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Proof step 3 from proof ex-p (previous step)

⊦ (∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ (_∃ xP)(H0) ∧ P(H1)) ↔ (∃ x. ((xP(x) * P))(H))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ (∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ (∃ x. xP(x)(H0)) ∧ P(H1)) ↔ (∃ x. ((xP(x) * P))(H))

Rule argument

Additional information

Used simplifier rules: ex: ⊦ (_∃ xP)(H) ↔ (∃ x. xP(x)(H));