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Proof step 4 from proof ex1-shift (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: ex1: ⊦ (_∃ rP)(H) ↔ (∃ r. rP(r)(H));