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Proof step 26 from proof newrefs-three (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ r ∈ st ↔ (∃ sk. r -- sk ∈ st)
Proof step 25:
⊦ ∀ sk. ¬ newref(st[newref(st) -- 0 ', noval][newref(st[newref(st) -- 0 ', noval]) -- 0 ', noval]) -- sk ∈ st

Rule argument

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

Additional information

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