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Proof step 4 from proof heap-order-pull-up-child (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
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])][m1 + 1].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, m1 + 1 < n, m < # kar, 2 * m + 1 < # kar, m1 + 1 < # kar, n ≤ # kar, ∀ m. m < n ∧ m ≠ 0 → ¬ kar[m].key < kar[(m - 1) / 2].key ⊦

Rule argument

Additional information

Used simplifier rules: fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≠ 0; (s)