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Proof step 1 from proof and-split (conclusion)

precise(P) ⊦ ((P * (P0 ∧ P1)))(H) ↔ ((P * P0))(H) ∧ ((P * P1))(H)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
((P * P0))(H), ((P * P1))(H), precise(P) ⊦ ((P * (P0 ∧ P1)))(H)

Rule argument

Additional information

Used simplifier rules: and-ass: ⊦ ((P0 ∧ P1) ∧ P2) = (P0 ∧ P1 ∧ P2);%and-com: ⊦ (P0 ∧ P1) = (P1 ∧ P0);%sep-ass: ⊦ (P0 * P1) * P2 = P0 * P1 * P2;%sep-com: ⊦ P0 * P1 = P1 * P0;%and-imp: ⊦ ((P * (P1 ∧ P2)))(H) → ((P * P1))(H);