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Proof step 8 from proof firstn-twice (previous step)

n ≤ 1 + # x0, ∀ n. n ≤ # x0 → firstn(n, firstn(n, x0)) = firstn(n, x0) ⊦ firstn(n, a ' + firstn(n - 1, x0)) = a ' + firstn(n - 1, x0), n = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 9:
n = m + 1, n ≤ 1 + # x0, ∀ n. n ≤ # x0 → firstn(n, firstn(n, x0)) = firstn(n, x0) ⊦ firstn(n, a ' + firstn(m, x0)) = a ' + firstn(m, x0), n = 0

Rule argument

spec: nat name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [n, m]

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n, m