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Proof step 5 from proof subset-delete (previous step)

st[rk] = (st0 -- rk .ref)[rk], rk ∈ st, rk ∈ st0 -- rk .ref, ∀ rk0. rk0 ∈ st → rk0 ∈ st0 -- rk .ref ∧ st[rk0] = (st0 -- rk .ref)[rk0] ⊦ rk .ref ∈ st, rk ∈ st0 ∧ st[rk] = st0[rk]

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
⊦ r -- sk ∈ st -- r0 ↔ r ≠ r0 ∧ r -- sk ∈ 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)]]