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Proof step 7 from proof newrefs-three (previous step)

⊦ newref_list(3, st) = newref(st) ' + newref(st[newref(st) -- 0 ', noval]) ' + newref(st[newref(st) -- 0 ', noval][newref(st[newref(st) -- 0 ', noval]) -- 0 ', noval]) '

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ 0 < n → newref_list(n, st) = newref(st) + newref_list(n -1, st[newref(st) -- 0 ', noval])
Proof step 6:
0 < 3 ⊦ newref(st) + newref_list(3 -1, st[newref(st) -- 0 ', noval]) = newref(st) ' + newref(st[newref(st) -- 0 ', noval]) ' + newref(st[newref(st) -- 0 ', noval][newref(st[newref(st) -- 0 ', noval]) -- 0 ', noval]) '

Rule argument

[crewrite-arg: newrefs-rec-01, F, [st, n] → [st, 3]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st, 3)]]