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Proof step 1 from proof dups-03 (conclusion)

⊦ ¬ dups(x) ∧ ¬ dups(y) → (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) = ∅, dups(x + y) ⊦ dups(y), dups(x)

Rule argument

Additional information

Used simplifier rules: dups-02: ⊦ ¬ dups(x + y) → list2set(x) ∩ list2set(y) = ∅;% ⊦ s ∩ s0 = s0 ∩ s;% ⊦ (s ∩ s0) ∩ s1 = s ∩ s0 ∩ s1;% ⊦ (x + y) + z = x + y + z;