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Proof step 6 from proof valid-idx-imp-1 (previous step)

m < # kar, m < n, n < # kar, m + 1 < # kar, ∀ m. ¬ n < m → kar[m].idx < # ar ∧ ar[kar[m].idx].idx = m ⊦ kar[m].idx < # ar ∧ ar[kar[m].idx].idx = m

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 7:
kar[m] = key-index(key, m0), m + 1 < # kar, n < # kar, m < n, m < # kar, ∀ m. ¬ n < m → kar[m].idx < # ar ∧ ar[kar[m].idx].idx = m ⊦ m0 < # ar ∧ ar[m0].idx = m

Rule argument

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

Additional information

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