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Proof step 7 from proof restN-dups (previous step)

n ≤ # x0 + 1, dups(restn(n - 1, x0)), ∀ n0. ¬ dups(restn(n0, x0)) ∨ ¬ n0 ≤ # x0 ⊦ n = 0, a ∈ x0, dups(x0)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 8:
n = m + 1, n ≤ # x0 + 1, dups(restn(m, x0)), ∀ n0. ¬ dups(restn(n0, x0)) ∨ ¬ n0 ≤ # x0 ⊦ n = 0, a ∈ x0, dups(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