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Proof step 1 from proof dups-02 (conclusion)

⊦ ¬ dups(x + y) → list2set(x) ∩ list2set(y) = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ #(list2set(x) ∩ list2set(y)) = 0, list2set(x) ∩ list2set(y) = ∅, dups(y), dups(x), dups(x + y)

Rule argument

Additional information

Used simplifier rules: ⊦ # s ≠ 0 → s ≠ ∅; (forward)% ⊦ (x + y) + z = x + y + z;% ⊦ ¬ dups(x + y) → ¬ dups(x) ∧ ¬ dups(y); (forward)% ⊦ s ≠ ∅ → # s ≠ 0; (forward)