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Proof step 4 from proof dups-appex (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 5:
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

Rule argument

pos: right 1

Additional information

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