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Proof step 7 from proof septract-dual (previous step)

⊦ (∀ H1. H1 ⊥ H ∧ P0(H1) → P1(H1 ∪ H)) ↔ ¬(∃ H0, H1. H0 ⊥ H ∧ H1 = H0 ∪ H ∧ P0(H0) ∧ (¬ P1)(H1))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
⊦ (∀ H1. H1 ⊥ H ∧ P0(H1) → P1(H1 ∪ H)) ↔ ¬(∃ H0. H0 ⊥ H ∧ P0(H0) ∧ ¬ P1(H0 ∪ H))

Rule argument

Additional information

Used simplifier rules: not: ⊦ (¬ P0)(H) ↔ ¬ P0(H);