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Proof step 7 from proof validrefkeynotnull-refval-in (previous step)

r ∈ st, validrefnotnull(str, st[r -- sk] .refval, ty, st), is_referencevalue(st[r -- sk]), r -- sk ∈ st ⊦ st[r -- sk] .refval ∈ st

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Rule argument

[speclemmaarg: ,is_referencevalue(val) → (r = val .refval ↔ val = refval(r)) ,[val, r] → [st[r -- sk], r0] ,T ,javavalue , ,elim-13 ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ is_referencevalue(val) → ( r = val .refval ↔ val = refval(r))])], [EXTRATERMS : (st[r -- sk], r0)]]