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Proof step 34 from proof ri-range-ext (previous step)

scan(key, zbrar, 0, nused) = m + 1, scan(key, zbrar, 0, nused) < nused, nused < # zbrar, scan(key, zbrar, 0, nused) < # zbrar, scan(key, zbrar, 0, nused) + 1 < # zbrar, sorted(zbrar, 0, nused), ∀ 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, zbrar[scan(key, zbrar, 0, nused)].key < key, key ∈ ri1 ∧ zbrar[scan(key, zbrar, 0, nused) - 1].key < key, key ∈ ri0 ∧ zbrar[m].key < key, (key ∈ ri0 ↔ key ∈ ri1) ∧ (key ∈ ri0 → ri0[key] = ri1[key])

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 35:
scan(key, zbrar, 0, nused) = m + 1, scan(key, zbrar, 0, nused) = m0 + 1, scan(key, zbrar, 0, nused) + 1 < # zbrar, scan(key, zbrar, 0, nused) < # zbrar, nused < # zbrar, scan(key, zbrar, 0, nused) < nused, sorted(zbrar, 0, nused), ∀ n. n ≤ nused → key-range(zbrar, n, nused + 1, ri0) = key-range(zbrar, n, nused + 1, ri1), ∀ n0. n0 < scan(key, zbrar, 0, nused) → zbrar[n0].key < key, ∀ n1. scan(key, zbrar, 0, nused) < n1 ∧ n1 < nused → key < zbrar[n1].key ⊦ scan(key, zbrar, 0, nused) = 0, zbrar[scan(key, zbrar, 0, nused)].key < key, (key ∈ ri0 ↔ key ∈ ri1) ∧ (key ∈ ri0 → ri0[key] = ri1[key]), key ∈ ri0 ∧ zbrar[m].key < key, key ∈ ri1 ∧ zbrar[m0].key < key

Rule argument

spec: nat name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [scan(key, zbrar, 0, nused), m0]

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: scan(key, zbrar, 0, nused), m0