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Proof step 16 from proof dups-def (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
a = (a ' + x0)[n], n ≤ # x0 ⊦ 0 = n, a ∈ x0, dups(x0), ∃ m, n. m < # x0 ∧ n < # x0 ∧ m ≠ n ∧ x0[m] = x0[n], 0 - 1 < # x0 ∧ n - 1 < # x0 ∧ 0 - 1 ≠ n - 1 ∧ x0[0 - 1] = x0[n - 1]

Rule argument

Additional information

Used simplifier rules: get-zero-01: ⊦ (a ' + x)[0] = a;%lez: ⊦ 0 ≤ m; from specification nat%a: ⊦ (x + y) + z = x + y + z; from specification list