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

list2set(x) ∩ s = ∅, a ∈ s, a ∈ x ⊦ s = ∅, # s = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ s ∩ s0 = ∅ ↔ (∀ a. ¬ a ∈ s ∨ ¬ a ∈ s0)
Proof step 4:
∀ a. ¬ a ∈ list2set(x) ∨ ¬ a ∈ s, a ∈ s, a ∈ x ⊦ s = ∅, # s = 0

Rule argument

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

Additional information

Used terms: s, list2set(x)