6
$\begingroup$

I have a statement,

Either p or q

and I have to write it in terms of logical connectives but I don't get which logical connector should I be using? Here is what I did (I think there could have been a better way to do this)

$(p \lor q ) \land (\neg((p \Rightarrow q) \land (q \Rightarrow p)))$

$\endgroup$
  • 1
    $\begingroup$ If by "either or" you mean in particular not both, might as well say it directly, $(p \lor q)\land \lnot(p \land q)$. Complicated combinations that involve implication can be hard to read. I think. There are other sensible ways. More symmetric would be $(p\land \lnot q) \lor (q\land \lnot p)$. $\endgroup$ – André Nicolas Sep 28 '11 at 22:12
  • $\begingroup$ (Added) If, again, you mean exclusive or, then your sentence works, but even better, part of it works. Look at $\lnot((p \to q)\land (q\to p))$. This holds if $p$ is true and $q$ is false, or vice-versa. So well done, except for the unnecessary $p \lor q$ part. But it took me a few seconds of looking to realize this, because the connective $\to$ is somehow less intuitive. (The connectives $\lor$ and $\land$ are closely connected to the geometric notions of union and intersection.) $\endgroup$ – André Nicolas Sep 28 '11 at 22:31
12
$\begingroup$

To me, the word "either" is unnecessarily confusing, and should be avoided if possible (of course, since it is part of the problem we have no choice in this case).

If "either $p$ or $q$" means the same thing as "$p$ or $q$", then the answer is simply $p\vee q$ (by the definition of $\vee$).

However, if "either $p$ or $q$" means "either $p$ or $q$, but not both" then this is equivalent to "$p$ is true and $q$ is false, or $p$ is false and $q$ is true". Do you see how to write the logical expression for this?

$\endgroup$
  • 3
    $\begingroup$ Let me give it a try, its (p ^ ~q) V (~p ^ q). Right? $\endgroup$ – fuddin Sep 28 '11 at 22:25
  • 2
    $\begingroup$ @Akito: Exactly right! :) $\endgroup$ – Zev Chonoles Sep 28 '11 at 22:25
  • 1
    $\begingroup$ My favorite ways to write the exclusive or are $\neg(p \Leftrightarrow q)$ ("$p$ and $q$ have different truth values"), or $\neg p \Leftrightarrow q$ and $p \Leftrightarrow \neg q$ ("$p$ and $q$ have opposite truth values"). $\endgroup$ – Omar Antolín-Camarena Sep 28 '11 at 22:40
  • $\begingroup$ Interpreting either ... or ... to mean anything different than p or q, but not both, would render the word either utterly useless. There is a reason for the word to exist, so interpreting it in other ways does not make sense. It is exactly for the purpose of expressing p or q, but not both , so that we can talk about these things in a precise way. Of course it's going to be useless, if you interpret it to mean something it does not. Still does not mean it should not be used. You'd rather not use it and express things imprecisely? That seems strange. $\endgroup$ – Zelphir Kaltstahl May 1 '17 at 23:33
3
$\begingroup$

If "Either p or q" means "p or q", then logical disjunction $p \lor q$ would do it, if "Either p or q" means not both, then exclusive disjunction is needed: $p \oplus q = (p\land \neg q)\lor (\neg p\land q) = (\neg p\lor \neg q)\land (p\lor q)$.

$\endgroup$
  • $\begingroup$ +1, for mentioning XOR. I was wondering if that was going to be on this question. $\endgroup$ – Conor O'Brien Sep 24 '15 at 3:08
2
$\begingroup$

There are several symbols for exclusive or, including $\oplus$ and $\veebar$. However, while in classical logic such connectives are both easily defined in terms of existing connectives and by means of a truth-table, they are not commonly employed in mathematics.

It's worth noting that there are many minimal functionally complete sets of logical connectives, but the definition of xor in terms of formulae using predefined connectives will be different in each. An obvious one is replacing $\phi \oplus \psi$ with $(\phi \vee \psi) \wedge \neg (\phi \wedge \psi)$.

$\endgroup$
1
$\begingroup$

Assuming I understand the question correctly (you're looking to describe "exclusive or"), hint: either p or q means p is true and not q, or (inclusive or) ...

You could also try draw the truth table.

$\endgroup$

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.