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Proof step 18 from proof max-ordered (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
max(a ' + x0) = a, a < max(x1), x1 ⊑ x0, ordered≤(x0) ⊦ x1 = [], max(x1) = a

Rule argument

Additional information

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