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Proof step 5 from proof lift-mapex (previous step)

⊦ (∃ o. ((r => o))(H)) ↔ !(me, O(r; me), _∃(λ o. r => o))(H)

User applied rule case distinction

And obtained the following new goals

Proof step 6:
∃ o. ((r => o))(H) ⊦ !(me, O(r; me), _∃(λ o. r => o))(H)
Proof step 15:
!(me, O(r; me), _∃(λ o. r => o))(H) ⊦ ∃ o. ((r => o))(H)

Rule argument

pos: right 1

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