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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information