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Proof step 2 from proof last-notin-butlast-nodups (previous step)

nx.last ∈ nx.butlast ⊦ nx = [], dups(nx)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
nx.last ∈ nx.butlast ⊦ nx = [], dups(nx)

Rule argument

Additional information

Used simplifier rules: ⊦ nx.last ∈ nx ↔ nx ≠ []; (s)%bin-01: nx ≠ [] ⊦ n ∈ nx.butlast → n ∈ nx; (forward) from specification list-last[nat-list-set]