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Proof step 24 from proof heap-order-min (previous step)

kar[m6 / 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), m0 = m6 + 1, m0 + 1 < # kar, m3 / 2 < # kar, m3 / 2 < n0, m3 + 1 < # kar, n0 < # kar, key < key0, m3 < n0, m3 < # kar, m3 / 2 + 1 < # kar, m0 < n0, key1 < key2, m0 < # kar, ∀ m0. ¬ n0 < m0 ∧ m0 ≠ 0 → ¬ kar[m0].key < kar[(m0 - 1) / 2].key ⊦ m0 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
kar[m6 / 2] = key-index(key2, m4), kar[0] = key-index(key0, m2), kar[m3 + 1] = key-index(key, m1), kar[m6 + 1] = key-index(key1, m), kar[m3 / 2] = key-index(key3, m5), m6 + 1 < # kar, m3 / 2 < # kar, m3 / 2 < n0, m3 + 1 < # kar, n0 < # kar, key < key0, m3 < n0, m3 < # kar, m3 / 2 + 1 < # kar, m6 + 1 < n0, key1 < key2, 2 + m6 < # kar, ∀ m0. ¬ n0 < m0 ∧ m0 ≠ 0 → ¬ kar[m0].key < kar[(m0 - 1) / 2].key ⊦

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%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≠ 0; (s)