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Proof step 8 from proof dups-appex (previous step)

dups(x0 + y), dups(x0 + y) ↔ (∃ a. a ∈ x0 ∧ a ∈ 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
a0 ∈ x0, a0 ∈ 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

Rule argument

Additional information

Used simplifier rules: in-06: ⊦ a ∈ x ∧ a ∈ y → dups(x + y);%a: ⊦ (x + y) + z = x + y + z; from specification list