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Proof step 14 from proof dups-def (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
(a ' + x0)[m] = (a ' + x0)[n], n ≤ # x0, m ≤ # x0 ⊦ 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]

Rule argument

Additional information

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