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Proof step 15 from proof sublist-move (previous step)

m1 ≤ # x0, m ≤ m1 + 1, m ≤ # x0 + 1, ∀ n, m, m0, n0. n + m ≤ m0 ∧ n0 + m0 ≤ # x0 → firstn(m, restn(n, firstn(m0, restn(n0, x0)))) = firstn(m, restn(n0 + n, x0)) ⊦ firstn(m, a ' + firstn(m1, x0)) = firstn(m, a ' + x0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 16:
n ≠ 0 ⊦ firstn((* n), x) = x.first ' + firstn(n -1, x.rest)
Proof step 17:
m ≠ 0, m1 ≤ # x0, m ≤ m1 + 1, m ≤ # x0 + 1, ∀ n, m, m0, n0. n + m ≤ m0 ∧ n0 + m0 ≤ # x0 → firstn(m, restn(n, firstn(m0, restn(n0, x0)))) = firstn(m, restn(n0 + n, x0)) ⊦ (a ' + firstn(m1, x0)).first ' + firstn(m -1, (a ' + firstn(m1, x0)).rest) = firstn(m, a ' + x0)

Rule argument

spec: inst: name: first-num seq: n ≠ 0 ⊦ firstn((* n), x) = x.first ' + firstn(n -1, x.rest) subst: [x, n] → [a ' + firstn(m1, x0), m] flags: rewrite, current

Additional information

Used simplifier rules: firstN-zero: ⊦ firstn(0, x) = []; Used terms: a ' + firstn(m1, x0), m