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Proof step 12 from proof dups-def (previous step)

∃ m, n. ¬ # x0 < m ∧ ¬ # x0 < n ∧ m ≠ n ∧ (a ' + x0)[m] = (a ' + x0)[n] ⊦ a ∈ x0, dups(x0), ∃ 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 13:
(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]

Rule argument

Additional information

Used simplifier rules: lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat