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Proof step 2 from proof dups-appex (previous step)

∀ y. dups(x0 + y) ↔ dups(x0) ∨ dups(y) ∨ (∃ a. a ∈ x0 ∧ a ∈ y) ⊦ dups((a + x0) + y) ↔ dups(a + x0) ∨ dups(y) ∨ (∃ a0. a0 ∈ a + x0 ∧ a0 ∈ y)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ y. dups(x0 + y) ↔ dups(x0) ∨ dups(y) ∨ (∃ a. a ∈ x0 ∧ a ∈ y) ⊦ a ∈ y ∨ dups(x0 + y) ↔ dups(x0) ∨ dups(y) ∨ (∃ a0. (a0 = a ∨ a0 ∈ x0) ∧ a0 ∈ y), 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)