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Proof step 2 from proof sublist-num-01 (previous step)

⊦ n = 0, sublist(0, n, a ' + x) = a ' + sublist(0, n -1, x)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ firstn(n, a ' + x) = a ' + firstn(n - 1, x), n = 0

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)