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Proof step 1 from proof dups-01 (conclusion)

⊦ ¬ dups(x + set2list(s ∪ s0)) → ¬ dups(x + set2list(s)) ∧ ¬ dups(x + set2list(s0))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ dups(x), dups(x + set2list(s ∪ s0)), ¬ dups(x + set2list(s)) ∧ ¬ dups(x + set2list(s0))

Rule argument

Additional information

Used simplifier rules: dups: ⊦ ¬ dups(set2list(s));% ⊦ (x + y) + z = x + y + z;% ⊦ ¬ dups(x + y) → ¬ dups(x) ∧ ¬ dups(y); (forward)