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Proof step 2 from proof and-weaken (
previous step
)
∀ H. P1(H) → P0(H) ⊦ (P0 ∧ P1) = P1
User applied rule expand right
And obtained the following new goals
Proof step 3
:
∀ H. P1(H) → P0(H) ⊦ ((P0 ∧ P1))(H) ↔ P1(H)
Rule argument
pos: right 1
Additional information