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Proof step 58 from proof newrefs-three (previous step)

newref(st[newref(st) -- 0 ', noval][newref(st[newref(st) -- 0 ', noval]) -- 0 ', noval]) = newref(st[newref(st) -- 0 ', noval]) ⊦ ∃ r, r0, r1. newref_list(3, st) = r ' + r0 ' + r1 ' ∧ newref(r1, st) ∧ newref(r, st) ∧ newref(r0, st) ∧ r ≠ r0 ∧ r ≠ r1 ∧ r0 ≠ r1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 56:
⊦ r ∈ st → newref(st) ≠ r
Proof step 57:
⊦ r ∈ st[r0 -- sk, val] ↔ r = r0 ∨ r ∈ st

Rule argument

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

Additional information

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