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Proof step 30 from proof no-dups-idx-xchg (previous step)

kar[m1] = key-index(key0, m3), kar[m] = key-index(key, m3), kar[m0] = key-index(key1, m2), m1 < # kar, m1 < n, m < # kar, m < n, m0 < # kar, m0 < n, ∀ m0, m1. m0 < # kar ∧ m0 < n ∧ m1 < # kar ∧ m1 < n ∧ m0 ≠ m1 → kar[m0].idx ≠ kar[m1].idx ⊦ m = m1, m = m0, m2 = m3

Heuristic quantifier applied rule all left

And closed the goal!

Rule argument

pos: left 10 terms: m, m1

Additional information

Used simplifier rules: ⊦ key-index(key, m).idx = m; (s) Used terms: m, m1