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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
⊦ (∃ H0. H0 ⊥ H ∧ P0(H0) ∧ P1(H0 ∪ H)) ↔ ¬(∀ H0. H0 ⊥ H ∧ P0(H0) → ¬ P1(H0 ∪ H))

Rule argument

Additional information

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