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Proof step 6 from proof in-butmin (previous step)

st .min = r0 -- sk, r -- sk0 ∈ st, r ∈ st ⊦ r = r0, r -- sk0 ∈ st .butmin

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ rk ∈ st .butmin ↔ st ≠ ∅ ∧ rk ∈ st ∧ rk ≠ st .min
Proof step 5:
⊦ ¬ r ∈ ∅

Rule argument

[speclemmaarg: ,rk ∈ st .butmin ↔ st ≠ ∅ ∧ rk ∈ st ∧ rk ≠ st .min ,[st, rk] → [st, r -- sk0] ,F ,ostore ,javastore ,butmin-in ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ st .min = rk ∧ st ≠ ∅ → rk ∈ st], [CSIMPSEQ : ⊦ r -- sk = r0 -- sk0 ↔ r = r0 ∧ sk = sk0])], [EXTRATERMS : (st, r -- sk0)]]