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Proof step 1 from proof ex2-shift (conclusion)

⊦ ((_∃ rrP * P))(H) ↔ (∃ r0, r1. ((rrP(r0, r1) * P))(H))

User applied rule case distinction

And obtained the following new goals

Proof step 2:
((_∃ rrP * P))(H) ⊦ ∃ r0, r1. ((rrP(r0, r1) * P))(H)
Proof step 6:
∃ r0, r1. ((rrP(r0, r1) * P))(H) ⊦ ((_∃ rrP * P))(H)

Rule argument

pos: right 1

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