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So, I often enjoy trying to prove "if and only if" statements by only using if and only if arguments.

i.e. RTP: $A \Leftrightarrow D$. Proof: $A \Leftrightarrow B \Leftrightarrow C \Leftrightarrow D$

My question is whether or not this is always possible? I'm aware that it's often easier to go one way then the other way, but is this ever the only way to do it?

(Let me know if I've not explained myself properly - found it quite a hard question to word!)

user37154
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    Not a full answer, but every time I've tried to do this (prove iff proofs using an iff argument) when its clearly easier to separate the directions, I have gotten small technical details wrong. For practicalities purpose I strongly suggest against trying to prove things this way! – DanZimm May 22 '14 at 20:50
  • I find this question interesting, although it may have a simple answer, I could imagine there being some crazy counterexample- I'm thinking of some sort of independence proof? – hmmmm May 22 '14 at 20:53
  • @DanZimm I often only have a quick go to see if I can find such a proof for that exact reason! I was just wondering whether, tricky or not, there had to be such a proof or if sometimes there simply wasn't. – user37154 May 22 '14 at 20:59
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    Try "a finite simple group has order $\le 60$ if and only if it's isomorphic to $A_5$." The two proof directions feel totally different to me, although I have no idea how I would prove that no "if and only if" proof exists, or even what that would mean exactly. – Qiaochu Yuan May 22 '14 at 23:28
  • I've seen proofs of the form that given conditions $X,Y,Z$, then $A\Rightarrow B\Rightarrow C\Rightarrow A$, which while not the same as what you're asking, seems to me a similar situation. – Pockets May 23 '14 at 08:25
  • In most cases, you would have to prove $A\implies D$ and then $D\implies A$. Or you could then prove the constrapositive, $\neg A \implies \neg D$. – Dan Christensen May 23 '14 at 17:35

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It is often that one of the directions is trivial, the other one hard.

If an "if and only if" argument is reasonably clean to obtain, it can be the nicest form of proof. But that isn't always easy to see how to do.

vonbrand
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  • Here's a concrete example I recently encountered on this site. $\Rightarrow$ is easy, but $\Leftarrow$ is more difficult: http://math.stackexchange.com/questions/799875/let-fx-in-fx-f-a-field-be-irreducible-and-let-alpha-be-a-root-of-f/799893 – Kaj Hansen May 22 '14 at 21:01
  • A rectangle can be exactly covered by squares (of arbitrary sizes) if and only if the ratio of the sides of the rectangle is a rational number. <= is quite trivial (if the ratio is p/q then the rectangle can be covered by p times q squares of equal size). – gnasher729 May 22 '14 at 21:59
  • There are solutions for the equation a^n + b^n = c^n with positive integers a, b and c <=> n = 1 or n = 2. – gnasher729 May 22 '14 at 22:00