What's the converse of this statement?

[TeethWhitener]

In general, for a statement $p\rightarrow q$, the accepted definitions are (or were when I was studying logic):
Converse: $q\rightarrow p$
Inverse: $\neg p \rightarrow \neg q$
Contrapositive: $\neg q \rightarrow \neg p$.

Edit: to actually answer OP's question: the converse of
$(a\wedge b)\rightarrow c$
is
$c \rightarrow (a\wedge b)$

 

[Adeimantus]

Yes, I do not doubt the linguistic, but what is which direction?

Let me summarize our question: $P$ if and only if $Q$.

Me: if case: $P \Rightarrow Q$ ; only-if case: $Q \Rightarrow P$
You: only-if case: $P \Rightarrow Q$ ; if case: $Q \Rightarrow P$

Is that correct? Not that we mean the same after all. Have you read my edit in the previous post? I guess those authors (I checked two native speakers, a British and an American) did well to avoid the entire situation by clearly repeating what they assume, resp. use $\Rightarrow$ or $\Leftarrow$ which seems more and more to be a good advice.

Yes, that is correct. And yes, I read your edit. It is absolutely the case that almost nobody ever actually bothers to dissect the meaning "if and only if". Although I am not a mathematician, I would imagine that many professional mathematicians don't bother about it, because it's not that important. They simply learn by repetition and association that this "iff" is what you have proved when you show that the implication goes both ways. It doesn't really matter much which way is which. They don't need to know exactly what they are saying when they say "iff". Just how to use it correctly.That being said, I'm pretty sure I know what the words mean, even though I am not a mathematician. But I don't care enough about it to argue for more than a couple of days.

Consider the example they give in the section on Euler diagrams. In the first diagram of this section, where A is a proper subset of B, we have A = {1, 9, 11} and B = A ∪ {4, 8}.

Let P be the statement x∈Ax∈Ax \in A, and let Q be the statement x∈Bx∈Bx \in B.

"If P then Q" is the implication x∈A⇒x∈Bx∈A⇒x∈Bx \in A \Rightarrow x \in B, which is clearly true. Every element of A is also an element of B.
The statement "P only if Q" is not true for all elements of B. As a counter example, let x = 4, an element of set B that is not also an element of set A.

x=4 is not a counter example. The "only if" just means that membership in B is a necessary condition for membership in A. The fact that it is in B is no guarantee that is in A. In this case, it is not in A. You are reading "only if" as if it meant "if and only if". It is hard to avoid this because in everyday language, the word "if" gets used for necessity, sufficiency, and causality depending on the context. Here, I am restricting myself to the technical meaning of "only if". It is another way of phrasing a necessary condition, and no more.

So when math people prove an if-and-only-if statement, they typically prove the "only-if" part first, whether they realize it or not. Since they are doing both directions anyway, it doesn't matter.

 

[Adeimantus]

In general, for a statement $p\rightarrow q$, the accepted definitions are (or were when I was studying logic):
Converse: $q\rightarrow p$
Inverse: $\neg p \rightarrow \neg q$
Contrapositive: $\neg q \rightarrow \neg p$.

Edit: to actually answer OP's question: the converse of
$(a\wedge b)\rightarrow c$
is
$c \rightarrow (a\wedge b)$

At last. Thank you.

 

[Mark44]

x=4 is not a counter example. The "only if" just means that membership in B is a necessary condition for membership in A. The fact that it is in B is no guarantee that is in A. In this case, it is not in A. You are reading "only if" as if it meant "if and only if".

I don't think I am. I showed that the implication $P \Rightarrow Q$ is true in the paragraph before the one you're talking about.

If you don't like "P only if Q" rephrase it as $Q \Rightarrow P$, which is false as my counterexample of x = 4 shows.

 

[Adeimantus]

I don't think I am. I showed that the implication $P \Rightarrow Q$ is true in the paragraph before the one you're talking about.

If you don't like "P only if Q" rephrase it as $Q \Rightarrow P$

Why would I do that? I've said over and over that it is the same thing as P->Q.

 

[Mark44]

Why would I do that? I've said over and over that it is the same thing as P->Q.

$Q \Rightarrow P$ is not the same thing as $P \Rightarrow Q$.

 

[Adeimantus]

No kidding. Thanks guys, it was fun.