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Proof step 26 from proof addobj-union (previous step)

newref(r, st), newref(r, st0) ⊦ addobj(r, class, fis, st) ∪ st0 = addobj(r, class, fis, st ∪ st0), r = jvmref

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 25:
newref(r, st), newref(r, st0) ⊦ ∀ rk. (rk ∈ addobj(r, class, fis, st) ∪ st0 ↔ rk ∈ addobj(r, class, fis, st ∪ st0)) ∧ (addobj(r, class, fis, st) ∪ st0)[rk] = (addobj(r, class, fis, st ∪ st0)[rk]), r = jvmref

Rule argument

[speclemmaarg: ,st1 = st2 ↔ (∀ rk. (rk ∈ st1 ↔ rk ∈ st2) ∧ st1[rk] = st2[rk]) ,[st2, st1] → [addobj(r, class, fis, st ∪ st0), addobj(r, class, fis, st) ∪ st0] ,F ,store ,javastore ,Extension ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (addobj(r, class, fis, st ∪ st0), addobj(r, class, fis, st) ∪ st0)]]