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Proof step 11 from proof ex-shift-o (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
P(H0), oP(o)(H1), H1 ⊥ H0, ∀ r. r ∈ H1 → O(r) = me ⊦ !(me, O, oP(o))(H1), ((!(me, O, oP(o)) * P))(H1 ∪ H0), ∃ o. ((!(me, O, oP(o)) * P))(H1 ∪ H0)

Proof step 11 from proof ex-shift-o (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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