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Proof step 3 from proof exin-uni (previous step)

⊦ r ∈ H ↔ (∃ H0, o. H = H0 ∪ (∅ ++ r)[r, o] ∧ ¬ r ∈ H0), r = null

User applied rule case distinction

And obtained the following new goals

Proof step 4:
r ∈ H ⊦ r = null, ∃ H0, o. H = H0 ∪ (∅ ++ r)[r, o] ∧ ¬ r ∈ H0
Proof step 15:
∃ H0, o. H = H0 ∪ (∅ ++ r)[r, o] ∧ ¬ r ∈ H0 ⊦ r = null, r ∈ H

Rule argument

pos: right 1

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