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Proof step 9 from proof exists-shift (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information