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Proof step 44 from proof max-ordered (previous step)

max(x2) < max(x2 + x0), ordered≤(x0), x1 ⊑ x0, max(x2 + x0) < max(x2 + x1), max(x2) < max(x2 + x1), ∀ x, x3. max(x2 + x) = max(x2) ∧ ordered≤(x) ∧ x3 ⊑ x → max(x2 + x3) = max(x2) ⊦ x2 + x0 = [], max(x2 + x1) = max(x2 + x0), x2 + x1 = [], x2 = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 45:
⊦ x ⊑ y ↔ (∃ z. x + z = y)

Rule argument

spec: list-last inst: name: prefix seq: ⊦ x ⊑ y ↔ (∃ z. x + z = y) subst: [y, x] → [x0, x1] flags: rewrite

Additional information

Used simplifier rules: max-ge: ⊦ x ≠ [] → (¬ ¬ max(x + x0) < max(x) ↔ false);%a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: x0, x1