Theorem reldisj

Description: Two ways of saying that two classes are disjoint, using the complement of B relative to a universe C . (Contributed by NM, 15-Feb-2007) (Proof shortened by Andrew Salmon, 26-Jun-2011)

	Ref	Expression
Assertion	reldisj	⊢ ( 𝐴 ⊆ 𝐶 → ( ( 𝐴 ∩ 𝐵 ) = ∅ ↔ 𝐴 ⊆ ( 𝐶 ∖ 𝐵 ) ) )

Proof

Step	Hyp	Ref	Expression
1		dfss2	⊢ ( 𝐴 ⊆ 𝐶 ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐶 ) )
2		pm5.44	⊢ ( ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐶 ) → ( ( 𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵 ) ↔ ( 𝑥 ∈ 𝐴 → ( 𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵 ) ) ) )
3		eldif	⊢ ( 𝑥 ∈ ( 𝐶 ∖ 𝐵 ) ↔ ( 𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵 ) )
4	3	imbi2i	⊢ ( ( 𝑥 ∈ 𝐴 → 𝑥 ∈ ( 𝐶 ∖ 𝐵 ) ) ↔ ( 𝑥 ∈ 𝐴 → ( 𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵 ) ) )
5	2 4	syl6bbr	⊢ ( ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐶 ) → ( ( 𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵 ) ↔ ( 𝑥 ∈ 𝐴 → 𝑥 ∈ ( 𝐶 ∖ 𝐵 ) ) ) )
6	5	sps	⊢ ( ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐶 ) → ( ( 𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵 ) ↔ ( 𝑥 ∈ 𝐴 → 𝑥 ∈ ( 𝐶 ∖ 𝐵 ) ) ) )
7	1 6	sylbi	⊢ ( 𝐴 ⊆ 𝐶 → ( ( 𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵 ) ↔ ( 𝑥 ∈ 𝐴 → 𝑥 ∈ ( 𝐶 ∖ 𝐵 ) ) ) )
8	7	albidv	⊢ ( 𝐴 ⊆ 𝐶 → ( ∀ 𝑥 ( 𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵 ) ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝑥 ∈ ( 𝐶 ∖ 𝐵 ) ) ) )
9		disj1	⊢ ( ( 𝐴 ∩ 𝐵 ) = ∅ ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵 ) )
10		dfss2	⊢ ( 𝐴 ⊆ ( 𝐶 ∖ 𝐵 ) ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝑥 ∈ ( 𝐶 ∖ 𝐵 ) ) )
11	8 9 10	3bitr4g	⊢ ( 𝐴 ⊆ 𝐶 → ( ( 𝐴 ∩ 𝐵 ) = ∅ ↔ 𝐴 ⊆ ( 𝐶 ∖ 𝐵 ) ) )