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Proof step 20 from proof ri-range-ext (previous step)

nused < # zbrar, scan(key, zbrar, 0, nused) < # zbrar, sorted(zbrar, 0, nused), scan(key, zbrar, 0, nused) ≤ nused, scan(key, zbrar, 0, nused) < nused → ¬ zbrar[scan(key, zbrar, 0, nused)].key < key, ∀ n1. scan(key, zbrar, 0, nused) < n1 ∧ n1 < nused → key < zbrar[n1].key, ∀ n0. n0 < scan(key, zbrar, 0, nused) → zbrar[n0].key < key, ∀ n. n ≤ nused → key-range(zbrar, n, nused + 1, ri0) = key-range(zbrar, n, nused + 1, ri1) ⊦ scan(key, zbrar, 0, nused) = 0, key ∈ key-range(zbrar, scan(key, zbrar, 0, nused), nused + 1, ri1), key ∈ key-range(zbrar, scan(key, zbrar, 0, nused), nused + 1, ri0), scan(key, zbrar, 0, nused) ≠ 0 ∧ scan(key, zbrar, 0, nused) < nused, (key ∈ ri0 ↔ key ∈ ri1) ∧ (key ∈ ri0 → ri0[key] = ri1[key])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
scan(key, zbrar, 0, nused) = nused, nused < # zbrar, sorted(zbrar, 0, nused), ∀ n0. n0 < nused → zbrar[n0].key < key, ∀ n. n ≤ nused → key-range(zbrar, n, nused + 1, ri0) = key-range(zbrar, n, nused + 1, ri1) ⊦ nused = 0, (key ∈ ri0 ↔ key ∈ ri1) ∧ (key ∈ ri0 → ri0[key] = ri1[key]), key ∈ ri1 ∧ zbrar[nused - 1].key < key, key ∈ ri0 ∧ zbrar[nused - 1].key < key

Rule argument

Additional information

Used simplifier rules: ⊦ n < n0 → (n0 < n ↔ false); (ss)% ⊦ ¬ n < n; (s)% ⊦ key0 ∈ ri |> key1 ↔ key0 ∈ ri ∧ key1 < key0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%⊦ n ≠ 0 ∧ n < # zbrar → key-range(zbrar, n, n + 1, ri) = ri |> zbrar[n - 1].key; (s)% ⊦ m ≤ m; (s)% ⊦ 0 ≤ m; (s)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%scan-min-eq: ⊦ m0 = scan(key, zbrar, m1, len) → m1 ≤ m0; (forward) from specification znode-ops% ⊦ m ≤ n → (n ≤ m ↔ m = n); (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)