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Proof step 5 from proof sublist-prefix (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
n ≠ 0 ⊦ firstn((* n), x) = x.head ' + firstn(n -1, x.tail)
Proof step 7:
n1 + m0 < # x, n0 < n1, ∀ n1, m0, x. n1 + m0 < # x ∧ n0 ≤ n1 → x = [] ∨ firstn(n0, restn(m0, x).tail) ⊑ firstn(n1, restn(m0, x).tail) ⊦ n1 = 0, x = [], restn(m0, x).tail.head ' + firstn(n0, restn(m0, x).tail.tail) ⊑ restn(m0, x).tail.head ' + firstn(n1 -1, restn(m0, x).tail.tail)

Rule argument

spec: <toplevel> name: first-num seq: n ≠ 0 ⊦ firstn((* n), x) = x.head ' + firstn(n -1, x.tail) subst: [x, n] → [restn(m0, x).tail, n1] flags: <all paths>

Additional information

Used simplifier rules: ⊦ ¬ m < 0; (s) Used terms: restn(m0, x).tail, n1