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Proof step 3 from proof heap-order-pull-up-child (previous step)

m < n, 2 * m + 1 < n, kar[m, (2 * m + 2 < n ∧ kar[2 * m + 2].key < kar[2 * m + 1].key ⊃ kar[2 * m + 2];kar[2 * m + 1])][m0].key < kar[m, (2 * m + 2 < n ∧ kar[2 * m + 2].key < kar[2 * m + 1].key ⊃ kar[2 * m + 2];kar[2 * m + 1])][(m0 - 1) / 2].key, m0 < n, m < # kar, 2 * m + 1 < # kar, m0 < # kar, n ≤ # kar, ∀ m. m < n ∧ m ≠ 0 → ¬ kar[m].key < kar[(m - 1) / 2].key ⊦ m0 = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
m0 = m1 + 1, m0 < # kar, 2 * m + 1 < # kar, m < # kar, m0 < n, kar[m, (2 * m + 2 < n ∧ kar[2 * m + 2].key < kar[2 * m + 1].key ⊃ kar[2 * m + 2];kar[2 * m + 1])][m0].key < kar[m, (2 * m + 2 < n ∧ kar[2 * m + 2].key < kar[2 * m + 1].key ⊃ kar[2 * m + 2];kar[2 * m + 1])][m1 / 2].key, 2 * m + 1 < n, m < n, n ≤ # kar, ∀ m. m < n ∧ m ≠ 0 → ¬ kar[m].key < kar[(m - 1) / 2].key ⊦ m0 = 0

Rule argument

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

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: m0, m1