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Proof step 2 from proof precise-disj (previous step)

H0 ∪ H2 = H1 ∪ H3, ((P0 * P1 * true))(H0 ∪ H2), P0(H0), P1(H1), precise(P1), precise(P0), H0 ⊥ H2, H1 ⊥ H3 ⊦ H0 ⊥ H1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information