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Proof step 1 from proof sublist-rec-01 (conclusion)

n ≤ # x, n ≠ 0 ⊦ sublist(0, n, x) = x.first ' + sublist(0, n -1, x.rest)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
n ≤ # x ⊦ n = 0, firstn(n, x) = x.first ' + firstn(n - 1, x.rest)

Rule argument

Additional information

Used simplifier rules: sublist-def: ⊦ sublist(m, n, x) = firstn(n, restn(m, x));%restN-zero: ⊦ restn(0, x) = x;%pred-03: ⊦ n -1 = n - 1; from specification nat