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Proof step 11 from proof sep-partial-mp (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 12:
P0(H2), P2(H0), H2 ⊥ H1, H0 ⊥ H2, H0 ⊥ H1, H2 ∪ H0 ⊥ H1 ∧ ((P0 * P2))(H2 ∪ H0) → P1((H2 ∪ H0) ∪ H1) ⊦ P1(H0 ∪ H2 ∪ H1)

Rule argument

pos: left 6 terms: H2 ∪ H0 abort: F discard: T computed: F

Additional information

Used terms: H2 ∪ H0