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Proof step 24 from proof sublist-move (previous step)

m2 + m ≤ # x0, m2 + m ≤ m1, m1 ≤ # x0, m2 + m ≤ # x0 ∧ m1 + 0 ≤ m → firstn(0, restn(m1, firstn(m, restn(m2, x0)))) = firstn(0, restn(m2 + m1, 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(m2, firstn(m1, x0))) = firstn(m, restn(m2, x0))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
m1 ≤ # x0, m2 + m ≤ m1, m2 + m ≤ # x0, m1 ≤ m → firstn(0, restn(m1, firstn(m, restn(m2, x0)))) = firstn(0, restn(m2 + m1, 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(m2, firstn(m1, x0))) = firstn(m, restn(m2, x0))

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ 0 + m = m; (s)