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Proof step 67 from proof prepost-bubble-up (previous step)

# kar0 = # kar, kar[(n1 - 1) / 2] = key-index(key0, m1), kar[n1] = key-index(key, m0), keyset(kar0, m2) = keyset(kar, m2), kar0[m / 2].idx < kar0[m + 1].idx, kar0[m / 2].idx < # ar, m + 1 < # kar, kar0[m + 1].key < kar0[m / 2].key, m + 1 < m2, m / 2 < # kar, n1 < # kar, n1 < m2, heap-order(kar0[m + 1, kar0[m / 2]], m2), id-idx(ar1, ar), no-dups-idx(kar, m2), no-dups-idx(kar0, m2), valid-idx(kar, ar1, m2), valid-idx(kar0, ar, m2), m2 ≤ # kar, # ar ≤ kar0[m + 1].idx, n1 ≠ 0 ∧ key < key0 → heap-order(kar[n1, key-index(key0, m1)], m2), n1 = 0 ∨ ¬ key < key0 → heap-order(kar, m2) ⊦ # kar = 1, m2 = 1

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: key-index name: elim seq: ⊦ key = ki.key ∧ m = ki.idx ↔ ki = key-index(key, m) subst: [ki, key, m] → [kar0[m + 1], key1, m3]

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n → (n ≤ m ↔ false); (s)%m < m0 ⊦ kar[m] = key-index(key, n) ∧ valid-idx(kar, ar, m0) → ¬ # ar ≤ n; (s)% ⊦ key = ki.key ∧ m = ki.idx ↔ ki = key-index(key, m); (elim) Used terms: kar0[m + 1], key1, m3