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) Avoid ax-12 . (Revised by GG, 28-Jun-2024)

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

Proof

Step	Hyp	Ref	Expression
1		df-ss	⊢ ( 𝐴 ⊆ 𝐶 ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐶 ) )
2		eleq1w	⊢ ( 𝑥 = 𝑦 → ( 𝑥 ∈ 𝐴 ↔ 𝑦 ∈ 𝐴 ) )
3		eleq1w	⊢ ( 𝑥 = 𝑦 → ( 𝑥 ∈ 𝐶 ↔ 𝑦 ∈ 𝐶 ) )
4	2 3	imbi12d	⊢ ( 𝑥 = 𝑦 → ( ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐶 ) ↔ ( 𝑦 ∈ 𝐴 → 𝑦 ∈ 𝐶 ) ) )
5	4	spw	⊢ ( ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐶 ) → ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐶 ) )
6		pm5.44	⊢ ( ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐶 ) → ( ( 𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵 ) ↔ ( 𝑥 ∈ 𝐴 → ( 𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵 ) ) ) )
7		eldif	⊢ ( 𝑥 ∈ ( 𝐶 ∖ 𝐵 ) ↔ ( 𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵 ) )
8	7	imbi2i	⊢ ( ( 𝑥 ∈ 𝐴 → 𝑥 ∈ ( 𝐶 ∖ 𝐵 ) ) ↔ ( 𝑥 ∈ 𝐴 → ( 𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵 ) ) )
9	6 8	bitr4di	⊢ ( ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐶 ) → ( ( 𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵 ) ↔ ( 𝑥 ∈ 𝐴 → 𝑥 ∈ ( 𝐶 ∖ 𝐵 ) ) ) )
10	5 9	syl	⊢ ( ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐶 ) → ( ( 𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵 ) ↔ ( 𝑥 ∈ 𝐴 → 𝑥 ∈ ( 𝐶 ∖ 𝐵 ) ) ) )
11	1 10	sylbi	⊢ ( 𝐴 ⊆ 𝐶 → ( ( 𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵 ) ↔ ( 𝑥 ∈ 𝐴 → 𝑥 ∈ ( 𝐶 ∖ 𝐵 ) ) ) )
12	11	albidv	⊢ ( 𝐴 ⊆ 𝐶 → ( ∀ 𝑥 ( 𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵 ) ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝑥 ∈ ( 𝐶 ∖ 𝐵 ) ) ) )
13		disj1	⊢ ( ( 𝐴 ∩ 𝐵 ) = ∅ ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵 ) )
14		df-ss	⊢ ( 𝐴 ⊆ ( 𝐶 ∖ 𝐵 ) ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝑥 ∈ ( 𝐶 ∖ 𝐵 ) ) )
15	12 13 14	3bitr4g	⊢ ( 𝐴 ⊆ 𝐶 → ( ( 𝐴 ∩ 𝐵 ) = ∅ ↔ 𝐴 ⊆ ( 𝐶 ∖ 𝐵 ) ) )