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Proof step 6 from proof first-mklist (previous step)

n0 + 1 ≠ 0, m ≤ n0 + 1, ∀ m, a. m ≤ n0 → firstn(m, mklist(a, n0)) = mklist(a, m) ⊦ firstn(m, a + mklist(a, n0 + 1 - 1)) = mklist(a, m)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
m ≤ n0 + 1, ∀ m, a. m ≤ n0 → firstn(m, mklist(a, n0)) = mklist(a, m) ⊦ firstn(m, a ' + mklist(a, n0)) = mklist(a, m)

Rule argument

Additional information

Used simplifier rules: oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%c: ⊦ a + x = a ' + x; from specification list%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%add-base: ⊦ n + 0 = n; from specification nat-basic2%add-01: ⊦ n ≤ m → m + n0 - n = (m - n) + n0; from specification nat