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Proof step 12 from proof disj-distr-uni (previous step)

H1 ⊥ H, ∀ r. ¬ r ∈ H0 ∨ ¬ r ∈ H1 ∧ ¬ r ∈ H ⊦ (¬ r ∈ H0 ∨ ¬ r ∈ H1) ∧ (¬ r0 ∈ H0 ∨ ¬ r0 ∈ H)

User applied rule case distinction

And obtained the following new goals

Proof step 13:
r ∈ H0, r ∈ H1, H1 ⊥ H, ∀ r. ¬ r ∈ H0 ∨ ¬ r ∈ H1 ∧ ¬ r ∈ H ⊦
Proof step 14:
r0 ∈ H0, r0 ∈ H, H1 ⊥ H, ∀ r. ¬ r ∈ H0 ∨ ¬ r ∈ H1 ∧ ¬ r ∈ H ⊦

Rule argument

pos: right 1

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