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Proof step 3 from proof sublist-move (previous step)

∀ 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))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
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(n0, a ' + x0)))) = firstn(m, restn(n0 + n, a ' + x0))

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list%succ: ⊦ m +1 = m + 1; from specification nat-basic2%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%c: ⊦ a + x = a ' + x; from specification list