[Imprint][Data Privacy Statement][E-Mail]

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Proof step 18 from proof prepost-bubble-up (previous step)

kar[(n1 - 1) / 2] = key-index(key0, m1), kar[n1] = key-index(key, m0), ar0 = ar1, bh0 = bin-heap(kar, m2), ki = bh.ar[n], ki0 = bh.ar[(n - 1) / 2], n = n0, (n - 1) / 2 < # (bh.ar:= bh.ar[n, ki0]).ar, ki0.idx < # ar, (n - 1) / 2 < # bh.ar, n < # bh.ar, bh.ar[n].key < bh.ar[(n - 1) / 2].key, n1 < # kar, n1 < m2, bubble-up-inv(n, bh, ar, bin-heap(kar, m2), ar0), no-dups-idx(kar, m2), valid-idx(kar, ar1, m2), m2 ≤ # kar, n ≥ 1, n ≠ 0 → n < # bh.ar ∧ (n - 1) / 2 < # bh.ar ∧ 2 ≠ 0 ∧ n ≥ 1, n1 ≠ 0 ∧ key < key0 → heap-order(kar[n1, key-index(key0, m1)], m2), n1 = 0 ∨ ¬ key < key0 → heap-order(kar, m2) ⊦ ⟪ar := ar[ki0.idx, ar[ki0.idx].idx:= n][ki.idx, ar[ki0.idx, ar[ki0.idx].idx:= n][ki.idx].idx:= ((n - 1) / 2)]⟫ ( ( ( (n - 1) / 2 ≠ 0 → (n - 1) / 2 < # (bh.ar:= bh.ar[n, ki0].ar:= (bh.ar:= bh.ar[n, ki0]).ar[(n - 1) / 2, ki]).ar ∧ ((n - 1) / 2 - 1) / 2 < # (bh.ar:= bh.ar[n, ki0].ar:= (bh.ar:= bh.ar[n, ki0]).ar[(n - 1) / 2, ki]).ar ∧ 2 ≠ 0 ∧ (n - 1) / 2 ≥ 1) ∧ ( (n - 1) / 2 ≠ 0 → (n - 1) / 2 < # (bh.ar:= bh.ar[n, ki0].ar:= (bh.ar:= bh.ar[n, ki0]).ar[(n - 1) / 2, ki]).ar ∧ ((n - 1) / 2 - 1) / 2 < # (bh.ar:= bh.ar[n, ki0].ar:= (bh.ar:= bh.ar[n, ki0]).ar[(n - 1) / 2, ki]).ar ∧ 2 ≠ 0 ∧ (n - 1) / 2 ≥ 1) ∧ ( (n - 1) / 2 ≠ 0 → (n - 1) / 2 < # (bh.ar:= bh.ar[n, ki0].ar:= (bh.ar:= bh.ar[n, ki0]).ar[(n - 1) / 2, ki]).ar ∧ ((n - 1) / 2 - 1) / 2 < # (bh.ar:= bh.ar[n, ki0].ar:= (bh.ar:= bh.ar[n, ki0]).ar[(n - 1) / 2, ki]).ar ∧ 2 ≠ 0 ∧ (n - 1) / 2 ≥ 1) ∧ ( (n - 1) / 2 ≠ 0 → (n - 1) / 2 < # (bh.ar:= bh.ar[n, ki0].ar:= (bh.ar:= bh.ar[n, ki0]).ar[(n - 1) / 2, ki]).ar ∧ ((n - 1) / 2 - 1) / 2 < # (bh.ar:= bh.ar[n, ki0].ar:= (bh.ar:= bh.ar[n, ki0]).ar[(n - 1) / 2, ki]).ar ∧ 2 ≠ 0 ∧ (n - 1) / 2 ≥ 1) ∧ ( (n - 1) / 2 ≠ 0 → (n - 1) / 2 < # (bh.ar:= bh.ar[n, ki0].ar:= (bh.ar:= bh.ar[n, ki0]).ar[(n - 1) / 2, ki]).ar ∧ ((n - 1) / 2 - 1) / 2 < # (bh.ar:= bh.ar[n, ki0].ar:= (bh.ar:= bh.ar[n, ki0]).ar[(n - 1) / 2, ki]).ar ∧ 2 ≠ 0 ∧ (n - 1) / 2 ≥ 1) ∧ bubble-up-inv((n - 1) / 2, bh.ar:= bh.ar[n, ki0].ar:= (bh.ar:= bh.ar[n, ki0]).ar[(n - 1) / 2, ki], ar, bin-heap(kar, m2), ar0)) ∧ (n - 1) / 2 < n0), 2 = 0, n = 0

Heuristic symbolic execution applied rule assign right

And obtained the following new goals

Proof step 19:
kar[n1] = key-index(key, m0), kar[(n1 - 1) / 2] = key-index(key0, m1), ar0 = ar1, bh0 = bin-heap(kar, m2), ki = bh.ar[n], ki0 = bh.ar[(n - 1) / 2], n = n0, ki.idx < # ar[ki0.idx, ar[ki0.idx].idx:= n], n1 < m2, n1 < # kar, bh.ar[n].key < bh.ar[(n - 1) / 2].key, n < # bh.ar, (n - 1) / 2 < # bh.ar, ki0.idx < # ar, (n - 1) / 2 < # (bh.ar:= bh.ar[n, ki0]).ar, bubble-up-inv(n, bh, ar, bin-heap(kar, m2), ar0), no-dups-idx(kar, m2), valid-idx(kar, ar1, m2), m2 ≤ # kar, n ≥ 1, n1 = 0 ∨ ¬ key < key0 → heap-order(kar, m2), n1 ≠ 0 ∧ key < key0 → heap-order(kar[n1, key-index(key0, m1)], m2), n ≠ 0 → n < # bh.ar ∧ (n - 1) / 2 < # bh.ar ∧ 2 ≠ 0 ∧ n ≥ 1 ⊦ 2 = 0, n = 0, ( ( (n - 1) / 2 ≠ 0 → (n - 1) / 2 < # (bh.ar:= bh.ar[n, ki0].ar:= (bh.ar:= bh.ar[n, ki0]).ar[(n - 1) / 2, ki]).ar ∧ ((n - 1) / 2 - 1) / 2 < # (bh.ar:= bh.ar[n, ki0].ar:= (bh.ar:= bh.ar[n, ki0]).ar[(n - 1) / 2, ki]).ar ∧ 2 ≠ 0 ∧ (n - 1) / 2 ≥ 1) ∧ ( (n - 1) / 2 ≠ 0 → (n - 1) / 2 < # (bh.ar:= bh.ar[n, ki0].ar:= (bh.ar:= bh.ar[n, ki0]).ar[(n - 1) / 2, ki]).ar ∧ ((n - 1) / 2 - 1) / 2 < # (bh.ar:= bh.ar[n, ki0].ar:= (bh.ar:= bh.ar[n, ki0]).ar[(n - 1) / 2, ki]).ar ∧ 2 ≠ 0 ∧ (n - 1) / 2 ≥ 1) ∧ ( (n - 1) / 2 ≠ 0 → (n - 1) / 2 < # (bh.ar:= bh.ar[n, ki0].ar:= (bh.ar:= bh.ar[n, ki0]).ar[(n - 1) / 2, ki]).ar ∧ ((n - 1) / 2 - 1) / 2 < # (bh.ar:= bh.ar[n, ki0].ar:= (bh.ar:= bh.ar[n, ki0]).ar[(n - 1) / 2, ki]).ar ∧ 2 ≠ 0 ∧ (n - 1) / 2 ≥ 1) ∧ ( (n - 1) / 2 ≠ 0 → (n - 1) / 2 < # (bh.ar:= bh.ar[n, ki0].ar:= (bh.ar:= bh.ar[n, ki0]).ar[(n - 1) / 2, ki]).ar ∧ ((n - 1) / 2 - 1) / 2 < # (bh.ar:= bh.ar[n, ki0].ar:= (bh.ar:= bh.ar[n, ki0]).ar[(n - 1) / 2, ki]).ar ∧ 2 ≠ 0 ∧ (n - 1) / 2 ≥ 1) ∧ ( (n - 1) / 2 ≠ 0 → (n - 1) / 2 < # (bh.ar:= bh.ar[n, ki0].ar:= (bh.ar:= bh.ar[n, ki0]).ar[(n - 1) / 2, ki]).ar ∧ ((n - 1) / 2 - 1) / 2 < # (bh.ar:= bh.ar[n, ki0].ar:= (bh.ar:= bh.ar[n, ki0]).ar[(n - 1) / 2, ki]).ar ∧ 2 ≠ 0 ∧ (n - 1) / 2 ≥ 1) ∧ bubble-up-inv ((n - 1) / 2, bh.ar:= bh.ar[n, ki0].ar:= (bh.ar:= bh.ar[n, ki0]).ar[(n - 1) / 2, ki], ar[ki0.idx, ar[ki0.idx].idx:= n][ki.idx, ar[ki0.idx, ar[ki0.idx].idx:= n][ki.idx].idx:= ((n - 1) / 2)], bin-heap(kar, m2), ar0)) ∧ (n - 1) / 2 < n0
Proof step 62:
kar[n1] = key-index(key, m0), kar[(n1 - 1) / 2] = key-index(key0, m1), ar0 = ar1, bh0 = bin-heap(kar, m2), ki = bh.ar[n], ki0 = bh.ar[(n - 1) / 2], n = n0, n1 < m2, n1 < # kar, bh.ar[n].key < bh.ar[(n - 1) / 2].key, n < # bh.ar, (n - 1) / 2 < # bh.ar, ki0.idx < # ar, (n - 1) / 2 < # (bh.ar:= bh.ar[n, ki0]).ar, bubble-up-inv(n, bh, ar, bin-heap(kar, m2), ar0), no-dups-idx(kar, m2), valid-idx(kar, ar1, m2), m2 ≤ # kar, n ≥ 1, n1 = 0 ∨ ¬ key < key0 → heap-order(kar, m2), n1 ≠ 0 ∧ key < key0 → heap-order(kar[n1, key-index(key0, m1)], m2), n ≠ 0 → n < # bh.ar ∧ (n - 1) / 2 < # bh.ar ∧ 2 ≠ 0 ∧ n ≥ 1 ⊦ 2 = 0, n = 0, ki.idx < # ar[ki0.idx, ar[ki0.idx].idx:= n] ∧ (ki0.idx < # ar ∧ ki0.idx < # ar) ∧ (ki.idx < # ar[ki0.idx, ar[ki0.idx].idx:= n] ∧ ki0.idx < # ar ∧ ki0.idx < # ar) ∧ 2 ≠ 0 ∧ n ≥ 1

Proof step 18 from proof prepost-bubble-up (previous step)

Heuristic symbolic execution applied rule assign right

And obtained the following new goals

Rule argument

Additional information