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Proof step 8 from proof max-ins (previous step)

∀ a. max(ins≤(a, x0)) = max(x0) ∨ x0 = [] ∨ max(x0) < a ⊦ x0 = [], max(ins≤(a, a0 ' + x0)) = max(a0 ' + x0), max(a0 ' + x0) < a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
∀ a1. max(ins≤(a1, x0)) = max(x0) ∨ max(x0) < a1 ⊦ x0 = [], max(ins≤(a, a0 ' + x0)) = (max(x0) < a0 ⊃ a0;max(x0)), (max(x0) < a0 ⊃ a0;max(x0)) < a

Rule argument

Additional information

Used simplifier rules: max-rec: ⊦ x ≠ [] → max(a ' + x) = (max(x) < a ⊃ a;max(x));%a: ⊦ (x + y) + z = x + y + z; from specification list