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Proof step 9 from proof get-nodups (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
n ≤ # x0, ∀ m0, n0. x0[m0] = x0[n0] ∧ n0 < # x0 ∧ m0 < # x0 → m0 = n0 ∨ x0 = [] ⊦ n = 0, dups(x0), x0[n - 1] ∈ x0

Rule argument

Additional information

Used simplifier rules: pred-03: ⊦ n -1 = n - 1; from specification nat