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Proof step 11 from proof lookup-range-rem-union (previous step)

key ∈ ks ∧ key ∈ ri ∧ adr = ri[key] ∨ key0 ∈ ri \ ks ∧ (ri \ ks)[key0] = adr ⊦ adr ∈ ri

User applied rule case distinction

And closed the goal!

Rule argument

pos: left 1

Additional information

Used simplifier rules: ⊦ key ∈ ri \ ks ↔ key ∈ ri ∧ ¬ key ∈ ks; (ws)% ⊦ ¬ key ∈ ks → (ri \ ks)[key] = ri[key]; (s)%key ∈ ri ⊦ ri[key] ∈ ri; (s)