This question has already been asked here twice, namely here and here, but none of the answers address my specific question, except probably for this answer, which comes close.

So, using the notation of the close answer, I don't understand why I have to rule out the tuples $(T,T)$ and $(F,F)$.

Let's call the sentence "Q is necessary but not sufficient for P" R.

As for $(F,F)$, if P is false when Q is also false, this should result in R = true; since Q is necessary for P, so the absence of Q should imply the absence of P. Why would I want R to be false in this case?

And for $(T,T)$, I will imagine a more complete picture. Let's say that P depends on Q and some other factors, collectively named W. Now, we should split the row $(T,T)$ into 2, one with W false, and another with W true. In the case with W true, R should evaluate to T, and in the case with W false, R should evaluate to false. On what basis, then, should we decide to rule out $(T,T)$ in the original statement! In my opinion, the row with $(T,T)$ should be undecidable.

I would be grateful if someone could explain to my why the correct answer is $¬(¬r∧¬p)→¬q∧¬((¬r∧¬p)→q)$ in a way other that

"is necessary" translates to so and so and "is sufficient" translates to so and so, so the conjunction of the first with the negation of the second gives the correct answer.


  • $\begingroup$ The $p$, $q$, $r$ in that answer don't match the $P$, $Q$, $R$ in your question, so you should probably define them if you want an explanation for them. $\endgroup$ – Misha Lavrov Oct 31 at 17:51
  • $\begingroup$ @MishaLavrov You're right, sorry I was confused. Fixed now. Thanks. $\endgroup$ – user401445 Oct 31 at 17:55

The statement "$P$ is necessary for $Q$" means "in order to have $Q$, we must have $P$" or $Q \to P$, though we can also write the contrapositive $\neg P \to \neg Q$.

The statement "$P$ is sufficient for $Q$" means "if we have $P$, we definitely have $Q$" or $P \to Q$, though we can also write the contrapositive $\neg Q \to \neg P$.

So, the statement "$P$ is necessary but not sufficient for $Q$" can be written as $$(\neg P \to \neg Q) \land \neg (P \to Q).$$ In the example you've given, $Q = q$ while $P = \neg r \land \neg p$ due to specifics of the other question, and if we substitute those for $P$ and $Q$, we get the statement you're quoting.

If you look at the logical statement carefully, it turns out that $(\neg P \to \neg Q) \land \neg (P \to Q)$ is only true in one case: when $P$ is true, but $Q$ is false. Why is that? Because in order to observe $P$ not being sufficient for $Q$, $P$ has to happen, and $Q$ has to still fail to happen.

This doesn't match our intuitions for what "$P$ is necessary but not sufficient for $Q$" means. We want to say something like:

There are some cases where $P$ happens, and $Q$ does not, because $P$ is not sufficient for $Q$. However, in all cases where $Q$ happens, $P$ also happens: $P$ is necessary for $Q$.

To say things like this, the language of logical statements doesn't suffice! We have to have quantifiers for talking about "some cases" and "all cases".

Let $P(x)$ and $Q(x)$ denote "in case $x$, $P$ holds" and "in case $x$, $Q$ holds". Then:

  • "$P$ is necessary for $Q$" means $\forall x\, Q(x) \to P(x)$.
  • "$P$ is sufficient for $Q$" means $\forall x\, P(x) \to Q(x)$. Its negation simplifies to $\exists x\, P(x) \land \neg Q(x)$.

The statement "$P$ is necessary but not sufficient for $Q$" has the more sophisticated interpretation $$ (\forall x\, Q(x) \to P(x)) \land (\exists x\, P(x) \land \neg Q(x)). $$ That is: "In all cases $x$ where $Q(x)$ holds, $P(x)$ also holds. However, there is a case where $P(x)$ holds, but $Q(x)$ does not".

A bare statement like $P(x) \land Q(x)$ is neither true nor false, because $x$ isn't quantified. The universal statement $\forall x\, P(x) \land Q(x)$ is false (because otherwise, $P$ would be sufficient for $Q$), but it's possible that $\exists x\,P(x) \land Q(x)$ is true. This is what you want to say when you say "In my opinion, the row with $(T,T)$ should be undecidable", but that's not a thing we can talk about without quantifiers: without quantifiers, rows aren't allowed to be undecidable.

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  • $\begingroup$ Thanks for the convincing answer. One interesting thing I’d like to point out, however, is stressing “the ability to observe P not being sufficient for Q”. I thought, while reading, that this is the exact opposite of what we do when we define the implication relationship. I mean, if x → y, and x is false whereas y is true, we say that x → y is true although “we didn’t observe x implying y”. Of course, I know that this is circular thinking because we would then want to know what “implying” means to judge if we have observed it or not, but this is what we were doing to begin with. (1) $\endgroup$ – user401445 Oct 31 at 19:49
  • $\begingroup$ What I want to say is for anyone who might have had the same thoughts. From en.wikipedia.org/wiki/… (Credit goes to Brian for pointing out the formal name “material implication”): the equivalence in the case of false P is only conventional, and hence the formal proof of equivalence is only partial. Other than those “axioms”, we have to see explicit evidence to judge truth values. Please correct me if I’m wrong. (2) $\endgroup$ – user401445 Oct 31 at 19:49

The problem here is that you’re treating material implication, symbolized by the logical connective $\to$, as if it were the everyday notion of implication; it isn’t.

$Q$ is sufficient for $P$’ means precisely that if $Q$ is true, then $P$ must be true; we formalize this as $Q\to P$. If $P$ and $Q$ are both false, this is vacuously true, since it says nothing about the truth of $P$ when $Q$ is false. That is simply how material implication works. Thus, $R$ must be false when $P$ and $Q$ are both false, since in that case $Q$ is sufficient for $P$.

The implication is also true when both $P$ and $Q$ are true, simply because $P$ is true; again, this is just how material implication works. In intuitive terms, the implication $Q\to P$ has the truth value $F$ only when its falsity can actually be demonstrated from the truth values of $P$ and $Q$, and that is the case only when $Q$ is true and $P$ is false. In all other cases its truth value is $T$. Thus, $R$ must also be false when $P$ and $Q$ are both true.

To say the same thing in slightly different words, neither $(F,F)$ nor $(T,T)$ rules out the possibility that $Q$ is sufficient for $P$, in the one case because we don’t have $Q$, and in the other case because we do have $P$. The only case that actually rules out the possibility that $Q$ is sufficient for $P$ is $(F,T)$. If we assert, as $R$ does, that $Q$ is not sufficient for $P$, we are ruling out the possibility that $Q$ is sufficient for $P$, so we are asserting that $P$ is false and $Q$ is true.

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