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

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]

User applied rule exists right

And closed the goal!

Rule argument

pos: right 3 terms: n0, n abort: F discard: T computed: T

Additional information

Used simplifier rules: lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic1%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1%ls-02: ⊦ m < n → ¬ n < m; from specification nat-basic1 Used terms: n0, n