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Proof step 11 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 all left
And closed the goal!
Rule argument
pos: left 4 terms: H0
Additional information
Used terms: H0