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Proof step 4 from proof dups-def (previous step)

dups(x0) ↔ (∃ m, n. m < # x0 ∧ n < # x0 ∧ m ≠ n ∧ x0[m] = x0[n]), dups(x0) ⊦ a ∈ x0 ∨ dups(x0) ↔ (∃ m, n. ¬ # x0 < m ∧ ¬ # x0 < n ∧ m ≠ n ∧ (a ' + x0)[m] = (a ' + x0)[n])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
x0[m] = x0[n], n < # x0, m < # x0, dups(x0) ⊦ m = n, x0 = [], ∃ m, n. ¬ # x0 < m ∧ ¬ # x0 < n ∧ m ≠ n ∧ (a ' + x0)[m] = (a ' + x0)[n]

Rule argument

Additional information

Used simplifier rules: ⊦ # x = 0 ↔ x = []; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic