The following sentence is known as the "drinker's paradox": "There is someone in the bar, so if he or she drinks, everyone in the bar drinks." It doesn't matter whether that one person encourages others to drink or whether another back door exists, as we'll see in a moment. It is a nice example from mathematical predicate 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 drinks in the bar, everyone in the bar drinks - because everyone drinks.
- At least one person doesn't drink. For any non-drinking person, it is true that if they drink, everyone in the pub is drinking - since the person is not drinking ( \(A \Rightarrow B\) is always true when \(A\) is false).
Formally, for an arbitrary predicate \(D\) and a non-empty set \(P\) the sentence can be represented as follows:
$$\exists x\in P.\ [D(x) \Rightarrow \forall y\in P.\ D(y)]$$