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Proof step 2 from proof adjunction (previous step)

⊦ (∀ H. P0(H) → ((P1 -* P2))(H)) ↔ (∀ H. ((P0 * P1))(H) → P2(H))

User applied rule case distinction

And obtained the following new goals

Proof step 3:
∀ H. P0(H) → ((P1 -* P2))(H) ⊦ ∀ H. ((P0 * P1))(H) → P2(H)
Proof step 15:
∀ H. ((P0 * P1))(H) → P2(H) ⊦ ∀ H. P0(H) → ((P1 -* P2))(H)

Rule argument

pos: right 1

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