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Proof step 3 from proof sublist-move (previous step)

n0 + m0 ≤ # x, n + m ≤ m0 ⊦ firstn(m, restn(n, firstn(m0, restn(n0, x)))) = firstn(m, restn(n0 + n, x))

User applied rule structural induction

And obtained the following new goals

Proof step 4:
n + m ≤ m0, n0 + m0 ≤ #(a + 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, restn(n0, a + x0)))) = firstn(m, restn(n0 + n, a + x0))

Rule argument

indvar: x

Additional information

Used simplifier rules: restN-zero: ⊦ restn(0, x) = x;%firstN-zero: ⊦ firstn(0, x) = [];%list freely generated by [], +;;% ⊦ # [] = 0; (s)% ⊦ 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 = m; (s)