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Proof step 17 from proof get-nodups (previous step)

x0[m1] = x0[n - 1], m1 < # x0, n ≤ # x0, ∀ m, n0. x0[m] = x0[n0] ∧ n0 < # x0 ∧ m < # x0 → m = n0 ⊦ m1 + 1 = n, x0 = [], n = 0, dups(x0), a ∈ x0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 18:
n = m0 + 1, x0[m1] = x0[m0], m1 < # x0, n ≤ # x0, ∀ m, n0. x0[m] = x0[n0] ∧ n0 < # x0 ∧ m < # x0 → m = n0 ⊦ m1 + 1 = n, x0 = [], n = 0, dups(x0), a ∈ x0

Rule argument

spec: nat inst: name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [n, m0] flags:

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n, m0