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Proof step 12 from proof max-ge (previous step)

max(x1 + x0) < a, a < (max(x1) < a ⊃ a;max(x1)), max(x1 + x0) < (max(x1) < a ⊃ a;max(x1)) ⊦ x1 + x0 = [], x1 = [], ∃ x0. max(x1 + x0) < max(x1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
max(x1 + x0) < a, a < (max(x1) < a ⊃ a;max(x1)), max(x1 + x0) < max(x1) ⊦ x1 + x0 = [], x1 = [], ∃ x0. max(x1 + x0) < max(x1)

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ a < a; (s)%transitivity: ⊦ a < b ∧ b < c → a < c; (forward) from specification oelem