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Proof step 1 from proof sharing-com (conclusion)

⊦ P0 ⊗ P1 = P1 ⊗ P0

User applied rule expand right

And obtained the following new goals

Proof step 2:
⊦ ((P0 ⊗ P1))(H) ↔ ((P1 ⊗ P0))(H)

Rule argument

pos: right 1

Additional information