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Proof step 4 from proof dups-sw (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
dups(x0 + y), ∀ y. dups(x0 + y) → dups(y + x0) ⊦ dups(y + x0), a ∈ y, a ∈ 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);% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a + x = a ' + x; (s)