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Proof step 9 from proof max-ins (previous step)

∀ 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 cut applied rule cut formula

And obtained the following new goals

Proof step 10:
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
Proof step 13:
∀ a1. max(ins≤(a1, x0)) = max(x0) ∨ max(x0) < a1 ⊦ max(x0) < a0, x0 = [], max(ins≤(a, a0 ' + x0)) = (max(x0) < a0 ⊃ a0;max(x0)), (max(x0) < a0 ⊃ a0;max(x0)) < a

Rule argument

fma: max(x0) < a0

Additional information

Used formulas: max(x0) < a0