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Proof step 5 from proof dups-appex (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
dups(x0 + y) ↔ (∃ a. a ∈ x0 ∧ a ∈ y), a ∈ y ∨ dups(x0 + y), ∀ y0. dups(x0 + y0) ↔ dups(y0) ∨ (∃ a. a ∈ x0 ∧ a ∈ y0) ⊦ dups(x0), dups(y), a ∈ x0, ∃ a0. (a0 = a ∨ a0 ∈ x0) ∧ a0 ∈ y

Proof step 5 from proof dups-appex (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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