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Proof step 24 from proof in-keyrange-in-ri-data (previous step)

m + 1 < # zbrar, 2 + m < # zbrar, m + 1 < nused, 2 + m < nused, nused ≤ # zbrar, ∃ key. key ∈ (ri |> zbrar[m].key) |≤ zbrar[m + 1].key ∧ ((ri |> zbrar[m].key) |≤ zbrar[m + 1].key)[key] = adr ⊦ 2 + m = nused, adr ∈ ri

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
2 + m < nused, m + 1 < nused, 2 + m < # zbrar, m + 1 < # zbrar, zbrar[m].key < key, key ∈ ri, nused ≤ # zbrar ⊦ 2 + m = nused, zbrar[m + 1].key < key, ((ri |> zbrar[m].key) |≤ zbrar[m + 1].key)[key] ∈ ri

Rule argument

Additional information

Used simplifier rules: ⊦ key0 ∈ ri |> key1 ↔ key0 ∈ ri ∧ key1 < key0; (s)% ⊦ key0 ∈ ri |≤ key1 ↔ key0 ∈ ri ∧ ¬ key1 < key0; (s)