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Proof step 29 from proof ex-shift-o-p (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 30:
⊦ !(id, O, P)(H) ↔ P(H) ∧ (∀ r. r ∈ H → O(r) = id)
Proof step 31:
((oP(o) * P0))(H1), P(H0), H1 ⊥ H0, ∀ r. r ∈ H1 → O(r) = me ⊦ ((!(me, O, _∃ oP * P0) * P))(H1 ∪ H0), ((_∃ oP * P0))(H1) ∧ (∀ r. r ∈ H1 → 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] → [H1, _∃ oP * P0, O, me] flags: <all paths>

Additional information

Used terms: H1, _∃ oP * P0, O, me