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Proof step 14 from proof ex-shift-o-p (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
P(H0), P0(H), oP(o)(H2), H0 ⊥ H, H0 ⊥ H2, H2 ⊥ H, ∀ r. r ∈ H2 ∨ r ∈ H → O(r) = me ⊦ !(me, O, oP(o) * P0)(H2 ∪ H), ((!(me, O, oP(o) * P0) * P))((H2 ∪ H) ∪ H0), ∃ o. ((!(me, O, oP(o) * P0) * P))((H2 ∪ H) ∪ H0)

Rule argument

Additional information

Used simplifier rules: ⊦ H1 ⊥ H2 → (H1 ∪ H2 ⊥ H0 ↔ H0 ⊥ H1 ∧ H0 ⊥ H2); (ws)