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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
max(x1 + x0) < max(a ' + x1), a < max(a ' + x1), max(x1 + x0) < a ⊦ x1 + x0 = [], x1 = [], ∃ x0. max(x1 + x0) < max(x1)

Rule argument

Additional information

Used simplifier rules: max-base: ⊦ max(a ') = a;% ⊦ ¬ a < a; (s)% ⊦ x + [] = x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ [] + x = x; (s)%transitivity: ⊦ a < b ∧ b < c → a < c; (forward) from specification oelem