Contraposition and law of excluded middle Does truth-equivalence of an $A \rightarrow B$ and contrapositive $\neg B \rightarrow \neg A$ rely on the law of excluded middle? 
 A: Consider the following proof:

$\quad A \to B\\
\quad\quad|\quad\neg B\\
\quad\quad|\quad\quad|\quad A\\
\quad\quad|\quad\quad|\quad B\\
\quad\quad|\quad\quad|\quad \bot\\
\quad\quad|\quad\neg A\\
\quad \neg B \to \neg A$

We assume $\neg B$ at the second line, then make new assumption $A$, and use modus ponens to get a contradiction. So the second assumption must be false. So the assumption $\neg B$ implies $\neg A$: which entitles us to conclude $\neg B \to \neg A$. That shows $A \to B$ implies $\neg B \to \neg A$, without any appeal to excluded middle or an equivalent.
But that is only half the job: what about the reverse implication? Let's start off trying to prove that:

$\quad \neg B \to \neg A\\
\quad\quad|\quad A\\
\quad\quad|\quad\quad|\quad \neg B\\
\quad\quad|\quad\quad|\quad \neg A\\
\quad\quad|\quad\quad|\quad \bot\\
\quad\quad|\quad\neg\neg B\\
\quad\quad|\quad B\\
\quad A \to B$

Now, that argument works in classical logic, but crucially we've used the rule that from $\neg\neg\varphi$ we can infer $\varphi$. And that rule is famously equivalent (on standard background assumptions) to the law of excluded middle.
OK that's just one attempt that fails without (an equivalent rule) to excluded middle. But indeed, you can't argue from $\neg B \to \neg A$ to $A \to B$ intuitionisitically, i.e. in the usual logic which drops excluded middle.
In summary, only one direction of the classical equivalence of $A \to B$ and $\neg B \to \neg A$ holds in the usual logic without excluded middle.
A: It would seem that it does not, but rather, on the law of non-contradiction. Assume 


*

*$A\implies B$

*$\neg B$ 


Now, to obtain a contradiction, suppose $A$. Then we have $B$ from (1). But we have $\neg B$ from (2), so by the law of non-contradiction, it follows that we have $\neg A$, and finally we have $\neg B \implies \neg A$.
The distinction between this and the law of the excluded middle is subtle but it's there. In order to prove via the law of excluded middle, we would knock out assumption (2), and instead assume $\neg (\neg B\implies \neg A)$, and from here derive a contradiction, yielding $\neg \neg (\neg B \implies\neg A)$, and by the law of the excluded middle, we would conclude that $\neg B \implies \neg A$.
For more on the distinction, see here.
A: No, it doesn't.  
In Lukasiewicz three-valued logic we have that Cab is truth-functionally equivalent to CNbNa and CNbNa is truth-functionally equivalent to Cba, but ApNp is not a tautology.  Here's the relevant three-valued truth-tables, where "f" is falsity, "n" indicates the third-truth value, and "t" truth.
C   f  n  t  N  A  f  n  t
f   t  t  t  t  f  f  n  t
n   n  t  t  n  n  n  n  t
t*  f  n  t  f  t  t  t  t

So, the truth tables for Cab and CNbNa we can write as follows:
a  b  Na  Nb  Cab  CNbNa
f  f  t   t   t    t
f  n  t   n   t    t
f  t  t   f   t    t
n  f  n   t   n    n
n  n  n   n   t    t
n  t  n   f   t    t
t  f  f   t   f    f
t  n  f   n   n    n
t  t  f   f   t    t

But, AnNn=Ann=n, and thus the law of the excluded middle is not a tautology.
One can also prove that from Wajsberg's basis under detachment and uniform substitution


*

*CCpqCCqrCpr

*CpCqp

*CCCpNppp

*CCNpNqCqp


that CCpqCNqNp follows.
