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Proof step 7 from proof validrefval-superset (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
is_referencevalue(val), validref(str, val .refval, ty, st), st ⊆ st0 ⊦ validref(str, val .refval, ty, st0)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]