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Proof step 11 from proof sublist-move (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
n + m ≤ # x0 + 1, n + m ≤ m1 + 1, m1 ≤ # 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, a ' + firstn(m1, x0))) = firstn(m, restn(n, a ' + x0))

Rule argument

Additional information

Used simplifier rules: lep-02: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2