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Proof step 1 from proof precise-split-r (conclusion)
precise(P1) ⊦ P1(H1) ∧ H0 ⊥ H1 → (((P0 * P1))(H0 ∪ H1) ↔ P0(H0))
System applied rule normalize
And obtained the following new goals
Proof step 2
:
P1(H1), precise(P1), H0 ⊥ H1 ⊦ ((P0 * P1))(H0 ∪ H1) ↔ P0(H0)
Rule argument
Additional information