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Proof step 10 from proof max-ins (previous step)

max(x0) < a0, ∀ 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
max(x0) < a0, ∀ a1. max(ins≤(a1, x0)) = max(x0) ∨ max(x0) < a1 ⊦ x0 = [], max(a ' + a0 ' + x0) = a0, a0 < a

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list%ins-y: ⊦ ¬ b < a → ins≤(a, b ' + x) = a + b + x; from specification olist