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Proof step 22 from proof newrefs-three (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 21:
¬ newref(st[newref(st) -- 0 ', noval][newref(st[newref(st) -- 0 ', noval]) -- 0 ', noval]) -- sk ∈ (st[newref(st) -- 0 ', noval][newref(st[newref(st) -- 0 ', noval]) -- 0 ', noval]) ⊦ ¬ newref(st[newref(st) -- 0 ', noval][newref(st[newref(st) -- 0 ', noval]) -- 0 ', noval]) -- sk ∈ st

Rule argument

[EXRARG : pos: left 1, [extquanttermlist: (sk)abort:F, discard:T, computed:F]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (sk)]]