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Proof step 4 from proof length-to (previous step)

n0 < # x, ∀ x. n0 ≤ # x → #(x to n0) = n0 ⊦ #(x to n0 + 1) = n0 + 1, x = []

User applied rule structural induction

And obtained the following new goals

Proof step 5:
∀ n0. n0 < # x0 ∧ (∀ x. n0 ≤ # x → #(x to n0) = n0) → #(x0 to n0 + 1) = n0 + 1 ∨ x0 = [], n0 < #(a + x0), ∀ x. n0 ≤ # x → #(x to n0) = n0 ⊦ #(a + x0 to n0 + 1) = n0 + 1, a + x0 = []

Rule argument

indvar: x

Additional information

Used simplifier rules: list freely generated by [], +;;