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Proof step 9 from proof validrefval-superset (previous step)

validrefval(str, val, ty, st), st ⊆ st0 ⊦ validrefval(str, val, ty, st0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ validrefval(str, val, ty, st) ↔ is_referencevalue(val) ∧ validref(str, val .refval, ty, st)
Proof step 8:
validrefval(str, val, ty, st), st ⊆ st0 ⊦ is_referencevalue(val) ∧ validref(str, val .refval, ty, st0)

Rule argument

[crewrite-arg: validrefval-def , F , [st, ty, val, str] → [st0, ty, val, str] ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st0, ty, val, str)]]