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Proof step 8 from proof restN-dups (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 9:
n = m + 1, dups(restn(m, x0)), n ≤ 1 + # x0, ∀ n0. ¬ dups(restn(n0, x0)) ∨ ¬ n0 ≤ # x0 ⊦ n = 0, dups(x0), a ∈ 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