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Proof step 4 from proof d (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 5:
dups(x0 + a ' + y), ∀ a, y. (dups(x0 + a ' + y) ↔ dups(x0 + y)) ∨ a ∈ y ∨ a ∈ x0 ⊦ dups(x0 + y), a = a0, a0 ∈ y, a0 ∈ x0, a ∈ y, a ∈ x0
Proof step 6:
dups(x0 + y), ∀ a, y. (dups(x0 + a ' + y) ↔ dups(x0 + y)) ∨ a ∈ y ∨ a ∈ x0 ⊦ dups(x0 + a ' + y), a = a0, a0 ∈ y, a0 ∈ x0, a ∈ y, a ∈ x0

Rule argument

pos: right 2

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