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Proof step 7 from proof and-combine (previous step)

P0(H0), P1(H0), precise(P0), H ⊥ H0, H0 ⊥ H ⊦ (((P0 ∧ P1) * P))(H0 ∪ H) ↔ P(H)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
P0(H0), P1(H0), precise(P0), H ⊥ H0 ⊦ (((P0 ∧ P1) * P))(H0 ∪ H) ↔ P(H)

Rule argument

Additional information

Used simplifier rules: ⊦ H0 ⊥ H1 → H1 ⊥ H0; (s)