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Proof step 6 from proof validrefkeynotnull-refval-in (previous step)

st[r -- sk] = refval(r0), r ∈ st, validrefnotnull(str, r0, ty, st), is_referencevalue(st[r -- sk]), r -- sk ∈ st ⊦ r0 ∈ st

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
st[r -- sk] = refval(r0), r -- sk ∈ st, validrefnotnull(str, r0, ty, st), r ∈ st ⊦ r0 ∈ st

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ is_referencevalue(refval(r))])]]