0
$\begingroup$

Reading my introduction to discrete math book comes with an introduction to logic. I understand the truth table for $p \rightarrow q$ but I am getting confused why $q\:unless\:\neg\:p$ is a similar statement because I'm not sure how to read it, specifically the word unless. How do these two compound propositions mean the same thing and can you give an example translated to English? Perhaps knowing how unless affects the statement can help me figure out the rest.

$\endgroup$
  • $\begingroup$ It is a little difficult to figure out what you are asking. $\endgroup$ – copper.hat Sep 25 '15 at 5:36
  • $\begingroup$ Yeah it's long, I'll try rephrasing. $\endgroup$ – Diehardwalnut Sep 25 '15 at 5:41
  • $\begingroup$ Wiktionary says "unless" means "1.Except on a specified condition; if not. ". That means that $q$ unless $\neg p$ means $q$ except if/if nat $\neg p$. It's indicates that the truth of $q$ should only be considered if $\neg p$ is true. $\endgroup$ – skyking Sep 25 '15 at 5:47
1
$\begingroup$
  1. You need to look at $p$, not $\lnot p$. In your case, if $q$ is false and $\lnot p$ is false (i.e. $p$ is true), $q$ unless $\lnot p$ is false, so in order of ($p$, $q$, statement) it is (true, false, false) which is the same as $p \rightarrow q$

  2. They are different in English but the same in math. If the politician is not elected ($p = F$), and he lowered tax ($q = T$), he will lower tax if he is elected is true while he will lower tax unless he is not elected sounds like a false. The confusion comes because in mathematics, given the conditions above, (he will lower tax unless he is elected) is true.

I think a better way to explain $q$ unless $\lnot p$ is "$q$ unless $\lnot p$, but if $\lnot p$ really happens then $q$ doesn't matter". (like what @skyking said)

Actually I believe Are these two statements equivalent? is the same question.

$\endgroup$
  • $\begingroup$ So I had to dig deeper to find out why $\neg\:p$, being false, made the compound proposition false when $q$ was false, in this case $p$ had to be true. $\endgroup$ – Diehardwalnut Sep 25 '15 at 6:00
  • $\begingroup$ @Diehardwalnut yes that's right. Otherwise it is not a fair comparison between the two statements. $\endgroup$ – dcstup Sep 25 '15 at 6:04

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.