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Proof step 5 from proof no-dups-idx-rem1 (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
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

Rule argument

spec: key-index name: elim seq: ⊦ key = ki.key ∧ m = ki.idx ↔ ki = key-index(key, m) subst: [ki, key, m] → [kar[m0], key0, m1]

Additional information

Used simplifier rules: ⊦ key = ki.key ∧ m = ki.idx ↔ ki = key-index(key, m); (elim) Used terms: kar[m0], key0, m1