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

Materialist
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  • Almost everything rely on this, what do you mean ? (As it mean something is false or true) – mesel Jun 18 '14 at 05:21
  • I think this is almost a duplicate of http://math.stackexchange.com/questions/838184/contraposition-in-intuitionistic-logic . Please try not to divide questions in that way; it leads to fragmented answers that are hard for others to follow – Carl Mummert Jun 18 '14 at 10:38

3 Answers3

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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.

Peter Smith
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  • Your first proof seems to rely on $(A \vdash \bot) \vdash \lnot A$. What is the difference between this and the law of the excluded middle? It seems to be just another way of declaring that $A$ is a two valued variable. Also, Why would your logic have an inference $(A \vdash \bot) \vdash \lnot A$ but not have an inference $(\lnot B \vdash \bot) \vdash B$? Is there a philosophical distinction between the two? Thanks~ – DanielV Jun 18 '14 at 06:57
  • @DanielV: in intuitionistic logic, $\lnot A$ is simply an abbreviation for $A \to \bot$. But from this, if we have $\lnot B \vdash \bot$, we only get $\lnot\lnot B$, not $B$. – Carl Mummert Jun 18 '14 at 10:40
  • @CarlMummert That explains it, thanks. – DanielV Jun 18 '14 at 21:20
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It would seem that it does not, but rather, on the law of non-contradiction. Assume

  1. $A\implies B$
  2. $\neg B$

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.

crf
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  • I get that $\neg B \rightarrow A \rightarrow B$ means both $B$ and $\neg B$ being true, leading to violation of law of non-contradiction. But what is the ground that THAT leads to $\neg B \rightarrow \neg A$? I do get that the law of excluded middle only applies for logical truth and not implication, but I don't still see the complete ground for concluding this way. – Materialist Jun 18 '14 at 06:09
  • @Materialist as far as I know (I am not an expert), the law of the excluded middle is formally equivalent to negation elimination. If we can obtain a statement without using negation elimination (i.e. without invoking $\neg\neg P \iff P$), we have proven it without making use of the law of the excluded middle. Here, we begin with, say, $(A\to B) & \neg B$. If we can now show $\neg A$, then we've shown that $(A\to B)\to (B\to A)$. Now, assume $A$; we show that this leads to $B&\neg B$, which allows us to conclude $\neg A$ without making use of the law of the excluded middle. – crf Jun 18 '14 at 06:20
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    This surely only deals with one direction of the equivalence: the other direction is more problematic. See my answer. – Peter Smith Jun 18 '14 at 20:07
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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

  1. CCpqCCqrCpr
  2. CpCqp
  3. CCCpNppp
  4. CCNpNqCqp

that CCpqCNqNp follows.

Doug Spoonwood
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