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Proof step 9 from proof wand-dual (previous step)

⊦ (∃ H0. H0 ⊥ H ∧ P0(H0) ∧ P1(H0 ∪ H)) ↔ ¬(∀ H0. H0 ⊥ H ∧ P0(H0) → ¬ P1(H0 ∪ H))

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

Additional information