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Proof step 13 from proof firstN-lastN (previous step)

1 + # x0 = n0 + n, ∀ n0, n. # x0 = n0 + n → firstn(n0, x0) + restn(n0, x0) = x0 ⊦ firstn(n0 - 1, x0) + restn(n0, a ' + x0) = x0, n0 = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 14:
1 + # x0 = n0 + n, n0 = m + 1, ∀ n0, n. # x0 = n0 + n → firstn(n0, x0) + restn(n0, x0) = x0 ⊦ firstn(m, x0) + restn(n0, a ' + x0) = x0, n0 = 0

Rule argument

spec: nat name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [n0, m]

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n0, m