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Proof step 2 from proof dups-05 (previous step)

⊦ dups(x + set2list(s)) ↔ s ∩ list2set(x) ≠ ∅, dups(x)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ ¬ dups(x) ∧ ¬ dups(y) → (dups(x + y) ↔ list2set(x) ∩ list2set(y) ≠ ∅)

Rule argument

name: dups-03 rotate: F subst: [y, x] → [set2list(s), x]

Additional information

Used simplifier rules: dups: ⊦ ¬ dups(set2list(s));%set2set: ⊦ list2set(set2list(s)) = s;% ⊦ s ∩ s0 = s0 ∩ s; Used terms: set2list(s), x