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Proof step 10 from proof max-ordered (previous step)

x2 ≠ [], max(x2 + x1) < a, max((a ' + x2) + x0) = (max(x2) < a ⊃ a;max(x2)), ordered≤(x0), x1 ⊑ x0, ∀ x0, x1. max(x2 + x0) = max(x2) ∧ ordered≤(x0) ∧ x1 ⊑ x0 → x2 = [] ∨ max(x2 + x1) = max(x2) ⊦ a = (max(x2) < a ⊃ a;max(x2)), x2 + x1 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
max((a ' + x2) + x0) = max(x2), x1 ⊑ x0, ordered≤(x0), max(x2 + x1) < a, ∀ x, x3. max(x2 + x) = max(x2) ∧ ordered≤(x) ∧ x3 ⊑ x → max(x2 + x3) = max(x2) ⊦ x2 = [], x2 + x1 = [], a = max(x2)

Rule argument

Additional information

Used simplifier rules: max-ge-first: ⊦ ¬ ¬ max(a ' + x) < a ↔ false;%a: ⊦ (x + y) + z = x + y + z; from specification list%lel: ⊦ a < b ∧ a = c → c < b; (forward) from specification oelem