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Proof step 10 from proof get-butlast (previous step)

n ≤ # x0, ∀ n0, a1. n0 < # x0 → (x0 + a1 ')[n0] = x0[n0] ⊦ (x0 + a ')[n - 1] = (a0 ' + x0)[n], n = 0, x0 = []

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 11:
n = m + 1, n ≤ # x0, ∀ n0, a1. n0 < # x0 → (x0 + a1 ')[n0] = x0[n0] ⊦ (x0 + a ')[m] = (a0 ' + x0)[n], n = 0, x0 = []

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