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Proof step 4 from proof newref-put-in (previous step)

rk ∈ st ⊦ newref(st[rk, val]) = newref(st), rk .ref ∈ st

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
⊦ r -- sk ∈ st → r ∈ st

Rule argument

[speclemmaarg: ,r = rk .ref ∧ sk = rk .key ↔ rk = r -- sk ,[rk, r, sk] → [rk, r, sk] ,T ,refkey , ,elim ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ r = rk .ref ∧ sk = rk .key ↔ rk = r -- sk])], [EXTRATERMS : (rk, r, sk)]]