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Proof step 11 from proof validreflist-rec (previous step)

validref(str, val .refval, ty, st), is_referencevalue(val), ∀ val. val ∈ vals → is_referencevalue(val) ∧ validref(str, val .refval, ty, st), val0 = val ∨ val0 ∈ vals ⊦ is_referencevalue(val0) ∧ validref(str, val0 .refval, ty, st)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 10:
val = refval(r), validref(str, r, ty, st), is_referencevalue(val), ∀ val. val ∈ vals → is_referencevalue(val) ∧ validref(str, val .refval, ty, st), val0 = val ∨ val0 ∈ vals ⊦ is_referencevalue(val0) ∧ validref(str, val0 .refval, ty, st)

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)]]