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Proof step 6 from proof ordered-shorter-sublist (previous step)

n1 + m0 < # x, n2 < n1, ordered≤(sublist(m0, n1 + 1, x)) ⊦ n1 = 0, x = [], ordered≤(sublist(m0, n2 + 1, x)), sublist(m0, n2 + 1, x) ⊑ sublist(m0, n1 + 1, x)

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: list-del inst: name: sublist-prefix seq: ⊦ n ≤ n1 ∧ n1 + m0 < # x → (¬ sublist(m0, n + 1, x) ⊑ sublist(m0, n1 + 1, x) ↔ false) subst: [n1, x, n, m0] → [n1, x, n2, m0] flags: rewrite

Additional information

Used simplifier rules: lels-03: ⊦ m < n → m ≤ n; from specification nat Used terms: n1, x, n2, m0