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Proof step 50 from proof prepost-insert (previous step)

kar0[m, key-index(key, m0)][m2 / 2] = key-index(key0, m1), m = m2 + 1, subarray(kar0, 0, m) = subarray(kar, 0, m), 2exp-size(# kar), 2exp-size(# kar0), m0 < # ar, m < # kar0, key < key0, heap-order(kar, m), heap-order(kar0, m), no-dups-idx(kar0[m, key-index(key, m0)], m + 1), no-dups-idx(kar, m), no-dups-idx(kar0, m), valid-idx(kar0, ar, m), valid-idx(kar, ar, m), m ≤ # kar, ∀ ki0. ki0 ∈ keyset(kar, m) → ki0.idx ≠ m0 ⊦ m = 0, heap-order(kar0[m, key-index(key0, m1)], m + 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 51:
kar0[m2 / 2] = key-index(key0, m1), subarray(kar0, 0, m2 + 1) = subarray(kar, 0, m2 + 1), 2exp-size(# kar), 2exp-size(# kar0), m2 < # kar, m0 < # ar, m2 + 1 < # kar0, key < key0, heap-order(kar, m2 + 1), heap-order(kar0, m2 + 1), no-dups-idx(kar0[m2 + 1, key-index(key, m0)], 2 + m2), no-dups-idx(kar, m2 + 1), no-dups-idx(kar0, m2 + 1), valid-idx(kar0, ar, m2 + 1), valid-idx(kar, ar, m2 + 1), ∀ ki. ki = kar[m2] ∨ ki ∈ keyset(kar, m2) → ki.idx ≠ m0 ⊦ heap-order(kar0[m2 + 1, key-index(key0, m1)], 2 + m2)

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ ki ∈ s ++ ki0 ↔ ki = ki0 ∨ ki ∈ s; (s)% ⊦ n < # kar → keyset(kar, n + 1) = keyset(kar, n) ++ kar[n]; (s)% ⊦ m + 1 ≤ n ↔ m < n; (s)% ⊦ m + 1 ≠ 0; (s)%m ≠ n ⊦ kar[m, ki][n] = kar[n]; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m / 2 ≠ m + 1; (s)