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Proof step 21 from proof subset-delete (previous step)

st[rk] = st0[rk], rk ∈ st, rk ∈ st0 ⊦ r = rk .ref, r ∈ st, rk ∈ st0 -- r

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 19:
⊦ r -- sk ∈ st -- r0 ↔ r ≠ r0 ∧ r -- sk ∈ st
Proof step 20:
⊦ r -- sk ∈ st → r ∈ st

Rule argument

[speclemmaarg: ,r = rk .ref ∧ sk = rk .key ↔ rk = r -- sk ,[rk, r, sk] → [rk, r0, sk] ,T ,refkey , ,elim ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ rk ∈ st → st[rk] = ∅[rk]], [CSIMPELIM : ⊦ r = rk .ref ∧ sk = rk .key ↔ rk = r -- sk])], [EXTRATERMS : (rk, r0, sk)]]