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Proof step 5 from proof in-prefixn-pos-smaller0 (previous step)

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)) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
n < pos(a, a0 ' + x0), n < m, a ∈ firstn(n, a0 ' + x0), m ≤ pos(a, a0 ' + x0), ∀ n0, m0, a1. ¬ n0 < m0 ∨ ¬ n0 < pos(a1, x0) ∨ ¬ a1 ∈ firstn(n0, x0) ∨ ¬ m0 ≤ pos(a1, x0) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ a + x = a ' + x; (s)