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Proof step 33 from proof heap-order-pull-up-child (previous step)

kar[3 + 2 * m1] = key-index(key, m3), kar[4 + 2 * m1] = key-index(key2, m4), kar[m1 + 1] = key-index(key3, m5), kar[m1 / 2] = key-index(key4, m6), 3 + 2 * m1 < # kar, m1 + 1 < # kar, m1 + 1 < n, key < key4, 3 + 2 * m1 < n, n ≤ # kar, ∀ m. m < n ∧ m ≠ 0 → ¬ kar[m].key < kar[(m - 1) / 2].key, ¬ 4 + 2 * m1 < n ∨ ¬ key2 < key ⊦ key3 < key4

Heuristic quantifier applied rule all left

And closed the goal!

Rule argument

pos: left 11 terms: 3 + 2 * m1

Additional information

Used simplifier rules: n < n0 ⊦ (* n) ≠ m0 + (* n0); (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (s)% ⊦ (2 * m + n) / 2 = m + n / 2; (s)%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult% ⊦ key-index(key, m).key = key; (s)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%nnn: ⊦ ¬ key < key0 ∧ ¬ key1 < key → ¬ key1 < key0; (forward) from specification oelem[binheap-key]%nll: ⊦ ¬ key < key0 ∧ key < key1 → key0 < key1; (forward) from specification oelem[binheap-key]% ⊦ m < n ↔ ¬ n ≤ m; (ss) Used terms: 3 + 2 * m1