Drinker Paradox - Proofs of The Paradox

Proofs of The Paradox

The proof begins by recognizing it is true that either everyone in the pub is drinking, or at least one person in the pub isn't drinking. Consequently, there are two cases to consider:

1. Suppose everyone is drinking. For any particular person, it can't be wrong to say that if that particular person is drinking, then everyone in the pub is drinking — because everyone is drinking. Because everyone is drinking, then that one person must drink because when ' that person ' drinks ' everybody ' drinks, everybody includes that person.
2. Suppose that at least one person is not drinking. For any particular nondrinking person, it still cannot be wrong to say that if that particular person is drinking, then everyone in the pub is drinking — because that person is, in fact, not drinking. In this case the condition is false, so the statement is vacuously true due to the nature of material implication in formal logic, which states that "If P, then Q" is always true if P (the condition or antecedent) is false.

Either way, there is someone in the pub such that, if he is drinking, everyone in the pub is drinking. A slightly more formal way of expressing the above is to say that if everybody drinks then anyone can be the witness for the validity of the theorem. And if someone doesn't drink, then that particular non-drinking individual can be the witness to the theorem's validity.

The proof above is essentially model-theoretic (can be formalized as such). A purely syntactic proof is possible and can even be mechanized (in Otter for example), but only for an equisatisfiable rather than a equivalent negation of the theorem. Namely, the negation of the theorem is

which is equivalent with the prenex normal form

By Skolemization the above is equisatisfiable with

The resolution of the two clauses and results in an empty set of clauses (i.e. a contradiction), thus proving the negation of the theorem is unsatisfiable. The resolution is slightly non-straightforward because it involves a search based on Herbrand's theorem for ground instances that are propositionally unsatisfiable. The bound variable x is first instantiated with a constant d (making use of the assumption that the domain is non-empty), resulting in the Herbrand universe:

One can sketch the following natural deduction:

Or spelled out:

1. Instantiating x with d yields which implies
2. x is then instantiated with f(d) yielding which implies .

Observe that and unify syntactically in their predicate arguments. An (automated) search thus finishes in two steps:

The proof by resolution given here uses the law of excluded middle, the axiom of choice, and non-emptiness of the domain as premises.