Theorem disj3

Description: Two ways of saying that two classes are disjoint. (Contributed by NM, 19-May-1998)

	Ref	Expression
Assertion	disj3	⊢ ( ( 𝐴 ∩ 𝐵 ) = ∅ ↔ 𝐴 = ( 𝐴 ∖ 𝐵 ) )

Proof

Step	Hyp	Ref	Expression
1		pm4.71	⊢ ( ( 𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵 ) ↔ ( 𝑥 ∈ 𝐴 ↔ ( 𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵 ) ) )
2		eldif	⊢ ( 𝑥 ∈ ( 𝐴 ∖ 𝐵 ) ↔ ( 𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵 ) )
3	2	bibi2i	⊢ ( ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ ( 𝐴 ∖ 𝐵 ) ) ↔ ( 𝑥 ∈ 𝐴 ↔ ( 𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵 ) ) )
4	1 3	bitr4i	⊢ ( ( 𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵 ) ↔ ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ ( 𝐴 ∖ 𝐵 ) ) )
5	4	albii	⊢ ( ∀ 𝑥 ( 𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵 ) ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ ( 𝐴 ∖ 𝐵 ) ) )
6		disj1	⊢ ( ( 𝐴 ∩ 𝐵 ) = ∅ ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵 ) )
7		dfcleq	⊢ ( 𝐴 = ( 𝐴 ∖ 𝐵 ) ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ ( 𝐴 ∖ 𝐵 ) ) )
8	5 6 7	3bitr4i	⊢ ( ( 𝐴 ∩ 𝐵 ) = ∅ ↔ 𝐴 = ( 𝐴 ∖ 𝐵 ) )