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Proof step 16 from proof heap-order-min (previous step)

kar[m3 / 2] = key-index(key3, m5), 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, key3 < 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 ⊦ (m0 < n0 ∧ m0 ≠ 0 → ¬ key1 < key2) ∧ m3 / 2 < n0 ∧ m3 ≠ 0 ∧ m3 ≠ 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
kar[(m0 - 1) / 2] = key-index(key2, m4), kar[0] = key-index(key0, m2), kar[m3 + 1] = key-index(key, m1), kar[m0] = key-index(key1, m), kar[m3 / 2] = key-index(key3, m5), m3 / 2 < # kar, m3 / 2 < n0, m3 + 1 < # kar, n0 < # kar, key < key0, m3 < n0, m3 < # kar, m3 / 2 + 1 < # kar, key3 < key0 ∧ (m0 < n0 ∧ m0 ≠ 0 → ¬ key1 < key2) → m3 = 0 ∨ m3 = 1, ∀ m0. ¬ n0 < m0 ∧ m0 ≠ 0 → ¬ kar[m0].key < kar[(m0 - 1) / 2].key ⊦ m3 ≠ 1 ∧ m3 ≠ 0 ∧ (m0 < n0 ∧ m0 ≠ 0 → ¬ key1 < key2)

Rule argument

Additional information

Used simplifier rules: ⊦ m +1 = m + 1; (s)%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic% ⊦ ¬ m < m / 2; (s)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat% ⊦ m < n ↔ ¬ n ≤ m; (ss)