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Proof step 3 from proof in-min (previous step)

st .min = r -- sk ⊦ st = ∅, r ∈ st

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ r ∈ st ↔ (∃ sk. r -- sk ∈ st)
Proof step 2:
st .min = r -- sk ⊦ st = ∅, ∃ sk. r -- sk ∈ st

Rule argument

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

Additional information

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