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Proof step 7 from proof subset-notin (previous step)

r ∈ st, ∀ rk. rk ∈ st → rk ∈ st0 ∧ st[rk] = st0[rk] ⊦ r ∈ st0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ r ∈ st ↔ (∃ sk. r -- sk ∈ st)
Proof step 6:
∃ sk. r -- sk ∈ st, ∀ rk. rk ∈ st → rk ∈ st0 ∧ st[rk] = st0[rk] ⊦ r ∈ st0

Rule argument

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

Additional information

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