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Proof step 4 from proof d (previous step)

∀ a, y. (dups(x0 + a ' + y) ↔ dups(x0 + y)) ∨ a ∈ x0 ∨ a ∈ y ⊦ dups((a0 + x0) + a ' + y) ↔ dups((a0 + x0) + y), a ∈ y, a ∈ a0 + x0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
∀ a, y. (dups(x0 + a ' + y) ↔ dups(x0 + y)) ∨ a ∈ x0 ∨ a ∈ y ⊦ dups(x0 + a ' + y) ↔ dups(x0 + y), a = a0, a ∈ y, a0 ∈ y, a0 ∈ x0, a ∈ x0

Rule argument

Additional information

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