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Proof step 4 from proof ex-p (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 5:
∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ (∃ x. xP(x)(H0)) ∧ P(H1) ⊦ ∃ x. ((xP(x) * P))(H)
Proof step 7:
∃ x. ((xP(x) * P))(H) ⊦ ∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ (∃ x. xP(x)(H0)) ∧ P(H1)

Rule argument

pos: right 1

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