Here’s a tautology in propositional logic:

⊨(P → Q) ∨ (Q → R)

Try throwing that into English. Here’s a reading using some propositions I just came up with:

“I’ll die if I’m immortal, or I’ll live if I die.”

Obviously, neither of those are the case. But this formula, (P → Q) ∨ (Q → R), is both provable and self-implied in classical propositional logic.

Here’s a syntactic proof by means of natural deduction using some basic rules of inference:

Ok, there are definitely shorter proofs, especially if you let other equivalencies into your rules of inference, but this spells out quite a bit and gets the job done.

So, ⊢ (P → Q) ∨ (Q → R)

The reason this is provable, despite being formally contingent in English, is that the material conditional, →, does not represent the English “if/then”. If you’ve taken a course in logic, you’ve probably heard this. The reason is that the truth conditions for the material conditional are a little different. Specifically, a formula of the form (α → β) is true on every valuation except those that make α true and β false.

Look at our formula again. It states, semantically, that either (P → Q) is true, *or *(Q → R) is. If (P → Q) happens to be true, then the whole formula is true, so that case is covered right off the bat.

If (P → Q) is false? Well, there’s only one valuation that makes (P → Q) false — the valuation where P is true, and Q is false. So, if (P → Q) is false, then Q is false as well. But, if Q is false, then (Q → R) is true, because (Q → R) can only be false when Q is true and R is false. If Q is false, then (Q → R) is true.

Going the other way, if (Q → R) is false, then Q is true and (P → Q) can’t be false, because it can only be false if Q is false.

As a result, whenever (P → Q) is false, (Q → R) is guaranteed to be true, and *vice versa*. So, (P → Q) ∨ (Q → R) is a tautology.

My point here is just a friendly reminder: be careful with translation. Expressing the English “if/then” with the material conditional (→) may sometimes get you into trouble. #