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Proof step 39 from proof prepost-insert (previous step)

kar0[m0 / 2] = key-index(key0, m1), key-index(key1, m) = kar[m0], subarray(kar0, 0, m0 + 1) = subarray(kar, 0, m0 + 1), 2exp-size(# kar), 2exp-size(# kar0), m0 < # kar, m0 + 1 < # kar0, m < # ar, m0 < # kar0, heap-order(kar, m0 + 1), heap-order(kar0, m0 + 1), no-dups-idx(kar, m0 + 1), no-dups-idx(kar0, m0 + 1), valid-idx(kar, ar, m0 + 1), valid-idx(kar0, ar, m0 + 1), ∀ ki. ki = key-index(key1, ki.idx) ∨ ki ∈ keyset(kar, m0) → ki.idx ≠ m ⊦ #(keyset(kar, m0) ++ key-index(key1, m)) = 0, no-dups-idx(kar0[m0 + 1, key-index(key, m)], 2 + m0) ∧ valid-idx(kar0[m0 + 1, key-index(key, m)], ar[m, ar[m].idx:= (m0 + 1)], 2 + m0) ∧ (¬ key < key0 → heap-order(kar0[m0 + 1, key-index(key, m)], 2 + m0)) ∧ (key < key0 → heap-order(kar0[m0 + 1, key-index(key0, m1)], 2 + m0))

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 16 terms: key-index(key1, m)

Additional information

Used simplifier rules: ⊦ key-index(key, m).idx = m; (s) Used terms: key-index(key1, m)