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Proof step 15 from proof ex-shift-o-p (previous step)

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)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 16:
⊦ !(id, O, P)(H) ↔ P(H) ∧ (∀ r. r ∈ H → O(r) = id)
Proof step 17:
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), ((oP(o) * P0))(H2 ∪ H) ∧ (∀ r. r ∈ H2 ∪ H → O(r) = me)

Rule argument

spec: <toplevel> name: owned seq: ⊦ !(id, O, P)(H) ↔ P(H) ∧ (∀ r. r ∈ H → O(r) = id) subst: [H, P, O, id] → [H2 ∪ H, oP(o) * P0, O, me] flags: <all paths>

Additional information

Used terms: H2 ∪ H, oP(o) * P0, O, me