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Proof step 16 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(x2 + x1) = max(a ' + x2), x2 + x1 = [], max(x2 + x1) < a

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 17:
⊦ x ≠ [] → max(a ' + x) = (max(x) < a ⊃ a;max(x))
Proof step 18:
¬ 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
Proof step 31:
x2 ≠ [], max((a ' + x2) + x0) = (max(x2) < a ⊃ a;max(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(x2) < a ⊃ a;max(x2)), x2 + x1 = [], max(x2 + x1) < a

Rule argument

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

Additional information

Used terms: x2, a