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Proof step 5 from proof max-ordered (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ x ≠ [] → max(a ' + x) = (max(x) < a ⊃ a;max(x))
Proof step 7:
max((a ' + x2) + x0) = max(a ' + 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) < a ⊃ a;max(x2 + x1)) = max(a ' + x2)

Rule argument

spec: <toplevel> name: max-rec seq: ⊦ x ≠ [] → max(a ' + x) = (max(x) < a ⊃ a;max(x)) subst: [x, a] → [x2 + x1, a] flags: <all paths>

Additional information

Used simplifier rules: max-base: ⊦ max(a ') = a;% ⊦ x + y = [] ↔ x = [] ∧ y = []; (as)% ⊦ [] + x = x; (s)% ⊦ x + [] = x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: x2 + x1, a