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Proof step 29 from proof addobj-delsw (previous step)

⊦ r = r0, addobj(r0, class, fis, st) -- r = addobj(r0, class, fis, st -- r)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ st1 = st2 ↔ (∀ rk. (rk ∈ st1 ↔ rk ∈ st2) ∧ st1[rk] = st2[rk])
Proof step 28:
⊦ r = r0, ∀ rk. (rk ∈ addobj(r0, class, fis, st) -- r ↔ rk ∈ addobj(r0, class, fis, st -- r)) ∧ (addobj(r0, class, fis, st) -- r)[rk] = (addobj(r0, class, fis, st -- r)[rk])

Rule argument

[speclemmaarg: ,st1 = st2 ↔ (∀ rk. (rk ∈ st1 ↔ rk ∈ st2) ∧ st1[rk] = st2[rk]) ,[st2, st1] → [addobj(r0, class, fis, st -- r), addobj(r0, class, fis, st) -- r] ,F ,store ,javastore ,Extension ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (addobj(r0, class, fis, st -- r), addobj(r0, class, fis, st) -- r)]]