0
$\begingroup$

I'm not sure if this question makes sense or not, but if one assesses the following question:

Are $\neg p \lor q \lor p$ and $p \lor \neg p$ logically equivalent?

How could they be? The question doesn't seem to make sense.

Let's assume a truth table:

My Truth Table

Is this what the question is asking? It's a very vague question. I am just wondering if it is asking that given $p$ and $q$, that the values are equivalent based on the values taken from the truth table. Thanks!

$\endgroup$
0
$\begingroup$

Yes, they are equivalent since the truth tables are identical. Another way to see it is that both expressions are disjunctions ("or" statements) which include both p and "not p". One of the other of "p" and "not p" will always be true, so the "or" expression overall will always be true. This is why your truth table is all t's.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Excellent, thanks Nick, btw is my truth table correct? lol $\endgroup$ – Linuxn00b Aug 31 '16 at 1:59
  • $\begingroup$ Yes, it looks correct. $\endgroup$ – Nick Aug 31 '16 at 2:05
  • $\begingroup$ Thanks Nick, marking your answer as the correct answer. $\endgroup$ – Linuxn00b Aug 31 '16 at 2:05
  • $\begingroup$ @Nick Is it correct to say that to be logically equivalent, all models of the one proposition must also be models of the other proposition (and vice versa)? Here an assignment of true to 'Q' is a model for the first proposition, but not for the second. Is that an issue? $\endgroup$ – Boolean_functions Aug 31 '16 at 15:13

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.