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Proof step 5 from proof wand-right (previous step)

⊦ (∀ H. ((P * Q0))(H) → Q1(H)) ↔ (∀ H. P(H) → ((Q0 -* Q1))(H))

User applied rule case distinction

And obtained the following new goals

Proof step 6:
∀ H. ((P * Q0))(H) → Q1(H) ⊦ ∀ H. P(H) → ((Q0 -* Q1))(H)
Proof step 11:
∀ H. P(H) → ((Q0 -* Q1))(H) ⊦ ∀ H. ((P * Q0))(H) → Q1(H)

Rule argument

pos: right 1

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