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From Wikipedia, we have:

In natural languages, an indicative conditional is the logical operation given by statements of the form "If A then B". Unlike the material conditional, an indicative conditional does not have a stipulated definition. The philosophical literature on this operation is broad, and no clear consensus has been reached.

http://en.wikipedia.org/wiki/Indicative_conditional

Is it not just a case of disallowing certain rules of inference in propositional logic, e.g. disallowing $\neg A \to A \implies B$ or some other rule(s) of inference that allows you do make such an inference?

I routinely use such rules (as above) in writing formal proofs. Would mathematics as we know it even be possible if we implemented such restrictions?

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It's not that simple. For starters, $\neg A\to A\implies B$ isn't a rule of inference in standard propositional logic. You may be thinking of $\neg A \to A\implies A$.

[EDIT after below exchange of comments: I thought the OP meant $\neg A \to A\vdash B$, but in fact the OP meant $\neg A \vdash A\to B$, which indeed is a rule.]

But also, it's not just a matter of disallowing certain rules. One also has to consider whether certain other rules should be allowed that aren't supported by the material conditional. For example, are $\neg(A\to B)$ and $A\to \neg B$ equivalent? Maybe; note that $1-P(B|A) = P(\neg B|A)$. But the material conditional doesn't support that equivalence.

Finally, there's the whole problem of the semantics for indicative conditionals. Two-valued truth-functional logic simply won't work. Should the logic be three-valued? Four-valued? Modal? No consensus exists.

I guess math as currently done would have to be pretty thoroughly reexamined if one were to replace the familiar (material) construal of if-then with something else. But to say for sure, one would have to have a definite "something else" proposal in hand.

For more on this, see also http://plato.stanford.edu/entries/conditionals/.

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  • $\begingroup$ People are always asking, how can we infer $A\implies B$ from $\neg A$ regardless of what $B$ may be? That's why I made above above suggestion. $\endgroup$ – Dan Christensen May 26 '14 at 4:56
  • $\begingroup$ Oh, we're reading the arrows differently. I thought you were describing a rule according to which "If not-A, then A" would imply B. But you meant a rule according to which not-A implies "If A, then B." Yes, that's a rule. We should probably be using the turnstile or the three-dot "therefore" symbol, but I haven't figured out the LaTeX codes for those yet. $\endgroup$ – StumpyLeg May 26 '14 at 5:39
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The indicative conditional describes an ordering of the truth values of statements: B is not less true than A. The various paradoxes and problems of the material conditional become perfectly reasonable with this interpretation, and no more mysterious than claiming that A <= B.

For classical two valued logic, the material conditional happens to describe the same relationship, and it is perfectly adequate for the job. But this success is misleading and even tends to obscure understanding of it.

In three valued logic, it turns out that doubts about the adequacy of the material conditional are well justified. It does not have the same useful, valuable properties in the three valued case that it does in the two valued case: and a conditional corresponding to an ordering of truth values has much better correspondence with the indicative conditional.

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  • $\begingroup$ The Drinker's paradox is true whether or not anyone at all in the pub is drinking, because a conditional with a false antecedent is technically true. It is also vacuous and gives no information. It would be a different matter if we could assume also that there is at least one person who is drinking (which there is a strong temptation to do anyway), but that would be a different, contingent statement and no longer a paradox. $\endgroup$ – Confutus May 28 '14 at 20:09
  • $\begingroup$ In the Drinker's Paradox, no one has to be drinking, but there has to be someone in the pub (i.e. a non-empty universe). $\endgroup$ – Dan Christensen May 28 '14 at 22:17

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