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Proof step 10 from proof firstn-at-in-dups (previous step)

x[n0] = x[n], n0 + 1 < # x, n < # x, n0 < # x, n0 < n, x[n] ∈ x, n ≤ # x ⊦ x = [], ∃ m, n. m < # x ∧ n < # x ∧ m ≠ n ∧ x[m] = x[n]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
x[n0] = x[n], n0 < n, n0 < # x, n < # x, n0 + 1 < # x, x[n] ∈ x ⊦ x = [], ∃ m, n. m < # x ∧ n < # x ∧ m ≠ n ∧ x[m] = x[n]

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → m ≤ n; (s)