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Proof step 4 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 → x2 = [] ∨ max(x2 + x1) = max(x2) ⊦ max((a ' + x2) + x1) = max(a ' + x2)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ x ≠ [] → max(a ' + x) = (max(x) < a ⊃ a;max(x))
Proof step 6:
x2 + x1 ≠ [], 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) < a ⊃ a;max(x2 + x1)) = max(a ' + x2)

Rule argument

spec: inst: name: max-rec seq: ⊦ x ≠ [] → max(a ' + x) = (max(x) < a ⊃ a;max(x)) subst: [x, a] → [x2 + x1, a] flags: rewrite, current

Additional information

Used simplifier rules: max-base: ⊦ max(a ') = a;%app-02-ant: ⊦ ¬ x + y ≠ [] ↔ x = [] ∧ y = []; from specification list%Nil: ⊦ [] + x = x; from specification list%app: ⊦ x + [] = x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: x2 + x1, a