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Proof step 12 from proof ri-range-ext (previous step)

key ∈ key-range(zbrar, scan(key, zbrar, 0, nused), nused + 1, ri0) ↔ key ∈ key-range(zbrar, scan(key, zbrar, 0, nused), nused + 1, ri1), key-range(zbrar, scan(key, zbrar, 0, nused), nused + 1, ri0)[key] = key-range(zbrar, scan(key, zbrar, 0, nused), nused + 1, ri1)[key], 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) ⊦ (key ∈ ri0 ↔ key ∈ ri1) ∧ (key ∈ ri0 → ri0[key] = ri1[key])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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