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Proof step 11 from proof get-nodups (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
x0[m - 1] = (a ' + x0)[n], n ≤ # x0, m ≤ # x0, ∀ m0, n0. x0[m0] = x0[n0] ∧ n0 < # x0 ∧ m0 < # x0 → m0 = n0 ∨ x0 = [] ⊦ m = n, m = 0, a ∈ x0, dups(x0)

Rule argument

Additional information

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