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Proof step 3 from proof dups-appex (previous step)

∀ 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 all left

And obtained the following new goals

Proof step 4:
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

Rule argument

pos: left 1 terms: y abort: F discard: F computed: T

Additional information

Used terms: y