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Proof step 21 from proof refin-del (previous step)

r ∈ st ⊦ r ∈ st -- r1, r = r1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 13:
⊦ r ∈ st ↔ (∃ sk. r -- sk ∈ st)
Proof step 20:
r ∈ st ⊦ ∃ sk. r -- sk ∈ st -- r1, r = r1

Rule argument

[crewrite-arg: instore, F, [st, r] → [st -- r1, r]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st -- r1, r)]]