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Proof step 6 from proof valid-idx-remlast (previous step)

kar[m] = key-index(key, n), m0 < # kar, m0 < m, m < # kar, m0 + 1 < # kar, ∀ m0. ¬ m < m0 → kar[m0].idx < # ar ∧ ar[kar[m0].idx].idx = m0 ⊦ kar[0, key-index(key, n)][m0].idx < # ar ∧ ar[n, ar[n].idx:= 0][kar[0, key-index(key, n)][m0].idx].idx = m0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 7:
kar[m] = key-index(key, n), kar[0, key-index(key, n)][m0] = key-index(key0, m1), m0 + 1 < # kar, m < # kar, m0 < m, m0 < # kar, ∀ m0. ¬ m < m0 → kar[m0].idx < # ar ∧ ar[kar[m0].idx].idx = m0 ⊦ m1 < # ar ∧ ar[n, ar[n].idx:= 0][m1].idx = 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[0, key-index(key, n)][m0], key0, m1]

Additional information

Used simplifier rules: ⊦ key = ki.key ∧ m = ki.idx ↔ ki = key-index(key, m); (elim) Used terms: kar[0, key-index(key, n)][m0], key0, m1