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Proof step 3 from proof rem-inter-empty3 (previous step)

(s \ s0) ∩ s1 = ∅, (s2 \ s) ∩ s1 = ∅ ⊦ #((s2 \ s0) ∩ s1) = 0, (s2 \ s0) ∩ s1 = ∅

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2)
Proof step 5:
(s2 \ s) ∩ s1 = ∅, (s \ s0) ∩ s1 = ∅ ⊦ #((s2 \ s0) ∩ s1) = 0, ∀ a. a ∈ (s2 \ s0) ∩ s1 ↔ a ∈ ∅

Rule argument

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

Additional information

Used terms: ∅, (s2 \ s0) ∩ s1