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Proof step 9 from proof restn-nonempty (previous step)

restn(n - 1, a0 ' + x0) = [], n ≤ # x0 ⊦ x0 = [], n = 0, ∃ n, a. restn(n, a ' + x0) = [] ∧ n < # x0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 10:
n = m + 1, restn(m, a0 ' + x0) = [], n ≤ # x0 ⊦ x0 = [], n = 0, ∃ n, a. restn(n, a ' + x0) = [] ∧ n < # x0

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