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Proof step 2 from proof sublist-rec (previous step)

⊦ n = 0, sublist(0, n, x) = x.head ' + sublist(0, n -1, x.tail), x = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ firstn(n, x) = x.head ' + firstn(n - 1, x.tail), n = 0, x = []

Rule argument

Additional information

Used simplifier rules: sublist-def: ⊦ sublist(m, n, x) = firstn(n, restn(m, x));%restN-zero: ⊦ restn(0, x) = x;% ⊦ n -1 = n - 1; (s)