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Proof step 9 from proof lookup-range-rem-union (previous step)

⊦ (∃ key. key ∈ ks ∧ key ∈ ri ∧ adr = ri[key]) ∨ (∃ key. key ∈ ri \ ks ∧ (ri \ ks)[key] = adr) ↔ adr ∈ ri

User applied rule case distinction

And obtained the following new goals

Proof step 10:
(∃ key. key ∈ ks ∧ key ∈ ri ∧ adr = ri[key]) ∨ (∃ key. key ∈ ri \ ks ∧ (ri \ ks)[key] = adr) ⊦ adr ∈ ri
Proof step 12:
adr ∈ ri ⊦ ∃ key. key ∈ ks ∧ key ∈ ri ∧ adr = ri[key], ∃ key. key ∈ ri \ ks ∧ (ri \ ks)[key] = adr

Rule argument

pos: right 1

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