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Proof step 7 from proof dups-appex (previous step)

a ∈ 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 quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 4 terms: a

Additional information

Used terms: a