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Proof step 10 from proof refin-del (previous step)

¬ r1 ∈ st -- r1, ∀ r0, sk. r1 ≠ r0 → (r0 -- sk ∈ st ↔ r0 -- sk ∈ st -- r1) ∧ st[r0 -- sk] = (st -- r1)[r0 -- sk], r ∈ st -- r1 ⊦ r ∈ st, r = r1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ r ∈ st ↔ (∃ sk. r -- sk ∈ st)
Proof step 9:
¬ r1 ∈ st -- r1, ∀ r0, sk. r1 ≠ r0 → (r0 -- sk ∈ st ↔ r0 -- sk ∈ st -- r1) ∧ st[r0 -- sk] = (st -- r1)[r0 -- sk], r ∈ st -- r1 ⊦ ∃ sk. r -- sk ∈ st, r = r1

Rule argument

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

Additional information

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