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Proof step 1 from proof ex-shift-o (conclusion)

⊦ ((!(me, O, _∃ oP) * P))(H) ↔ (∃ o. ((!(me, O, oP(o)) * P))(H))

User applied rule case distinction

And obtained the following new goals

Proof step 2:
((!(me, O, _∃ oP) * P))(H) ⊦ ∃ o. ((!(me, O, oP(o)) * P))(H)
Proof step 14:
∃ o. ((!(me, O, oP(o)) * P))(H) ⊦ ((!(me, O, _∃ oP) * P))(H)

Rule argument

pos: right 1

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