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Proof step 12 from proof and-combine (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
precise(P0), P1(H0), P0(H0), H ⊥ H0 ⊦ (P * (P0 ∧ P1))(H ∪ H0) ↔ P(H)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]