# How to decompose A if and only if B?

I've read that the formula A if and only if B is a formulation of the equivalence that can be decomposed in A if B (the implication B -> A) and in A only if B (the implication A -> B).

I don't agree with the second part : for me it looks like that A only if B is the translation of ¬B -> ¬A (I know that it's the contraposition of A -> B and consequently it means the same but it's not the same formulation).

So can anyone tell me the correct interpretation please.

thanks,

Bruno

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"if A then B" and "if B then A" are two completely different beasts: witness the difference between "if it's nighttime, then the sky will go dark" and "if the sky will go dark, then it's nighttime". –  Ｊ. Ｍ. Dec 9 '10 at 20:08
You seem to understand what "if and only if" means. Please clarify what you are asking for. –  Rasmus Dec 9 '10 at 20:43
I am not quite sure if your question is one about language or one about logic. If it is about logic, the following may interest you: en.wikipedia.org/wiki/… –  Willie Wong Dec 9 '10 at 20:49
I would be inclined toward a hybrid of the answers thus far: "A if and only if B" means $A\Leftrightarrow B$, which means $A\implies B \wedge B\implies A$. –  Isaac Dec 9 '10 at 20:51
Thank you Ross for your commitment :-) and thank's everyone for answering my intricate question ! Thank you for the interesting link Willie. –  Bruno Dec 9 '10 at 21:31

If I understand your question correctly, you are bothered by the semantics of the statement "$A$ only if $B$". You can take this statement as idiomatic, that is, if taken literally it has one meaning, but social convention (in this case the social convention of the mathematics community) leads us to interpret it as always meaning $A \rightarrow B$. There is certainly room in English to take certain phrases as being somewhat illogical. My point is that "$A$ only if $B$" doesn't actually have to stand up to logical dissection of the semantics. Extracting meaning from English sentences is not linear.

However, I always look at that the statement "$A$ only if $B$" as meaning A is true only if B is true. If we look at a truth table for a statement where A is true only if B is true, then we see that this is equivalent to the statement $A\rightarrow B$.

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you need not to have a space after \ to get the symbols. –  Ross Millikan Dec 9 '10 at 21:08

For an if an only if (iff) theorem, you have $A \Leftrightarrow B$. This means that A only occurs when B occurs and B only occurs when A occurs. If you are trying to prove an iff theorem, you must prove it in both directions. That is, you must show that $A \Rightarrow B$ and then show that $B \Rightarrow B$.

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I would expand "A only if B" to "A can only be true if B is also true", then go to "if A then B" and represent that as $A \implies B$. As you say, that is English processing, not logic processing.

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So you go from "A can only be true if B is also true" to "if A then B" ? Isn't a step misssing ? Or you may be right and that's because of the language and not the mathematics. –  Bruno Dec 9 '10 at 20:27
And isn't this step : If B is not true then A isn't neither ? –  Bruno Dec 9 '10 at 20:29
I think when you go from "A can only be true if B is also true" to " If B is not true then A isn't neither" you are taking the contrapositive, which as you say is valid, but is not getting you to logical symbols as directly as possible. –  Ross Millikan Dec 9 '10 at 20:36

In classical logic, $B\rightarrow A$ is equivalent to $\lnot A \rightarrow \lnot B$, so you will simply use the simplest definition of $A\leftrightarrow B$, which is to use $B\rightarrow A$. In intuitionistic logic, the contraposition is not generally derivable. You have $(B\rightarrow A)\rightarrow(\lnot A\rightarrow \lnot B)$, but not $(\lnot A\rightarrow \lnot B)\rightarrow(B\rightarrow A)$. Hence, your definition of equivalence is weaker - and doesnt really make sense, it only makes sense if you assume $A\vee \lnot A$.

And if you look at it computationally, $A\leftrightarrow B$ means you have terms $u^{A\rightarrow B}$, $v^{B\rightarrow A}$, such that if you have an object $a^A$ you always get an object $b^B$ and vice versa. With your definition, you would have $u^{A\rightarrow B}$, but $v^{(A\rightarrow\bot)\rightarrow(B\rightarrow\bot)}$ where $\lnot A = A\rightarrow\bot$ and $\bot$ can be defined as $0=1$ (or any other type of which no term is derivable), hence, $v$ maps terms $t^{A\rightarrow\bot}$ that map terms of type $A$ on something not derivable onto terms $u^{B\rightarrow\bot}$ mapping terms of type $B$ on something not derivable.

This isnt what we usually want from equivalence.

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Thanks for your explanation ! –  Bruno Dec 12 '10 at 21:49