The following sentence is known as the "drinker's paradox": "There is someone in the pub, so if he or she is drinking, everyone in the pub is drinking." It doesn't matter if that one person encourages others to drink or some other back door exists, as we'll see in a moment. It is a nice example from mathematical first order logic.
We start by stating that either everyone in the pub drinks or at least one person in the pub does not drink. The following case distinction is therefore appropriate:
- Everyone drinks. Then if someone is drinking in the pub, everyone in the pub is drinking - because everyone is drinking.
- At least one person doesn't drink. For any non-drinking person, it is true that if they drink, everyone in the pub drinks - since the person does not drink ( \(A \Rightarrow B\) is always true when \(A\) is false).
Formally, the theorem can be formalized as follows for any predicate \(D\) and a non-empty set \(P\):
$$\exists x\in P.\ [D(x) \Rightarrow \forall y\in P.\ D(y)]$$