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Proof step 15 from proof sep-partial-mp (previous step)

P0(H2), P2(H0), H2 ⊥ H1, H0 ⊥ H2, H0 ⊥ H1, ((P0 * P2))(H2 ∪ H0) → P1((H2 ∪ H0) ∪ H1) ⊦ P1(H0 ∪ H2 ∪ H1)

User applied rule case distinction

And obtained the following new goals

Proof step 16:
P0(H2), P2(H0), H0 ⊥ H1, H0 ⊥ H2, H2 ⊥ H1 ⊦ ((P0 * P2))(H2 ∪ H0), P1(H0 ∪ H2 ∪ H1)
Proof step 18:
P0(H2), P1((H2 ∪ H0) ∪ H1), P2(H0), H0 ⊥ H1, H0 ⊥ H2, H2 ⊥ H1 ⊦ P1(H0 ∪ H2 ∪ H1)

Rule argument

pos: left 6

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