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Proof step 5 from proof ex-p (
previous step
)
∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ (∃ x. xP(x)(H0)) ∧ P(H1) ⊦ ∃ x. ((xP(x) * P))(H)
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 6
:
P(H0), xP(x)(H1), H1 ⊥ H0 ⊦ ∃ x. ((xP(x) * P))(H1 ∪ H0)
Rule argument
Additional information