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Proof step 17 from proof heap-order-overwrite-children-ok (previous step)

kar[m0] = key-index(key2, m4), kar[2 * m0 + 1] = key-index(key4, m6), kar[2 * m0 + 2] = key-index(key3, m5), kar[(m0 * 2 + n0) + 1] = key-index(key0, m2), (m0 * 2 + n0) + 1 < # kar, m0 < # kar, (m0 * 2 + n0) + 1 < n, key0 < key1, m0 < n, n0 < 2, 2 * m0 + 1 < n, 2 * m0 + 1 < # kar, n ≤ # kar, 2 * m0 + 2 < n → ¬ key3 < key1, ∀ m. m < n ∧ m ≠ 0 → ¬ kar[m].key < kar[(m - 1) / 2].key ⊦ key1 < key2, key4 < key1, key4 < key2

User applied rule constructor cut

And closed the goal!

Rule argument

var: n0

Additional information

Used simplifier rules: nat freely generated by 0, +1;;% ⊦ m +1 = m + 1; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%n < n0 ⊦ m + (* n) < (* n0) ↔ m < n0 - n; (s)% ⊦ m < 1 ↔ m = 0; (s)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%mult-c: ⊦ m * n = n * m; from specification nat-mult%nnn: ⊦ ¬ key < key0 ∧ ¬ key1 < key → ¬ key1 < key0; (forward) from specification oelem[binheap-key]% ⊦ key-index(key, m) = key-index(key0, m0) ↔ key = key0 ∧ m = m0; (ws)