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Proof step 8 from proof septract-dual (previous step)

⊦ (∀ H1. H1 ⊥ H ∧ P0(H1) → P1(H1 ∪ 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