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Proof step 6 from proof restn-empty (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 7:
n = m + 1, restn(m, x0) = [], n ≤ # x0, ∀ n0. n0 < # x0 → restn(n0, x0) = [] → n0 = # x0 ⊦ n = # x0 + 1, 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