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Proof step 11 from proof ex1-shiftw (previous step)

rP(r)(H0), H0 ⊥ H, ∀ H0. H0 ⊥ H ∧ (∃ r. rP(r)(H0)) → P(H0 ∪ H) ⊦ P(H0 ∪ H), ∃ r. rP(r)(H0)

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 2 terms: r

Additional information

Used terms: r