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Proof step 5 from proof first-mklist (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ m ≠ 0 → mklist(a, m) = a + mklist(a, m - 1)
Proof step 7:
m ≤ n0 + 1, ∀ m, a. m ≤ n0 → firstn(m, mklist(a, n0)) = mklist(a, m) ⊦ n0 + 1 = 0, firstn(m, a + mklist(a, n0 + 1 - 1)) = mklist(a, m)

Rule argument

spec: list-last name: mk-rec seq: ⊦ m ≠ 0 → mklist(a, m) = a + mklist(a, m - 1) subst: [m, a] → [n0 + 1, a] flags: <all paths>

Additional information

Used simplifier rules: ⊦ m + 1 ≠ 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: n0 + 1, a