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Proof step 7 from proof get-append-less (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 8:
n = m + 1, n ≤ # x0, ∀ n, y. n < # x0 → (x0 + y)[n] = x0[n] ∨ x0 = [] ⊦ n = 0, (x0 + y)[m] = (a ' + x0)[n]

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