17 at 20:17
1
$\begingroup$

Examine $\def\getsto{{\;\leftrightarrow\;}}\def\false{\underset{\tiny\text{false}}\bot}\def\true{\underset{\tiny\text{true}}\top} p \getsto (q \to r)$ and $(p \getsto q) \to r$

$\begin{align}p \getsto (q \to r) ~& \iff (\neg p\vee (q\to r)\wedge(p\vee\neg(q\to r)) \\ &\iff (\neg p\vee\neg q\vee r)\wedge(p\vee\neg(\neg q\vee r))) \\ &\iff (\neg p\vee\neg q\vee r)\wedge (p\vee(q\wedge\neg r)) \\[3ex](p \getsto q) \to r ~&\iff \neg(p\getsto q)\vee r \\ & \iff (p\oplus q)\vee r \\ & \iff ((\neg p\vee \neg q)\wedge(p\vee q))\vee r \\ & \iff (\neg p\vee \neg q\vee r)\wedge(p\vee q\vee r) \end{align}$

Test when $r$ true but $p,q$ both false.

$ \false\getsto(\false\to\true)$ is false but $(\false\getsto\false)\to\true$ is true.

|cite|improve this answer
$\endgroup$

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.