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Proof step 4 from proof in-06 (previous step)

a ∈ a0 + x0, a ∈ y, ∀ a, y. a ∈ y ∧ a ∈ x0 → dups(x0 + y) ⊦ dups((a0 + x0) + y)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
a ∈ y, a ∈ x0, ∀ a, y. a ∈ y ∧ a ∈ x0 → dups(x0 + y) ⊦ dups(x0 + y), a0 ∈ x0, a0 ∈ y

Rule argument

Additional information

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