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

P(H0), H1 ⊥ H0, ∀ r. r ∈ H1 → O(r) = me, ∃ H0, H2. H0 ⊥ H2 ∧ H1 = H0 ∪ H2 ∧ oP(o)(H0) ∧ P0(H2) ⊦ ∃ o. ((!(me, O, oP(o) * P0) * P))(H1 ∪ H0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
P(H0), P0(H), oP(o)(H2), H2 ⊥ H, H0 ⊥ H2, H0 ⊥ H, ∀ r. r ∈ H2 ∨ r ∈ H → O(r) = me ⊦ ∃ o. ((!(me, O, oP(o) * P0) * P))((H2 ∪ H) ∪ H0)

Rule argument

Additional information

Used simplifier rules: ⊦ H0 ⊥ H1 → (r ∈ H0 ∪ H1 ↔ r ∈ H0 ∨ r ∈ H1); (s)% ⊦ H1 ⊥ H2 → (H1 ∪ H2 ⊥ H0 ↔ H0 ⊥ H1 ∧ H0 ⊥ H2); (ws)