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Proof step 8 from proof exists-shiftw (previous step)

∀ H0. H0 ⊥ H ∧ (∃ a. Pa(a)(H0)) → P(H0 ∪ H) ⊦ ∀ H1. H1 ⊥ H ∧ Pa(a)(H1) → P(H1 ∪ H)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
Pa(a)(H0), H0 ⊥ H, ∀ H0. H0 ⊥ H ∧ (∃ a. Pa(a)(H0)) → P(H0 ∪ H) ⊦ P(H0 ∪ H)

Rule argument

Additional information