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Proof step 15 from proof newrefs-two (previous step)

¬ newref(st[newref(st) -- 0 ', noval]) ∈ st[newref(st) -- 0 ', noval], newref(st[newref(st) -- 0 ', noval]) -- sk ∈ st, newref(st[newref(st) -- 0 ', noval]) ∈ st ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ r ∈ st ↔ (∃ sk. r -- sk ∈ st)
Proof step 14:
∀ sk. ¬ newref(st[newref(st) -- 0 ', noval]) -- sk ∈ st[newref(st) -- 0 ', noval], newref(st[newref(st) -- 0 ', noval]) -- sk ∈ st, newref(st[newref(st) -- 0 ', noval]) ∈ st ⊦

Rule argument

[crewrite-arg: instore , F , [st, r] → [st[newref(st) -- 0 ', noval], newref(st[newref(st) -- 0 ', noval])] ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st[newref(st) -- 0 ', noval], newref(st[newref(st) -- 0 ', noval]))]]