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

n0 < n, m + n ≤ # x ⊦ sublist(m, n, x)[n0] = x[m + n0]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n0 < n, m + n ≤ # x ⊦ firstn(n, restn(m, x))[n0] = x[m + n0]

Rule argument

Additional information

Used simplifier rules: sublist-def: ⊦ sublist(m, n, x) = firstn(n, restn(m, x));% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic