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Proof step 11 from proof max-ordered (previous step)

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)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 12:
⊦ x ≠ [] → max(a ' + x) = (max(x) < a ⊃ a;max(x))
Proof step 13:
x2 + x0 ≠ [], (max(x2 + x0) < a ⊃ a;max(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

spec: inst: name: max-rec seq: ⊦ x ≠ [] → max(a ' + x) = (max(x) < a ⊃ a;max(x)) subst: [x, a] → [x2 + x0, a] flags: rewrite, current

Additional information

Used simplifier rules: app-02-ant: ⊦ ¬ x + y ≠ [] ↔ x = [] ∧ y = []; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: x2 + x0, a