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

∀ y. dups(x0 + y) → dups(y + x0), dups((a + x0) + y) ⊦ dups(y + (a + x0))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: rec: ⊦ dups(a ' + x) ↔ ¬ ¬(a ∈ x ∨ dups(x));%f-01: ⊦ dups(x + a ' + y) ↔ a ∈ x ∨ a ∈ y ∨ dups(x + y);%in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y); from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list