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Proof step 22 from proof range-write-in (previous step)

key ∈ ri, ∀ key0, key1. key0 ∈ ri ∧ key1 ∈ ri ∧ ri[key0] = ri[key1] → key0 = key1, adr0 = adr ∨ ri[key0] ≠ ri[key] ∧ key0 ∈ ri ∧ ri[key0] = adr0 ⊦ ∃ key0. key0 ∈ ri[key, adr] ∧ ri[key, adr][key0] = adr0

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 3

Additional information