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Proof step 2 from proof sublist-move (previous step)

n + m ≤ m0, n0 + m0 ≤ # x ⊦ 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 3:
∀ 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)), n + m ≤ m0, n0 + m0 ≤ #(a + 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 [], +;;%size-01: ⊦ # [] = 0; from specification list-data%lez-01: ⊦ m ≤ 0 ↔ m = 0; from specification nat%z-01: ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%add-base: ⊦ n + 0 = n; from specification nat-basic2