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Proof step 15 from proof first-mklist (previous step)

m0 ≤ n0, ∀ m, a. m ≤ n0 → firstn(m, mklist(a, n0)) = mklist(a, m) ⊦ a ' + firstn(m0, mklist(a, n0)) = a + mklist(a, m0 + 1 - 1), m0 + 1 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
m0 ≤ n0, ∀ m, a. m ≤ n0 → firstn(m, mklist(a, n0)) = mklist(a, m) ⊦ firstn(m0, mklist(a, n0)) = mklist(a, m0)

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ x + y = x + z ↔ y = z; (s)% ⊦ a + x = a ' + x; (s)% ⊦ n + 0 = n; (s)% ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≠ 0; (s)