I What's the converse of this statement?

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The discussion revolves around determining the converse of the statement "if A is an additive subgroup of the additive group B and G is B-magic then G is A-magic." Participants express confusion over the logical structure and implications of the statement, particularly regarding the role of the subgroup relationship as background knowledge. The proposed converse, which suggests that if G is A-magic then it must also be B-magic, is debated, with some arguing it lacks mathematical sense. Clarifications are made about the definitions of "converse" and "negation," emphasizing that the original statement's truth does not guarantee the truth of its converse. Ultimately, the conversation highlights the complexities of interpreting mathematical implications and the necessity of clear definitions.
  • #31
Adeimantus said:
If you use the IF-THEN construction, you change the order of P and Q to distinguish directions. If you use the IF and ONLY-IF constructions, you leave P and Q in the same order. In case this is a point of confusion, "P if Q" is not the same as "if P, then Q". Instead, it's actually the same as "if Q, then P". To recapitulate,

"if P, then Q" is the same as "P only if Q"
"if Q, then P" is the same as "P if Q"

Putting them together,

"(if Q, then P) AND (if P, then Q)" is the same as "P if and only if Q"

I hope this is clear. If not, you might google it. If you find that I have it backwards, show me, and I will gladly take back everything I said.
Yes, but allow me not to follow such a - in my view artificial - construction. To attach meaning to the order of where P and where Q has to stand would imply to make the entire wording even more context sensitive as it already is given by the language itself. I don't think it is actually used this way. I think it's the other way around.

I would search for an example, but before I put effort in it, since I still think your position has to be proven not mine, I like to ask you whom would you accept as an authority to decide this question?

We say "P is valid then and only then if Q is valid" so in this case there is actually a difference in language.

Edit: You may be right. I haven't found a direct example, as the authors of my books were smart enough to avoid a reference to "the if-part" or "to the only-if-part". They either use arrows, or simply recall the condition they start with. The closest I came on a quick search was the following proposition stated by Peter Hilton:
Statement: ## \mu## is monomorphic if and only of it is injective.
Proof: If ##\mu ## is injective then ... Conversely, suppose ##\mu## monomorphic

It doesn't finally answer the question, as I said, they avoid ambiguities, but it tends a little bit towards your interpretation.
 
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  • #32
Here's the first sentence of the Wikipedia article on "if and only if". https://en.wikipedia.org/wiki/If_and_only_if
Wikipedia said:
In logic and related fields such as mathematics and philosophy, if and only if (shortened iff) is a biconditional logical connective between statements.

In that it is biconditional (a statement of material equivalence),[1] the connective can be likened to the standard material conditional ("only if", equal to "if ... then") combined with its reverse ("if"); hence the name.
 
  • #33
Adeimantus said:
Here's the first sentence of the Wikipedia article on "if and only if". https://en.wikipedia.org/wiki/If_and_only_if
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.
 
  • #34
@Mark44 can you decide this? Post #33
 
  • #35
Adeimantus said:
The "con" in "converse" makes it sound like it should have some sense of disputing the original claim, or contradicting it.
The prefix would be "contra" to negate what it applies to, not "con".

Adeimantus said:
"if P, then Q" is the same as "P only if Q"
This is what the wikipedia page says, and I agree somewhat, but their explanation isn't as clear as it should be, because it neglects to mention that P and Q have to be equivalent statements. 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 \in A##, and let Q be the statement ##x \in B##.

"If P then Q" is the implication ##x \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. Here P and Q are not equivalent statements, being related to two sets, one of which is a proper subset of the other.

To my way of thinking "if and only if" implications apply only to equivalent sets. "P if and only if Q" can be resolved into "P if Q", which is the same as "If Q then P", and "P only if Q", which is the same as "If P then Q", provided that we're dealing with equivalent statements or sets that are identical.

BTW, their example of Madison's taste in fruits does more to confuse than to clarify.
 
  • #36
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)##
 
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  • #37
fresh_42 said:
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.

Mark44 said:
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.
 
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  • #38
TeethWhitener said:
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.
 
  • #39
Adeimantus said:
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.
 
  • #40
Mark44 said:
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.
 
  • #41
Adeimantus said:
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##.
 
  • #42
No kidding. Thanks guys, it was fun.
 

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