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Proof step 6 from proof no-dups-idx-rem1 (previous step)

kar[m] = key-index(key, m1), kar[m0] = key-index(key0, m1), m < # kar, m < n, m0 < # kar, m0 < n, ∀ m, m0. m < # kar ∧ ¬ n < m ∧ m0 < # kar ∧ ¬ n < m0 ∧ m ≠ m0 → kar[m].idx ≠ kar[m0].idx ⊦ m = m0

User applied rule all left

And closed the goal!

Rule argument

pos: left 7 terms: m, m0 abort: F discard: F computed: T

Additional information

Used simplifier rules: ⊦ m < n → ¬ n < m; (s)% ⊦ key-index(key, m).idx = m; (s) Used terms: m, m0