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Proof step 5 from proof eqrefval-01 (conclusion)

⊦ val = refval(r) ↔ r = val .refval ∧ is_referencevalue(val)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
⊦ refval(r) .refval = r
Proof step 2:
⊦ is_referencevalue(refval(r))
Proof step 4:
is_referencevalue(val) ⊦ val = refval(val .refval)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]