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Proof step 2 from proof dups-02 (previous step)

⊦ #(list2set(x) ∩ list2set(y)) = 0, list2set(x) ∩ list2set(y) = ∅, dups(y), dups(x), dups(x + y)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ s ∩ s0 = ∅ ↔ (∀ a. ¬ a ∈ s ∨ ¬ a ∈ s0)

Rule argument

seq: ⊦ s ∩ s0 = ∅ ↔ (∀ a. ¬ a ∈ s ∨ ¬ a ∈ s0) subst: [s0, s] → [list2set(y), list2set(x)] rotate: F spec: set-union inst: name: int-empty-01

Additional information

Used simplifier rules: in: ⊦ a ∈ list2set(x) ↔ a ∈ x;% ⊦ a ∈ x ∧ a ∈ y → dups(x + y);% ⊦ (x + y) + z = x + y + z; Used terms: list2set(y), list2set(x)