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

∀ x0. max(x1 + x0) < max(x1) → x1 = [], max((a + x1) + x0) < max(a + x1) ⊦ a + x1 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
max((a ' + x1) + x0) < max(a ' + x1), (∃ x0. max(x1 + x0) < max(x1)) → x1 = [] ⊦

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