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Proof step 8 from proof newref-same (previous step)

st .refs = st0 .refs ⊦ r ∈ st ↔ r ∈ st0, newref(st) = newref(st0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ rset = rset0 ↔ (∀ r. r ∈ rset ↔ r ∈ rset0)
Proof step 7:
∀ r. r ∈ st .refs ↔ r ∈ st0 .refs ⊦ r ∈ st ↔ r ∈ st0, newref(st) = newref(st0)

Rule argument

[speclemmaarg: ,rset = rset0 ↔ (∀ r. r ∈ rset ↔ r ∈ rset0) ,[rset0, rset] → [st0 .refs, st .refs] ,F ,set-basic ,references ,Extension ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st0 .refs, st .refs)]]