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Proof step 21 from proof heap-order-pull-up-child (previous step)

m1 % 2 = 1, kar[(m1 - 1) + 2] = key-index(key0, m4), kar[(m1 - 1) + 1] = key-index(key1, m3), kar[m1 + 1] = key-index(key3, m5), kar[m1 / 2] = key-index(key4, m6), key3 < key1, key4 < key1, key0 < key1, (m1 - 1) + 2 < n, m1 / 2 < n, (m1 - 1) + 1 < n, key3 < key0, m1 + 1 < n, m1 + 1 < # kar, m1 / 2 < # kar, (m1 - 1) + 1 < # kar, key4 < key0, (m1 - 1) + 2 < # kar, n ≤ # kar, ∀ m. m < n ∧ m ≠ 0 → ¬ kar[m].key < kar[(m - 1) / 2].key ⊦ key3 < key4

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: nat name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [m1, m0]

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%m0 < n, n ≠ 0 ⊦ (m + (* m0)) % (* n) = (* m0) ↔ m % n = 0; (s)% ⊦ m + 1 ≠ 0; (s)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat% ⊦ key-index(key, m) = key-index(key0, m0) ↔ key = key0 ∧ m = m0; (ws)%lel: ⊦ key < key0 ∧ key = key1 → key1 < key0; (forward) from specification oelem[binheap-key]% ⊦ ¬ key < key; (s)%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: m1, m0