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Proof step 6 from proof validrefval-superset (previous step)

is_referencevalue(val), validref(str, val .refval, ty, st), st ⊆ st0 ⊦ validref(str, val .refval, ty, st0)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 5:
val = refval(r), is_referencevalue(val), validref(str, r, ty, st), st ⊆ st0 ⊦ validref(str, r, ty, st0)

Rule argument

[speclemmaarg: ,is_referencevalue(val) → (r = val .refval ↔ val = refval(r)) ,[val, r] → [val, r] ,T ,javavalue , ,elim-13 ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ is_referencevalue(val) → ( r = val .refval ↔ val = refval(r))])], [EXTRATERMS : (val, r)]]