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Proof step 1 from proof introduce-sharing (conclusion)

⊦ ((P0 ⊗ P1))(H) ↔ (∃ P. (((P -* P0) * (P -* P1) * P))(H))

User applied rule case distinction

And obtained the following new goals

Proof step 2:
((P0 ⊗ P1))(H) ⊦ ∃ P. (((P -* P0) * (P -* P1) * P))(H)
Proof step 23:
∃ P. (((P -* P0) * (P -* P1) * P))(H) ⊦ ((P0 ⊗ P1))(H)

Rule argument

pos: right 1

Additional information