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Proof step 6 from proof heap-order-min (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 7:
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 < kar[0].key ∧ (m - 1) / 2 < n0 ∧ (∀ m. m < n0 ∧ m ≠ 0 → ¬ kar[m].key < kar[(m - 1) / 2].key) ∧ n0 ≤ # kar → (m - 1) / 2 = 0, ∀ m0. ¬ n0 < m0 ∧ m0 ≠ 0 → ¬ kar[m0].key < kar[(m0 - 1) / 2].key ⊦ m = 0

Rule argument

pos: left 7 terms: kar, (m - 1) / 2 abort: F discard: T computed: F

Additional information

Used terms: kar, (m - 1) / 2