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Proof step 19 from proof max-ordered (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
max(a ' + x0) = a, a < max(x1), ordered≤(x0), x1 ⊑ x0 ⊦ max(x1) = a, x1 = []

Rule argument

Additional information

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