# Can you do without conditional proof if you have indirect proof? **URL:** \ **Category:** Factual Questions\ **Created:** [October 5, 2012, 4:17pm UTC](https://boards.straightdope.com/t/can-you-do-without-conditional-proof-if-you-have-indirect-proof/636820 "2012-10-05T16:17:28Z")\ **Posts on this page:** 3\ **Page:** 3
**Author:** ![Frylock](https://avatars.discourse-cdn.com/v4/letter/f/ce7236/32.png) [@Frylock](https://boards.straightdope.com/u/Frylock)\ **Post date:** [October 9, 2012, 6:51pm UTC](https://boards.straightdope.com/t/can-you-do-without-conditional-proof-if-you-have-indirect-proof/636820/41 "2012-10-09T18:51:31Z")
> [@chrisk](#): > > Okay, then, consider this… > > A: All the barbers in this town are Chinese. > > B: You’re lying. I’ve met everybody in town, and I didn’t see any Chinese barbers. > > A: Did you see any barbers at all? > > B: Well, no, I guess not. But if there aren’t any barbers, then you can’t say anything about them. > > A: No, it’s true. All the barbers of this town are Chinese, and what’s more, they’re all missing a finger. > Do you agree with B, or can you see from A’s (slightly whimsical) point of view? Speaking logically, ‘all of group A can be described by B’ is true if group A is empty - just like the empty set is a subset of every other set. That’s similar to a demonstration I give to students that when there are no vases on the table, all the vases on the table are red. (I ask them to bring me all the vases, then I ask them what proportion of the vases they brought me were red…) But the fact that the members of an empty set satisfy all predicates doesn’t seem to show that there is a 0-ary case of & and that it is the same thing as “true.” For one thing, why believe there is a 0-ary case of &? And then, if there is, why not say it’s the same thing as “false” since when there are no conjuncts, all of the conjuncts are false? But having said that, I do see how the reasoning is supposed to work–and you and **Indistinguishable** have made it clear how it’s useful to do things the way you’re describing. ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [October 9, 2012, 10:38pm UTC](https://boards.straightdope.com/t/can-you-do-without-conditional-proof-if-you-have-indirect-proof/636820/42 "2012-10-09T22:38:33Z")
> [@Indistinguishable](#): > > Think of A v B v C v D v … as meaning “There exists a conjunct which is true”. If there are no conjuncts, this is vacuously false. Er, replace “conjunct” with “disjunct” throughout, of course… ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [October 9, 2012, 11:05pm UTC](https://boards.straightdope.com/t/can-you-do-without-conditional-proof-if-you-have-indirect-proof/636820/43 "2012-10-09T23:05:31Z")
> [@Frylock](#): > > But the fact that the members of an empty set satisfy all predicates doesn’t seem to show that there is a 0-ary case of & and that it is the same thing as “true.” For one thing, why believe there is a 0-ary case of &? And then, if there is, why not say it’s the same thing as “false” since when there are no conjuncts, all of the conjuncts are false? I know you understand already, but just to clarify this particular argument: The reason is that multi-ary conjunction is a form of universal quantification (“For all X in the set of conjuncts, X”), and universal quantification over an empty domain is true, not false. Similarly, multi-ary disjunction is a form of existential quantification (“There exists an X in the set of disjuncts such that X”), and existential quantification over an empty domain is false, not true. The point isn’t “All the conjuncts/disjuncts have the same truth value”; the point is that it’s just quantification, and we already know how quantification over an empty domain is most cleanly defined, as you’ve been teaching your students. > [@Frylock](#): > > That’s similar to a demonstration I give to students that when there are no vases on the table, all the vases on the table are red. (I ask them to bring me all the vases, then I ask them what proportion of the vases they brought me were red…) Of course, it would be equally correct to say 100%, 0%, or 22% of the vases are red, which may be confusing for students who are used to identifying “0% [or any proportion less than 100%] are red” with “Not all of them are red”]. But I imagine you deal with this potential confusion by a similar sort of explanation to what was just noted above. [Previous page](https://boards.straightdope.com/t/can-you-do-without-conditional-proof-if-you-have-indirect-proof/636820.md?page=2)