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Proof step 12 from proof exists-shift (previous step)

P(H0), Pa(a)(H1), H1 ⊥ H0 ⊦ ((_∃ Pa * P))(H1 ∪ H0), (_∃ Pa)(H1) ∧ P(H0) ∧ H1 ⊥ H0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
P(H0), Pa(a)(H1), H1 ⊥ H0 ⊦ ((_∃ Pa * P))(H1 ∪ H0), ∃ a. Pa(a)(H1)

Rule argument

Additional information

Used simplifier rules: exists: ⊦ (_∃ Pa)(H) ↔ (∃ a. Pa(a)(H));