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Proof step 11 from proof heap-order-min (previous step)

kar[0] = key-index(key0, m2), kar[m3 + 1] = key-index(key, m1), kar[m0] = key-index(key1, m), m3 + 1 < # kar, n0 < # kar, key < key0, m3 < n0, m3 < # kar, kar[m3 / 2].key < key0 ∧ m3 / 2 < n0 ∧ (m0 < n0 ∧ m0 ≠ 0 → ¬ key1 < kar[(m0 - 1) / 2].key) → m3 = 0 ∨ m3 = 1, ∀ m0. ¬ n0 < m0 ∧ m0 ≠ 0 → ¬ kar[m0].key < kar[(m0 - 1) / 2].key ⊦

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 12:
kar[m0] = key-index(key1, m), kar[m3 + 1] = key-index(key, m1), kar[0] = key-index(key0, m2), kar[(m0 - 1) / 2] = key-index(key2, m4), m3 < # kar, m3 < n0, key < key0, n0 < # kar, m3 + 1 < # kar, kar[m3 / 2].key < key0 ∧ m3 / 2 < n0 ∧ (m0 < n0 ∧ m0 ≠ 0 → ¬ key1 < key2) → m3 = 0 ∨ m3 = 1, ∀ m0. ¬ n0 < m0 ∧ m0 ≠ 0 → ¬ kar[m0].key < kar[(m0 - 1) / 2].key ⊦

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 - 1) / 2], key2, m4]

Additional information

Used simplifier rules: ⊦ key = ki.key ∧ m = ki.idx ↔ ki = key-index(key, m); (elim) Used terms: kar[(m0 - 1) / 2], key2, m4