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Proof step 1 from proof and-combine (conclusion)
precise(P0) ⊦ H ⊥ H0 ∧ P0(H0) ∧ P1(H0) → (((P * (P0 ∧ P1)))(H ∪ H0) ↔ P(H))
System applied rule normalize
And obtained the following new goals
Proof step 2
:
P0(H0), P1(H0), precise(P0), H ⊥ H0 ⊦ ((P * (P0 ∧ P1)))(H ∪ H0) ↔ P(H)
Rule argument
Additional information