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Proof step 8 from proof newref-newreflist-subset (previous step)

st0 ⊆ st, r ∈ newref_list(m, st[newref(st) -- 0 ', noval]) ⊦ st0 ⊆ st[newref(st) -- 0 ', noval], r = newref(st), newref(r, st0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ newref(r, st) → (st ⊆ st0[r -- sk, val] ↔ st ⊆ st0)
Proof step 7:
⊦ ¬ newref(newref(st), st0) → ¬ st0 ⊆ st

Rule argument

[crewrite-arg: stsubset-new , F , [val, sk, r, st0, st] → [noval, 0 ', newref(st), st, st0] ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (noval, 0 ', newref(st), st, st0)]]