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Proof step 13 from proof wand-dual (previous step)

P0(H0), P1(H0 ∪ H), H0 ⊥ H ⊦ ∃ H0. H0 ⊥ H ∧ P0(H0) ∧ P1(H0 ∪ H)

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: H0

Additional information

Used terms: H0