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Proof step 8 from proof in-keyrange-in-ri (previous step)

key ∈ (nused = 1 ⊃ ri;ri |≤ zbrar[0].key), nused ≤ # zbrar ⊦ # zbrar = 0, nused = 0, key ∈ ri

User applied rule case distinction

And closed the goal!

Rule argument

pos: left 1

Additional information

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