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Proof step 25 from proof heap-order-min (previous step)

kar[m6 / 2] = key-index(key2, m4), kar[0] = key-index(key0, m2), kar[m3 + 1] = key-index(key, m1), kar[m6 + 1] = key-index(key1, m), kar[m3 / 2] = key-index(key3, m5), m6 + 1 < # kar, m3 / 2 < # kar, m3 / 2 < n0, m3 + 1 < # kar, n0 < # kar, key < key0, m3 < n0, m3 < # kar, m3 / 2 + 1 < # kar, m6 + 1 < n0, key1 < key2, 2 + m6 < # kar, ∀ m0. ¬ n0 < m0 ∧ m0 ≠ 0 → ¬ kar[m0].key < kar[(m0 - 1) / 2].key ⊦

User applied rule all left

And closed the goal!

Rule argument

pos: left 18 terms: m6 + 1 abort: F discard: T computed: F

Additional information

Used simplifier rules: ⊦ m < n + 1 ↔ ¬ n < m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m + 1 ≠ 0; (s)% ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (s)% ⊦ n + 0 = n; (s)% ⊦ key-index(key, m).key = key; (s)% ⊦ ¬ m < n → ¬ m + m0 < n; (s) Used terms: m6 + 1