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Proof step 15 from proof newref-02 (previous step)

rk ∈ st, ∀ rk0. rk0 ∈ st → rk0 .ref ≠ rk .ref ⊦ rk = r0 -- sk, rk .ref = r0, rk .ref = jvmref

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 14:
rk = r -- sk0, rk ∈ st, ∀ rk0. rk0 ∈ st → rk0 .ref ≠ r ⊦ rk = r0 -- sk, r = r0, r = jvmref

Rule argument

[speclemmaarg: ,r = rk .ref ∧ sk = rk .key ↔ rk = r -- sk ,[rk, r, sk] → [rk, r, sk0] ,T ,refkey , ,elim ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ r = rk .ref ∧ sk = rk .key ↔ rk = r -- sk])], [EXTRATERMS : (rk, r, sk0)]]