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Proof step 2 from proof less-same (previous step)

n + 1 < # x ⊦ less#(a, x[n, x[n + 1]][n + 1, x[n]]) = less#(a, x), x = []

User applied rule structural induction

And obtained the following new goals

Proof step 3:
n + 1 < #(a0 + x0), ∀ n, a. n + 1 < # x0 → less#(a, x0[n, x0[n + 1]][n + 1, x0[n]]) = less#(a, x0) ∨ x0 = [] ⊦ a0 + x0 = [], less#(a, (a0 + x0)[n, (a0 + x0)[n + 1]][n + 1, (a0 + x0)[n]]) = less#(a, a0 + x0)

Rule argument

indvar: x

Additional information

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