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

kar[m0, kar[m]][m1] = key-index(key, kar[m0, kar[m]][m2].idx), m2 < m, m1 < m, m1 < # kar, m < # kar, m2 + 1 < # kar, m2 < # kar, m1 + 1 < # kar, ∀ m1, m2. m1 < # kar ∧ ¬ m < m1 ∧ m2 < # kar ∧ ¬ m < m2 ∧ m1 ≠ m2 → kar[m1].idx ≠ kar[m2].idx ⊦ m1 = m2

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
kar[m0, kar[m]][m1] = key-index(key, m3), kar[m0, kar[m]][m2] = key-index(key0, m3), m1 + 1 < # kar, m2 < # kar, m2 + 1 < # kar, m < # kar, m1 < # kar, m1 < m, m2 < m, ∀ m1, m2. m1 < # kar ∧ ¬ m < m1 ∧ m2 < # kar ∧ ¬ m < m2 ∧ m1 ≠ m2 → kar[m1].idx ≠ kar[m2].idx ⊦ m1 = 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[m0, kar[m]][m2], key0, m3]

Additional information

Used simplifier rules: ⊦ key = ki.key ∧ m = ki.idx ↔ ki = key-index(key, m); (elim) Used terms: kar[m0, kar[m]][m2], key0, m3