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Proof step 13 from proof get-nodups (previous step)

x0[m - 1] = (a ' + x0)[n], m ≤ # x0, n ≤ # x0, ∀ m0, n0. x0[m0] = x0[n0] ∧ m0 < # x0 ∧ n0 < # x0 → m0 = n0 ∨ x0 = [] ⊦ m = 0, m = n, dups(x0), a ∈ x0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 14:
x0[m1] = (a ' + x0)[n], m = m1 + 1, n ≤ # x0, m ≤ # x0, ∀ m0, n0. x0[m0] = x0[n0] ∧ m0 < # x0 ∧ n0 < # x0 → m0 = n0 ∨ x0 = [] ⊦ m = n, m = 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] → [m, m1]

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: m, m1