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Proof step 6 from proof newreflist-dups-01 (previous step)

¬ dups(newref_list(m, st[newref(st) -- 0 ', noval])), ∀ st. ¬ dups(newref_list(m, st)) ⊦ ¬ newref(st) ∈ newref_list(m, st[newref(st) -- 0 ', noval]) ∧ ¬ dups(newref_list(m, st[newref(st) -- 0 ', noval]))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
newref(st) ∈ newref_list(m, st[newref(st) -- 0 ', noval]), ∀ st. ¬ dups(newref_list(m, st)) ⊦ dups(newref_list(m, st[newref(st) -- 0 ', noval]))

Rule argument

[emptyarg]

Additional information

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