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Proof step 2 from proof sharing-com (previous step)

⊦ ((P0 ⊗ P1))(H) ↔ ((P1 ⊗ P0))(H)

User applied rule case distinction

And obtained the following new goals

Proof step 3:
((P0 ⊗ P1))(H) ⊦ ((P1 ⊗ P0))(H)
Proof step 17:
((P1 ⊗ P0))(H) ⊦ ((P0 ⊗ P1))(H)

Rule argument

pos: right 1

Additional information