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Proof step 3 from proof a (previous step)

dups(x) ⊦ dups(x + y), dups(y)

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 4:
dups(a + x0), ∀ y. dups(x0) → dups(x0 + y) ∨ dups(y) ⊦ dups(y), dups((a + x0) + y)

Rule argument

term: x

Additional information

Used simplifier rules: e-01: ⊦ ¬ dups([]);%list freely generated by [], +;;