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Proof step 4 from proof app-01-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
dups(y), ∀ y. dups(y) → dups(x0 + y) ⊦ dups(x0 + y), a ∈ x0, a ∈ y

Rule argument

Additional information

Used simplifier rules: dups-cons: ⊦ dups(a ' + x) ↔ dups(x) ∨ a ∈ x;% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a + x = a ' + x; (s)