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Proof step 17 from proof sharing-emp (previous step)

P(H2 ∪ H0), H2 ⊥ H0, H2 ⊥ H1, H1 ⊥ H0, ∀ r. ¬ r ∈ H1 ∧ ¬ r ∈ H0 ⊦ P(H2), ¬ r ∈ H1 ∧ ¬ r0 ∈ H0

User applied rule case distinction

And obtained the following new goals

Proof step 18:
P(H2 ∪ H0), r ∈ H1, H1 ⊥ H0, H2 ⊥ H1, H2 ⊥ H0, ∀ r. ¬ r ∈ H1 ∧ ¬ r ∈ H0 ⊦ P(H2)
Proof step 20:
P(H2 ∪ H0), r0 ∈ H0, H1 ⊥ H0, H2 ⊥ H1, H2 ⊥ H0, ∀ r. ¬ r ∈ H1 ∧ ¬ r ∈ H0 ⊦ P(H2)

Rule argument

pos: right 2

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