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Proof step 1 from proof sublist-append-04 (conclusion)

⊦ n + n0 + n1 ≤ # x → sublist(n, n0, x) + sublist(n + n0, n1, x) = sublist(n, n0 + n1, x)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
n + n0 + n1 ≤ # x ⊦ firstn(n0, restn(n, x)) + firstn(n1, restn(n + n0, x)) = firstn(n0 + n1, restn(n, x))

Rule argument

Additional information

Used simplifier rules: sublist-def: ⊦ sublist(m, n, x) = firstn(n, restn(m, x));