• If $p\implies q$ ("$p$ implies $q$"), then $p$ is a sufficient condition for $q$.

  • If $\lnot p\implies \lnot q$ ("not $p$ implies not $q$"), then $p$ is a necessary condition for $q$.

I don't understand what sufficient and necessary mean in this case. How do you know which one is necessary and which one is sufficient?

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    $\begingroup$ Those are the definitions of necessary and of sufficient. $\endgroup$ – Zhen Lin Dec 11 '12 at 9:37
  • $\begingroup$ Are you aware that $\bar p \implies \bar q$ is the logical equivalent of $q \implies p$ ? Do you know what "necessary" and "sufficient" mean in the English language? $\endgroup$ – DanielV Nov 11 '14 at 8:55

Suppose first that $p$ implies $q$. Then knowing that $p$ is true is sufficient (i.e., enough evidence) for you to conclude that $q$ is true. It’s possible that $q$ could be true even if $p$ weren’t, but having $p$ true ensures that $q$ is also true.

Now suppose that $\text{not-}p$ implies $\text{not-}q$. If you know that $p$ is false, i.e., that $\text{not-}p$ is true, then you know that $\text{not-}q$ is true, i.e., that $q$ is false. Thus, in order for $q$ to be true, $p$ must be true: without that, you automatically get that $q$ is false. In other words, in order for $q$ to be true, it’s necessary that $p$ be true; you can’t have $q$ true while $p$ is false.

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    $\begingroup$ As usual, I will point out that a downvote not accompanied by an explanatory comment is less than useful. $\endgroup$ – Brian M. Scott Dec 11 '12 at 12:46
  • $\begingroup$ how do you know that p is true or not? $\endgroup$ – cloud9resident Dec 12 '12 at 8:44
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    $\begingroup$ @cloud9resident: You don’t know. Whether $p$ is true or not is a completely separate issue from whether it is a necessary or sufficient condition for $q$. The latter does not depend in any way on the truth or falsity of $p$. $\endgroup$ – Brian M. Scott Dec 12 '12 at 8:46
  • $\begingroup$ But for not p implies not q, in terms of that reasoning, not q doesn't tell you anything about not p because here are different paths to not q? $\endgroup$ – committedandroider Jan 25 '15 at 22:32
  • $\begingroup$ @committedandroider: That’s correct. Knowing that $\neg q$ is true tells you nothing about whether $\neg p$ is true or not. $\endgroup$ – Brian M. Scott Jan 25 '15 at 22:34

I always think of it in terms of sets.

enter image description here

In the picture above, for an element to be purple, it's necessary to be red, but it is not sufficient.

The same holds for the blue set, to be in the blue set is a necessary condition in order to be purple, but it is not enough, it's not sufficient.

A sufficient condition is stronger than a necessary condition. If you tell me that you have a red or blue element I can't say for sure if it is in the purple set, but if you tell me that you have a purple element I now for sure that it is in the red and blue sets.

  • $\begingroup$ But purple is neither red nor blue… $\endgroup$ – Geremia Aug 30 '18 at 15:31

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