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Proof step 7 from proof subseteq-inter-empty-empty (previous step)

∀ a. a ∈ s0 → a ∈ s1, ∀ a. ¬ a ∈ s1 ∨ ¬ a ∈ s2 ⊦ #(s0 ∩ s2) = 0, s0 ∩ s2 = ∅

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2)
Proof step 9:
∀ a. a ∈ s0 → a ∈ s1, ∀ a. ¬ a ∈ s1 ∨ ¬ a ∈ s2 ⊦ #(s0 ∩ s2) = 0, ∀ a. a ∈ s0 ∩ s2 ↔ a ∈ ∅

Rule argument

spec: set-basic inst: name: Extension seq: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2) subst: [s2, s1] → [∅, s0 ∩ s2] flags: rewrite

Additional information

Used terms: ∅, s0 ∩ s2