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Proof step 8 from proof heap-order-min (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 9:
kar[m] = key-index(key, m1), kar[0] = key-index(key0, m2), m = m3 + 1, m < # kar, key < key0, n0 < # kar, m ≤ n0, kar[m3 / 2].key < key0 ∧ m3 / 2 < n0 ∧ (m0 < n0 ∧ m0 ≠ 0 → ¬ kar[m0].key < kar[(m0 - 1) / 2].key) → m3 = 0 ∨ m3 = 1, ∀ m0. ¬ n0 < m0 ∧ m0 ≠ 0 → ¬ kar[m0].key < kar[(m0 - 1) / 2].key ⊦ m = 0

Rule argument

spec: nat name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [m, m3]

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: m, m3