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Proof step 10 from proof ex3-shiftw (previous step)

oP(o)(H0), H0 ⊥ H, H0 ⊥ H ∧ (∃ o. oP(o)(H0)) → P(H0 ∪ H), ∀ H1. H1 ⊥ H ∧ (∃ o. oP(o)(H1)) → P(H1 ∪ H) ⊦ P(H0 ∪ H)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information