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Proof step 10 from proof in-butmin (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ r ∈ st ↔ (∃ sk. r -- sk ∈ st)
Proof step 9:
st .min = r0 -- sk, ∃ sk. r -- sk ∈ st ⊦ r = r0, r ∈ st .butmin

Rule argument

[crewrite-arg: instore, F, [st, r] → [st, r]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st, r)]]