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Proof step 1 from proof exists-shift (conclusion)

⊦ ((_∃ Pa * P))(H) ↔ (∃ a. ((Pa(a) * P))(H))

User applied rule case distinction

And obtained the following new goals

Proof step 2:
((_∃ Pa * P))(H) ⊦ ∃ a. ((Pa(a) * P))(H)
Proof step 6:
∃ a. ((Pa(a) * P))(H) ⊦ ((_∃ Pa * P))(H)

Rule argument

pos: right 1

Additional information