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Proof step 6 from proof restn-first-get (previous step)

n ≤ # x0, ∀ n. n < # x0 → restn(n, x0).first = x0[n] ⊦ restn(n - 1, x0).first = (a ' + x0)[n], n = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 7:
n = m + 1, n ≤ # x0, ∀ n. n < # x0 → restn(n, x0).first = x0[n] ⊦ restn(m, x0).first = (a ' + x0)[n], n = 0

Rule argument

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

Additional information

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