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Proof step 4 from proof exists-shift (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
P(H0), Pa(a)(H1), H1 ⊥ H0 ⊦ ∃ a. ((Pa(a) * P))(H1 ∪ H0)

Rule argument

Additional information

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