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Proof step 10 from proof septract-dual (previous step)

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: H0

Additional information

Used terms: H0