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Proof step 14 from proof exists-shiftw (previous step)

∀ a. ((Pa(a) -* P))(H) ⊦ ∀ H1. H1 ⊥ H ∧ (_∃ Pa)(H1) → P(H1 ∪ H)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
Pa(a)(H0), H0 ⊥ H, ∀ a. ((Pa(a) -* P))(H) ⊦ P(H0 ∪ H)

Rule argument

Additional information

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