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Proof step 21 from proof ri-range-ext (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 22:
nused = m + 1, scan(key, zbrar, 0, nused) = nused, nused < # zbrar, sorted(zbrar, 0, nused), ∀ n. n ≤ nused → key-range(zbrar, n, nused + 1, ri0) = key-range(zbrar, n, nused + 1, ri1), ∀ n0. n0 < nused → zbrar[n0].key < key ⊦ nused = 0, key ∈ ri0 ∧ zbrar[m].key < key, key ∈ ri1 ∧ zbrar[m].key < key, (key ∈ ri0 ↔ key ∈ ri1) ∧ (key ∈ ri0 → ri0[key] = ri1[key])

Rule argument

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

Additional information

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