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Proof step 1 from proof And-prop-inj (conclusion)

⊦ (⌜bv⌝ ∧ P0) = (⌜bv⌝ ∧ P1) ↔ bv → P0 = P1

User applied rule case distinction

And obtained the following new goals

Proof step 2:
(⌜bv⌝ ∧ P0) = (⌜bv⌝ ∧ P1), bv ⊦ P0 = P1
Proof step 5:
bv → P0 = P1 ⊦ (⌜bv⌝ ∧ P0) = (⌜bv⌝ ∧ P1)

Rule argument

pos: right 1

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