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Proof step 20 from proof max-ins (previous step)

a0 < a, ∀ a1. max(ins≤(a1, x0)) = max(x0) ∨ max(x0) < a1 ⊦ max(a0 + ins≤(a, x0)) = max(x0), x0 = [], max(x0) < a, max(x0) < a0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
a0 < a, ∀ a1. max(ins≤(a1, x0)) = max(x0) ∨ max(x0) < a1 ⊦ max(a0 ' + ins≤(a, x0)) = max(x0), x0 = [], max(x0) < a0, max(x0) < a

Rule argument

Additional information

Used simplifier rules: ⊦ a + x = a ' + x; (s)