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Proof step 10 from proof ex-p (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information