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Proof step 3 from proof sublist-append-04 (previous step)

n + n0 + n1 ≤ # x ⊦ firstn(n0, restn(n, x)) + firstn(n1, restn(n + n0, x)) = firstn(n0 + n1, restn(n, x))

User applied rule structural induction

And obtained the following new goals

Proof step 4:
n + n0 + n1 ≤ #(a + x0), ∀ n, n0, n1. n + n0 + n1 ≤ # x0 → firstn(n0, restn(n, x0)) + firstn(n1, restn(n + n0, x0)) = firstn(n0 + n1, restn(n, x0)) ⊦ firstn(n0, restn(n, a + x0)) + firstn(n1, restn(n + n0, a + x0)) = firstn(n0 + n1, restn(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% ⊦ [] + x = x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list