I will use "*p*→*q*" for the indicative conditional "if *p*, then *q*". I will use "*p*⊃*q*" for the material conditional "(not *p*) or *q*". I will say that "indicatives are material" providing that *p*→*q* and *p*⊃*q* are logically equivalent for all *p* and *q*, where *a* and *b* are logically equivalent if and only if it is necessary that (*a* if and only if *b*). I will say that *p* entails *q* provided that it is necessary that *p*⊃*q*.

Almost no philosopher thinks indicatives are material. There are very plausible counterexamples. For instance, suppose it is lightly raining in Seattle and Seattle is not having a drought. Let *p* be "Seattle is having heavy rain" and let *q* be "Seattle is having a drought". Then *p*⊃*q*, since *p* is false. But it seems quite wrong to say that if Seattle is having heavy rain, then Seattle is having a drought, so *p*→*q* doesn't seem to be true.

I am going to offer some arguments that indicatives are material. Say that → is non-hyperintensional provided *p*→*q* and *p**→*q** are logically equivalent whenever *p* and *p** are logically equivalent and *q* and *q** are logically equivalent. Consider the following two theses:

- For any possible world
*w*: (*p*at*w*) → (*q*at*w*) if and only if (*p*→*q*at w). - For any predicates
*F*and*G*, from "Every*F*is a*G*" (where "*x*is an*F*" is more euphonious way of saying that*x*satisfies*F*) together with the assumption that*c*exists, it logically follows that if*c*is an*F*, then*c*is a*G*.

- If (1) is true and → is non-hyperintensional, then indicatives are material.
- If (2) is true and → is non-hyperintensional, then indicatives are material.

- If (2) is true, then one has to assign the same truth value as the material conditional does to a number of paradoxical-sounding examples of indicative conditional sentences that are relevantly just like the standard alleged counterexamples to the thesis that all indicatives are material.

**Argument for (5):** Take my heavy rain and drought in Seattle case. Suppose that as it happens, there is no place where there presently is heavy rain. Let *F**x* say that *x* is having heavy rain. Let *G**x* say that *x* is having drought. Then all *F*s are *G*s. (If you think, with Aristotle, that "All *F*s are *G*s" requires there to be an *F*, then add the premise that on Venus somewhere right now there is a drought but a very, very brief heavy rain is currently occurring. I will leave out such modifications in the future.) Then by (2), we have to say that if *F*(Seattle), then *G*(Seattle):

- If Seattle is having heavy rain, then Seattle is having drought.

We can also use (2) to manufacture a true-antecedent, true-consequent case. Suppose that it is raining in both Seattle and the Sahara. Then the following is a standard alleged counterexample of the true-antecedent, true-consequent sort:

- If it's raining in Seattle, then it's raining in the Sahara.

*F*

*x*say that

*x*is a planet on which it is raining in Seattle, and let

*G*

*x*say that

*x*is a planet on which it is raining in the Sahara. Then every

*F*is a

*G*, since the only

*F*is earth. By (2):

- If earth is a planet on which it is raining in Seattle, then earth is a planet on which it is raining in the Sahara.

**Argument for (3):** First we need a special case:

- If
*p*and*q*are non-contingent and → is non-hyperintensional, then*p*→*q*is logically equivalent to*p*⊃*q*.

- 2+2=4→2+3=5. (necessary, necessary)
- 2+2=5→2+3=6. (impossible, impossible)
- 2+2=5→ (2+2=5 or 1+1=2 or both). (impossible, necessary)
- 2+2=4→2+2=5. (necessary, impossible)

*p*and

*q*are non-contingent, then they are respectively logically equivalent to the antecedent and consequent of exactly one of (10)-(13). By non-hyperintensionality of →, it follows

*p*→

*q*must have the same truth value as the conditional in that line. But the truth values of (10)-(13) are just as the material conditional says they are: thus, clearly, (10)-(12) are (necessarily) true and (13) is (necessarily) false. So,

*p*→

*q*must have the same truth value as

*p*⊃

*q*, assuming

*p*and

*q*are non-contingent.

The argument for (3) is now easy. Observe that (*p* at *w*) and (*q* at *w*) are non-contingent, even if *p* and *q* are contingent. So,

- (
*p*at*w*) → (*q*at*w*) is logically equivalent to (*p*at*w*) ⊃ (*q*at*w*).

- (
*p*⊃*q*at*w*) is logically equivalent to (*p*at*w*) ⊃ (*q*at*w*).

- (
*p*→*q*at*w*) is logically equivalent to (*p*⊃*q*at*w*)

**Argument for (4):** The most intuitive form of the argument is to assume theism, and let *F**x* say that *x* is an omniscient being that knows that *p*, and let *G**x* say that *x* knows that *q*. Then as long as *p*⊃*q*, it will be the case that every *F* is a *G* (just think about the four possible truth-value combinations). Hence:

- If God is an omniscient being that knows that
*p*, then God knows that*q*.

*p*and

*q*, respectively, and so we get

*p*→

*q*by non-hyperintensionality.

If we don't want to suppose there is a God, let's suppose that numbers and sets exist necessarily. Let *P* be the singleton set whose only member is *p*. Let *Q* be the singleton set whose only members is *q*. Then, let *F**x* say that *x* is greater than zero and *x* equals the number of truths in *P*. Let *G**x* say that *x* is greater than zero and *x* equals the number of truths in *Q*. Then, if *p*⊃*q*, it is easy to see that all *F*s are *G*s, so:

- If one is greater than zero and one equals the number of truths in
*P*, then one is greater than zero and one equals the number of truths in*Q*.

*p*and

*q*respectively, so by non-hyperintensionality we get

*p*→

*q*, once again.

## 1 comment:

It’s been a while since I looked at this literature, but I’m not sure your sociological claim is true. You may have more company than you think.

On the argument for (5):

The most serious competitor view to the material indicatives view, I think, is that “If p then q” expresses the view that P(q|p) is high. Then (2) looks like this: Given that Every F is a G, and that c exists, it follows that P(c is a G|c is an F) is high.

In one sense yes, in another sense no. In one sense, the probability is 1. But one might want to say that the “givens” have to be incorporated into the probability measure: so that what follows is P(c is a G | c is an F & Every F is a G & c exists) is high; but it does not follow that P(c is a G | c is an F) is high absolutely. Or maybe one might want to resist the idea of “the” P() function, making it context-relative somehow, in which case the competing view of conditionals will not assign them a single meaning.

Maybe there is a standard way to think about this; I don’t know enough about probability to say.

Still thinking about the other arguments.

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