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Proof step 3 from proof max-ins (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ a. max(ins≤(a, x0)) = max(x0) ∨ x0 = [] ∨ max(x0) < a ⊦ max(ins≤(a, a0 ' + x0)) = max(a0 ' + x0), max(a0 ' + x0) < a

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list%e-01: ⊦ a ' + x ≠ []; from specification list%c: ⊦ a + x = a ' + x; from specification list