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Proof step 5 from proof heap-order-min (previous step)

kar[m] = key-index(key, m1), n0 < # kar, key < kar[0].key, m < # kar, m ≤ n0, ∀ m0. ¬ n0 < m0 ∧ m0 ≠ 0 → ¬ kar[m0].key < kar[(m0 - 1) / 2].key, ∀ kar0, m0. kar0[m0].key < kar0[0].key ∧ m0 < n0 ∧ (∀ m. m < n0 ∧ m ≠ 0 → ¬ kar0[m].key < kar0[(m - 1) / 2].key) ∧ n0 ≤ # kar0 → m0 = 0 ⊦ m = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
kar[m] = key-index(key, m1), kar[0] = key-index(key0, m2), m < # kar, key < key0, n0 < # kar, m ≤ n0, ∀ kar0, m0. kar0[m0].key < kar0[0].key ∧ m0 < n0 ∧ (∀ m. m < n0 ∧ m ≠ 0 → ¬ kar0[m].key < kar0[(m - 1) / 2].key) ∧ n0 ≤ # kar0 → m0 = 0, ∀ m0. ¬ n0 < m0 ∧ m0 ≠ 0 → ¬ kar[m0].key < kar[(m0 - 1) / 2].key ⊦ m = 0

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], key0, m2]

Additional information

Used simplifier rules: ⊦ key = ki.key ∧ m = ki.idx ↔ ki = key-index(key, m); (elim) Used terms: kar[0], key0, m2