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Proof step 25 from proof sublist-move (previous step)

n0 = m1 + 1, n + m ≤ m0, n0 + m0 ≤ # 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, restn(n, firstn(m0, restn(m1, x0)))) = firstn(m, restn(n0 + n, a ' + x0)), n0 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 26:
n + m ≤ m0, m1 + m0 ≤ # x0, ∀ 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, restn(n, firstn(m0, restn(m1, x0)))) = firstn(m, restn(m1 + n, x0))

Rule argument

Additional information

Used simplifier rules: restN-rec: ⊦ restn(n + 1, x) = restn(n, x.rest);%lep-02: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; from specification nat%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%a: ⊦ (x + y) + z = x + y + z; from specification list%c-07: ⊦ (a ' + x).rest = x; from specification list