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Proof step 6 from proof sublist-append-04 (previous step)

¬ 0 < n, 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))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
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, a ' + x0) + firstn(n1, restn(n0, a ' + x0)) = firstn(n0 + n1, a ' + x0)

Rule argument

Additional information

Used simplifier rules: restN-zero: ⊦ restn(0, x) = x;%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%add-base: ⊦ n + 0 = n; from specification nat-basic2%zero: ⊦ 0 < m ↔ m ≠ 0; from specification nat-basic1