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Suppose, for the sake of discussion, that I am given that the following implication is true: $$q

Here is my question:

Does it follow (by Exportation) that the implication $$\sigma(q)<\sigma(n) \implies q

It is known that $$q

I guess it all boils down as to whether $$qExportation.

MY ATTEMPT

By Exportation, we get that $$q

IS THIS PROOF CORRECT?

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    Are you aware that $a \iff b \iff c$ isn't the same thing as $(a \iff b) \land (b \iff c)$ ? – 2017-01-10
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    @DanielV, thank you for your comment. Yes, I am aware of that. Perhaps I was not too accurate with my question. I meant to have the biconditional $$q – 2017-01-10
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    @DanielV, isn't that biconditional (in spelled-out implication form) the same as $$q – 2017-01-10
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    Are you aware that $a \implies b \implies c \implies d$ is not the same as $(a \implies b) \land (b \implies c) \land (c \implies d)$, but it is actually the same as $(a \land b \land c) \implies d$ ? – 2017-01-10
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    Not too familiar with it, but it seems the rule you are trying to invoke is $(a \iff b) \land (b \iff c) \land (c \iff d)$ is equal to $(a \implies b) \land (b \implies c) \land (c \implies d) \land (d \implies a)$. – 2017-01-10
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    Also, unless the actual definitions of $\sigma$, $q$, $<$, etc are relevant to the proof, you could clean the whole thing up by replacing $q < n$ with $A$, $\sigma(q) < \sigma(n)$ with $B$, and $\sigma(q)/n < \sigma(n)/q$ with $C$. Basically, if $\sigma$ is relevant, then define it, if it isn't relevant, then leave it out. – 2017-01-10

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Your proof cannot be correct, because the result does not follow.

For a counterexample, take $q=\sigma(q) = n =1$, and $\sigma(n)=2$. Then $q

Of course, this only tells us that your proof is not correct, and not what is wrong with your proof.

One thing wrong with your proof is that $P \leftrightarrow Q \leftrightarrow R$ is not equivalent to $P \rightarrow Q \rightarrow R \rightarrow Q \rightarrow P$: First of all, the conditional is not associative, so you need parentheses to indicate what that second statement even means, but going by the rest of your proof, you must mean $P \rightarrow (Q \rightarrow (R \rightarrow (Q\rightarrow P)))$. But that is a tautology, and hence not equivalent to $P \leftrightarrow Q \leftrightarrow R$, which clearly is not a tautology.