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Proof step 3 from proof dups-01 (previous step)

b ∈ y -1l b, b ∈ y ⊦ dups(y)

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 4:
b ∈ a + y0, b ∈ (a + y0) -1l b, ∀ b. b ∈ y0 -1l b ∧ b ∈ y0 → dups(y0) ⊦ dups(a + y0)

Rule argument

term: y

Additional information

Used simplifier rules: del1-e: ⊦ [] -1l a = [];%list freely generated by [], +;;% ⊦ ¬ a ∈ []; (s)