# The Definition of the Indicative Conditional

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?

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.

• 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. – Dan Christensen May 26 '14 at 4:56