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Proof step 3 from proof okval-refval-01 (previous step)

is_referencevalue(val) ⊦ okval(val, ty) ↔ ty = void_type ∨ is_reftype(ty)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ is_referencevalue(val) ↔ (∃ r. val = refval(r))
Proof step 2:
⊦ okval(refval(r), ty) ↔ ty = void_type ∨ is_reftype(ty)

Rule argument

[speclemmaarg: ,is_referencevalue(val) ↔ (∃ r. val = refval(r)) ,[val] → [val] ,F ,javavalue , ,ex-04 ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (val)]]