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Proof step 19 from proof heap-order-min (previous step)

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

User applied rule all left

And closed the goal!

Rule argument

pos: left 8 terms: 2 abort: F discard: T computed: F

Additional information

Used simplifier rules: ⊦ m ≠ 0 → (m < 2 ↔ m = 1); (s)% ⊦ key-index(key, m).key = key; (s) Used terms: 2