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Proof step 5 from proof subseteq-inter-empty-empty (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2)
Proof step 7:
∀ a. a ∈ s0 → a ∈ s1, ∀ a. a ∈ s1 ∩ s2 ↔ a ∈ ∅ ⊦ #(s0 ∩ s2) = 0, s0 ∩ s2 = ∅

Rule argument

spec: set-basic name: Extension seq: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2) subst: [s2, s1] → [∅, s1 ∩ s2] flags: <all paths>

Additional information

Used terms: ∅, s1 ∩ s2