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Proof step 28 from proof newrefs-two (previous step)

⊦ ∃ r, r0. newref_list(2, st) = r ' + r0 ' ∧ newref(r, st) ∧ newref(r0, st) ∧ r ≠ r0

User applied rule exists right

And obtained the following new goals

Proof step 27:
⊦ newref_list(2, st) = newref(st) ' + newref(st[newref(st) -- 0 ', noval]) ' ∧ newref(newref(st), st) ∧ newref(newref(st[newref(st) -- 0 ', noval]), st) ∧ newref(st) ≠ newref(st[newref(st) -- 0 ', noval])

Rule argument

[EXRARG : pos: right 1, [extquanttermlist: (newref(st), newref(st[newref(st) -- 0 ', noval])) abort:F , discard:T , computed:F ]]

Additional information

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