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Proof step 13 from proof range-del-in (previous step)

key ∈ ri, ∀ key0, key1. key0 ∈ ri ∧ key1 ∈ ri ∧ ri[key0] = ri[key1] → key0 = key1 ⊦ (∃ key0. key0 ≠ key ∧ key0 ∈ ri ∧ ri[key0] = adr) ↔ adr ≠ ri[key] ∧ (∃ key. key ∈ ri ∧ ri[key] = adr)

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

Additional information