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Proof step 14 from proof no-dups-idx-xchg (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 15:
kar[m0] = key-index(key0, m3), kar[m] = key-index(key, m3), kar[m1] = key-index(key1, m2), m0 < # kar, m0 < n, m < # kar, m < n, m1 < # kar, m1 < n, ∀ m0, m1. m0 < # kar ∧ m0 < n ∧ m1 < # kar ∧ m1 < n ∧ m0 ≠ m1 → kar[m0].idx ≠ kar[m1].idx ⊦ m = m0, m3 = m2

Rule argument

spec: key-index name: elim seq: ⊦ key = ki.key ∧ m = ki.idx ↔ ki = key-index(key, m) subst: [ki, key, m] → [kar[m1], key1, m2]

Additional information

Used simplifier rules: ⊦ key = ki.key ∧ m = ki.idx ↔ ki = key-index(key, m); (elim) Used terms: kar[m1], key1, m2