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Proof step 1 from proof sublist-mklist (conclusion)

⊦ n + m ≤ m0 → sublist(n, m, mklist(a, m0)) = mklist(a, m)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
n + m ≤ m0 ⊦ firstn(m, mklist(a, m0 - n)) = mklist(a, m)

Rule argument

Additional information

Used simplifier rules: sublist-def: ⊦ sublist(m, n, x) = firstn(n, restn(m, x));%rest-mklist-01: ⊦ m + m0 ≤ n → restn(m, mklist(a, n)) = mklist(a, n - m);%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2