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Proof step 5 from proof precise-split-r (previous step)

((P0 * P1))(H1 ∪ H0), P1(H1), precise(P1), H0 ⊥ H1, H1 ⊥ H0 ⊦ P0(H0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
((P0 * P1))(H1 ∪ H0), P1(H1), precise(P1), H0 ⊥ H1 ⊦ P0(H0)

Rule argument

Additional information

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