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Proof step 3 from proof ordered-shorter-sublist (previous step)

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

User applied rule insert lemma

And obtained the following new goals

Proof step 4:
⊦ x ⊑ x0 ∧ ordered≤(x0) → (ordered≤(x) ↔ true)
Proof step 5:
n1 + m0 < # x, n2 < n1, ordered≤(sublist(m0, n1 + 1, x)), sublist(m0, n2 + 1, x) ⊑ sublist(m0, n1 + 1, x) ∧ ordered≤(sublist(m0, n1 + 1, x)) → (ordered≤(sublist(m0, n2 + 1, x)) ↔ true) ⊦ x = [], ordered≤(sublist(m0, n2 + 1, x))

Rule argument

spec: <toplevel> name: ordered-prefix seq: ⊦ x ⊑ x0 ∧ ordered≤(x0) → (ordered≤(x) ↔ true) subst: [x, x0] → [sublist(m0, n2 + 1, x), sublist(m0, n1 + 1, x)]

Additional information

Used terms: sublist(m0, n2 + 1, x), sublist(m0, n1 + 1, x)