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Proof step 1 from proof f-01 (conclusion)

⊦ dups(x + a ' + y) ↔ a ∈ x ∨ a ∈ y ∨ dups(x + y)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ ¬ a ∈ x ∧ ¬ a ∈ y → (dups(x + a ' + y) ↔ dups(x + y))
Proof step 3:
¬(¬ a ∈ x ∧ ¬ a ∈ y) ⊦ dups(x + a ' + y) ↔ a ∈ x ∨ a ∈ y ∨ dups(x + y)

Rule argument

spec: inst: name: d seq: ⊦ ¬ a ∈ x ∧ ¬ a ∈ y → (dups(x + a ' + y) ↔ dups(x + y)) subst: [y, a, x] → [y, a, x] flags: rewrite, current

Additional information

Used terms: y, a, x