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Proof step 6 from proof add-refs (previous step)

⊦ r0 ∈ addobj(r, class, fis, st) ↔ r0 = r ∨ r0 ∈ st .refs

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ r ∈ st .refs ↔ r ∈ st
Proof step 4:
⊦ r ≠ r0 → (r ∈ addobj(r0, class1, fis, st) ↔ r ∈ st)
Proof step 5:
⊦ r ∈ addobj(r0, class1, fis, st) ↔ r = r0 ∨ r ∈ st

Rule argument

[speclemmaarg: ,r ∈ st .refs ↔ r ∈ st ,[st, r] → [st, r0] ,F ,storefunsx , ,refs-def ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st, r0)]]