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Proof step 23 from proof no-dups-idx-xchg (previous step)

kar[m0] = key-index(key, m3), kar[m2] = key-index(key0, m3), m0 < # kar, m0 < n, m1 < # kar, m1 < n, m2 < # kar, m2 < n, ∀ m0, m1. m0 < # kar ∧ m0 < n ∧ m1 < # kar ∧ m1 < n ∧ m0 ≠ m1 → kar[m0].idx ≠ kar[m1].idx ⊦ m1 = m2, m2 = m0

Heuristic quantifier applied rule all left

And closed the goal!

Rule argument

pos: left 9 terms: m2, m0

Additional information

Used simplifier rules: ⊦ key-index(key, m).idx = m; (s) Used terms: m2, m0