 Solution

ID: past-exam-of-the-logic-and-set-theory-course-of-the-university-of-oxford/2007/a/2/ii/solution

β

. The two instances of L1 are

α \to (β \to α), α \to (\neg β \to α) .

The instance of L2 with

γ = α

(α \to (β \to α)) \to ((α \to (\neg β \to α)) \to (α \to α)) .

Apply modus ponens rule first with the first L1 instance and then with the second. This is an assumption-free logical derivation, so

⊢ α \to α .

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