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Proof step 2 from proof dups-firstoc-rew (previous step)

dups(x) ⊦ ∃ y, a, z. x = y + a ' + z ∧ a ∈ y ∧ ¬ dups(y)

User applied rule structural induction

And obtained the following new goals

Proof step 3:
dups(x0) → (∃ y, a, z. x0 = y + a ' + z ∧ a ∈ y ∧ ¬ dups(y)), dups(a + x0) ⊦ ∃ y, a0, z. a + x0 = y + a0 ' + z ∧ a0 ∈ y ∧ ¬ dups(y)

Rule argument

indvar: x

Additional information

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