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Proof step 9 from proof firstn-at-in-dups (previous step)

n0 < n ∧ n ≤ # x, x[n0] = x[n], n0 < # x, n0 < n, n < # x, x[n] ∈ x, n0 + 1 < # x ⊦ n = 0, 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 10:
x[n0] = x[n], n0 < # x, n0 < n, n < # x, x[n] ∈ x, n0 + 1 < # x ⊦ n = 0, x = [], ∃ m, n. m < # x ∧ n < # x ∧ m ≠ n ∧ x[m] = x[n]

Rule argument

Additional information

Used simplifier rules: lels-03: ⊦ m < n → m ≤ n; from specification nat