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Proof step 2 from proof a (previous step)

⊦ dups(x + y), ¬ dups(x) ∧ ¬ dups(y)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
dups(x) ⊦ dups(x + y), dups(y)

Rule argument

Additional information

Used simplifier rules: app-01-02: ⊦ dups(y) → dups(x + y);%a: ⊦ (x + y) + z = x + y + z; from specification list