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Proof step 4 from proof f-01 (
previous step
)
⊦ dups(x + a ' + y) ↔ a ∈ x ∨ a ∈ y ∨ dups(x + y), ¬ a ∈ x ∧ ¬ a ∈ y
User applied rule case distinction
And obtained the following new goals
Proof step 5
:
a ∈ x ∨ a ∈ y ∨ dups(x + y) ⊦ dups(x + a ' + y), ¬ a ∈ x ∧ ¬ a ∈ y
Rule argument
pos: right 1
Additional information