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Proof step 5 from proof sharing-emp (previous step)

⊦ (∃ H0, H1, H2. H0 ⊥ H2 ∧ H0 ⊥ H1 ∧ H1 ⊥ H2 ∧ H = H0 ∪ H1 ∪ H2 ∧ P(H0 ∪ H2) ∧ H1 ∪ H2 = ∅) ↔ P(H)

User applied rule case distinction

And obtained the following new goals

Proof step 6:
∃ H0, H1, H2. H0 ⊥ H2 ∧ H0 ⊥ H1 ∧ H1 ⊥ H2 ∧ H = H0 ∪ H1 ∪ H2 ∧ P(H0 ∪ H2) ∧ H1 ∪ H2 = ∅ ⊦ P(H)
Proof step 22:
P(H) ⊦ ∃ H0, H1, H2. H0 ⊥ H2 ∧ H0 ⊥ H1 ∧ H1 ⊥ H2 ∧ H = H0 ∪ H1 ∪ H2 ∧ P(H0 ∪ H2) ∧ H1 ∪ H2 = ∅

Rule argument

pos: right 1

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