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Proof step 24 from proof no-dups-idx-rec (previous step)

kar[m0] = key-index(key0, m3), m0 < n, n < # kar, m0 < # kar, m0 + 1 < # kar, ∀ m0, m1. m0 < # kar ∧ m0 < n ∧ m1 < # kar ∧ m1 < n ∧ m0 ≠ m1 → kar[m0].idx ≠ kar[m1].idx ⊦ m0 = n, ∃ m. m < n ∧ key-index(key0, m3) = kar[m]

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 2 terms: m0

Additional information

Used terms: m0