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Proof step 106 from proof prepost-remove (previous step)

ar[m] = elemindex(a, m1), kar[m1] = key-index(key, m), m0 = m2 + 1, 2exp-size(# kar), m1 < m0, m1 < # kar, heap-order(kar, m0), no-dups-idx(kar, m0), valid-idx(kar, ar, m0), key-index(key, m) ∈ keyset(kar, m0), m0 ≤ # kar, # ar ≤ kar[m2].idx ⊦ # keyset(kar, m0) = 0, keyset(kar, m0) = ∅, m0 = 1, m0 = 0, m2 = m1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 107:
ar[m] = elemindex(a, m1), kar[m1] = key-index(key, m), 2exp-size(# kar), m1 < m2, m1 < # kar, m2 < # kar, m1 + 1 < # kar, heap-order(kar, m2 + 1), no-dups-idx(kar, m2 + 1), valid-idx(kar, ar, m2 + 1), # ar ≤ kar[m2].idx, key-index(key, m) = kar[m2] ∨ key-index(key, m) ∈ keyset(kar, m2) ⊦ #(keyset(kar, m2) ++ kar[m2]) = 0, m2 = m1

Rule argument

Additional information

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