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Proof step 8 from proof sublist-move (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
n ≠ 0 ⊦ firstn((* n), x) = x.head ' + firstn(n -1, x.tail)
Proof step 10:
n + m ≤ 1 + # x0, m0 ≤ 1 + # x0, n + m ≤ m0, ∀ n0, m0, n, m. n0 + m0 ≤ # x0 ∧ n + m ≤ m0 → firstn(m, restn(n, firstn(m0, restn(n0, x0)))) = firstn(m, restn(n0 + n, x0)) ⊦ firstn(m, restn(n, (a ' + x0).head ' + firstn(m0 -1, (a ' + x0).tail))) = firstn(m, restn(n, a ' + x0)), m0 = 0

Rule argument

spec: <toplevel> name: first-num seq: n ≠ 0 ⊦ firstn((* n), x) = x.head ' + firstn(n -1, x.tail) subst: [x, n] → [a ' + x0, m0] flags: <all paths>

Additional information

Used simplifier rules: firstN-zero: ⊦ firstn(0, x) = [];%restN-zero: ⊦ restn(0, x) = x;% ⊦ m ≤ 0 ↔ m = 0; (s)% ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; (ws)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ 0 ≤ m; (s) Used terms: a ' + x0, m0