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Proof step 2 from proof dups-03 (previous step)

list2set(x) ∩ list2set(y) = ∅, dups(x + y) ⊦ dups(y), dups(x)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ dups(x + y) ↔ dups(x) ∨ dups(y) ∨ (∃ a. a ∈ x ∧ a ∈ y)
Proof step 4:
list2set(x) ∩ list2set(y) = ∅, dups(x) ∨ dups(y) ∨ (∃ a. a ∈ x ∧ a ∈ y) ⊦ dups(y), dups(x)

Rule argument

seq: ⊦ dups(x + y) ↔ dups(x) ∨ dups(y) ∨ (∃ a. a ∈ x ∧ a ∈ y) subst: [y, x] → [y, x] rotate: F spec: list-dup inst: name: dups-appex

Additional information

Used terms: y, x