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Proof step 4 from proof sublist-prefix (previous step)

n1 + m0 < # x, n0 < n1, ∀ n1, m0, x. n1 + m0 < # x ∧ n0 ≤ n1 → x = [] ∨ firstn(n0, restn(m0, x).rest) ⊑ firstn(n1, restn(m0, x).rest) ⊦ x = [], n1 = 0, restn(m0, x).rest.first ' + firstn(n0, restn(m0, x).rest.rest) ⊑ firstn(n1, restn(m0, x).rest)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
n ≠ 0 ⊦ firstn((* n), x) = x.first ' + firstn(n -1, x.rest)
Proof step 6:
n1 ≠ 0, n1 + m0 < # x, n0 < n1, ∀ n1, m0, x. n1 + m0 < # x ∧ n0 ≤ n1 → x = [] ∨ firstn(n0, restn(m0, x).rest) ⊑ firstn(n1, restn(m0, x).rest) ⊦ x = [], n1 = 0, restn(m0, x).rest.first ' + firstn(n0, restn(m0, x).rest.rest) ⊑ restn(m0, x).rest.first ' + firstn(n1 -1, restn(m0, x).rest.rest)

Rule argument

spec: inst: name: first-num seq: n ≠ 0 ⊦ firstn((* n), x) = x.first ' + firstn(n -1, x.rest) subst: [x, n] → [restn(m0, x).rest, n1] flags: rewrite, current

Additional information

Used terms: restn(m0, x).rest, n1