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Proof step 6 from proof newrefs-three (previous step)

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]) '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⊦ newref_list(2, st[newref(st) -- 0 ', noval]) = newref(st[newref(st) -- 0 ', noval]) ' + newref(st[newref(st) -- 0 ', noval][newref(st[newref(st) -- 0 ', noval]) -- 0 ', noval]) '

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPSEQ : ⊦ refs + refs0 = refs + refs1 ↔ refs0 = refs1], [CSIMPSEQ : ⊦ r + refs = r ' + refs])]]