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Proof step 18 from proof ex-shift-o (previous step)

!(me, O, oP(o))(H1), P(H0), H1 ⊥ H0 ⊦ ((!(me, O, _∃ oP) * P))(H1 ∪ H0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 19:
⊦ !(id, O, P)(H) ↔ P(H) ∧ (∀ r. r ∈ H → O(r) = id)
Proof step 20:
P(H0), oP(o)(H1), H1 ⊥ H0, ∀ r. r ∈ H1 → O(r) = me ⊦ ((!(me, O, _∃ oP) * P))(H1 ∪ H0)

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(o), O, me] flags: <all paths>

Additional information

Used terms: H1, oP(o), O, me