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Proof step 8 from proof prepost-update (previous step)

kar[m1] = key-index(key0, m), ar[m] = elemindex(a, m1), ar0 = ar, bh = bin-heap(kar, m0), n = m1, 2exp-size(# kar), heap-order(kar, m0), no-dups-idx(kar, m0), valid-idx(kar, ar, m0), key-index(key0, m) ∈ keyset(kar, m0), m0 ≤ # kar ⊦ ⟪if ¬ n < bh.size then abort; let key0 = bh.ar[n].key in {bh.ar[n].key := key; if key < key0 then bin_min_heap_bubble_up#(n; bh, ar) else bin_min_heap_bubble_down#(n; bh, ar)}⟫ ( bh.size ≤ # bh.ar ∧ no-dups-idx(bh.ar, bh.size) ∧ valid-idx(bh.ar, ar, bh.size) ∧ 2exp-size(# bh.ar) ∧ heap-order(bh.ar, bh.size) ∧ id-idx(ar0, ar) ∧ keyset(bh.ar, bh.size) = (keyset(kar, m0) -- key-index(key0, m)) ++ key-index(key, m)), # keyset(kar, m0) = 0, keyset(kar, m0) = ∅

Heuristic conditional right split applied rule if right

And obtained the following new goals

Rule argument

pos: right 1

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