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Proof step 3 from proof sublist-append-04 (previous step)

∀ n, n0, n1. n + n0 + n1 ≤ # x0 → firstn(n0, restn(n, x0)) + firstn(n1, restn(n + n0, x0)) = firstn(n0 + n1, restn(n, x0)), n + n0 + n1 ≤ #(a + x0) ⊦ firstn(n0, restn(n, a + x0)) + firstn(n1, restn(n + n0, a + x0)) = firstn(n0 + n1, restn(n, a + x0))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
n + n0 + n1 ≤ # x0 + 1, ∀ n, n0, n1. n + n0 + n1 ≤ # x0 → firstn(n0, restn(n, x0)) + firstn(n1, restn(n + n0, x0)) = firstn(n0 + n1, restn(n, x0)) ⊦ firstn(n0, restn(n, a ' + x0)) + firstn(n1, restn(n + n0, a ' + x0)) = firstn(n0 + n1, restn(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