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Proof step 4 from proof in-prefixn-pos-smaller0 (previous step)

n < m, n < pos(a, x), a ∈ firstn(n, x), m ≤ pos(a, x) ⊦

User applied rule structural induction

And obtained the following new goals

Proof step 5:
n < pos(a, a0 + x0), n < m, a ∈ firstn(n, a0 + x0), m ≤ pos(a, a0 + x0), ∀ n, m, a. ¬(n < m ∧ n < pos(a, x0) ∧ a ∈ firstn(n, x0) ∧ m ≤ pos(a, x0)) ⊦

Rule argument

indvar: x

Additional information

Used simplifier rules: list freely generated by [], +;;% ⊦ pos(a, []) = 0; (s)% ⊦ m ≤ 0 ↔ m = 0; (s)% ⊦ m < n → n ≠ 0; (s)