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Proof step 7 from proof valid-idx-imp-1 (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 8:
kar[m] = key-index(key, m0), ar[m0] = elemindex(a, m1), m < # kar, m < n, n < # kar, m + 1 < # kar, ∀ m. ¬ n < m → kar[m].idx < # ar ∧ ar[kar[m].idx].idx = m ⊦ m0 < # ar ∧ m1 = m

Rule argument

spec: elem-index name: elim seq: ⊦ a = ei.elem ∧ m = ei.idx ↔ ei = elemindex(a, m) subst: [ei, a, m] → [ar[m0], a, m1]

Additional information

Used simplifier rules: ⊦ a = ei.elem ∧ m = ei.idx ↔ ei = elemindex(a, m); (elim) Used terms: ar[m0], a, m1