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Proof step 11 from proof sublist-prefix (previous step)

n0 ≤ m, (m + 1) + m0 < # x, m + m0 < # x, firstn(n0, restn(m0, x).rest) ⊑ firstn(m, restn(m0, x).rest), ∀ n1, m0, x. n1 + m0 < # x ∧ n0 ≤ n1 → x = [] ∨ firstn(n0, restn(m0, x).rest) ⊑ firstn(n1, restn(m0, x).rest) ⊦ x = [], firstn(n0, restn(m0, x).rest.rest) ⊑ firstn(m, restn(m0, x).rest.rest)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 12:
⊦ x.rest = restn(1, x)
Proof step 13:
n0 ≤ m, (m + 1) + m0 < # x, m + m0 < # x, firstn(n0, restn(1, restn(m0, x))) ⊑ firstn(m, restn(1, restn(m0, x))), ∀ n1, m0, x. n1 + m0 < # x ∧ n0 ≤ n1 → x = [] ∨ firstn(n0, restn(m0, x).rest) ⊑ firstn(n1, restn(m0, x).rest) ⊦ x = [], firstn(n0, restn(1, restn(m0, x)).rest) ⊑ firstn(m, restn(1, restn(m0, x)).rest)

Rule argument

spec: inst: name: rest-rest1 seq: ⊦ x.rest = restn(1, x) subst: [x] → [restn(m0, x)] flags: rewrite, current

Additional information

Used terms: restn(m0, x)