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Proof step 3 from proof And-prop-inj (previous step)

(⌜true⌝ ∧ P0) = (⌜true⌝ ∧ P1) ⊦ P0 = P1

User applied rule expand right

And obtained the following new goals

Proof step 4:
(⌜true⌝ ∧ P0) = (⌜true⌝ ∧ P1) ⊦ P0(H) ↔ P1(H)

Rule argument

pos: right 1

Additional information