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Proof step 93 from proof prepost-update (previous step)

ar[m] = elemindex(a, m1), kar[m1] = key-index(key1, m), kar[2 * m1 + 2] = key-index(key0, m2), kar[2 * m1 + 1] = key-index(key2, m3), ar0 = ar, bh = bin-heap(kar[m1, key-index(key, m)], m0), 2exp-size(# kar), m1 < # kar, m1 < m0, heap-order(kar, m0), no-dups-idx(kar, m0), valid-idx(kar, ar, m0), key-index(key1, m) ∈ keyset(kar, m0), m0 ≤ # kar, (2 * m1 + 1 < m0 → ¬ key2 < key) ∧ (2 * m1 + 2 < m0 → ¬ key0 < key) → heap-order(kar[m1, key-index(key, m)], m0), 2 * m1 + 1 < m0 ∧ key2 < key ∨ 2 * m1 + 2 < m0 ∧ key0 < key → heap-order(kar[m1, (2 * m1 + 2 < m0 ∧ key0 < key2 ⊃ key-index(key0, m2);key-index(key2, m3))], m0) ⊦ # keyset(kar, m0) = 0, keyset(kar, m0) = ∅, key < key1, ∀ bh, ar. bh.size = m0 ∧ # bh.ar = # kar ∧ id-idx(ar0, ar) ∧ no-dups-idx(bh.ar, bh.size) ∧ valid-idx(bh.ar, ar, bh.size) ∧ heap-order(bh.ar, bh.size) ∧ (keyset(kar, bh.size) -- kar[m1]) ++ key-index(key, m) = keyset(bh.ar, bh.size) → bh.size ≤ # bh.ar ∧ 2exp-size(# bh.ar) ∧ keyset(bh.ar, bh.size) = (keyset(kar, bh.size) -- key-index(key1, m)) ++ key-index(key, m)

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic[keyindex-set]%ne-01: ⊦ ki ∈ s → s ≠ ∅; (forward) from specification set-basic[keyindex-set]%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic