Theorem difuncomp

Description: Express a class difference using unions and class complements. (Contributed by Thierry Arnoux, 21-Jun-2020)

	Ref	Expression
Assertion	difuncomp	⊢ ( 𝐴 ⊆ 𝐶 → ( 𝐴 ∖ 𝐵 ) = ( 𝐶 ∖ ( ( 𝐶 ∖ 𝐴 ) ∪ 𝐵 ) ) )

Proof

Step	Hyp	Ref	Expression
1		sseqin2	⊢ ( 𝐴 ⊆ 𝐶 ↔ ( 𝐶 ∩ 𝐴 ) = 𝐴 )
2	1	biimpi	⊢ ( 𝐴 ⊆ 𝐶 → ( 𝐶 ∩ 𝐴 ) = 𝐴 )
3		incom	⊢ ( 𝐶 ∩ 𝐴 ) = ( 𝐴 ∩ 𝐶 )
4	2 3	eqtr3di	⊢ ( 𝐴 ⊆ 𝐶 → 𝐴 = ( 𝐴 ∩ 𝐶 ) )
5	4	difeq1d	⊢ ( 𝐴 ⊆ 𝐶 → ( 𝐴 ∖ 𝐵 ) = ( ( 𝐴 ∩ 𝐶 ) ∖ 𝐵 ) )
6		difundi	⊢ ( 𝐶 ∖ ( ( 𝐶 ∖ 𝐴 ) ∪ 𝐵 ) ) = ( ( 𝐶 ∖ ( 𝐶 ∖ 𝐴 ) ) ∩ ( 𝐶 ∖ 𝐵 ) )
7		dfss4	⊢ ( 𝐴 ⊆ 𝐶 ↔ ( 𝐶 ∖ ( 𝐶 ∖ 𝐴 ) ) = 𝐴 )
8	7	biimpi	⊢ ( 𝐴 ⊆ 𝐶 → ( 𝐶 ∖ ( 𝐶 ∖ 𝐴 ) ) = 𝐴 )
9	8	ineq1d	⊢ ( 𝐴 ⊆ 𝐶 → ( ( 𝐶 ∖ ( 𝐶 ∖ 𝐴 ) ) ∩ ( 𝐶 ∖ 𝐵 ) ) = ( 𝐴 ∩ ( 𝐶 ∖ 𝐵 ) ) )
10	6 9	syl5eq	⊢ ( 𝐴 ⊆ 𝐶 → ( 𝐶 ∖ ( ( 𝐶 ∖ 𝐴 ) ∪ 𝐵 ) ) = ( 𝐴 ∩ ( 𝐶 ∖ 𝐵 ) ) )
11		indif2	⊢ ( 𝐴 ∩ ( 𝐶 ∖ 𝐵 ) ) = ( ( 𝐴 ∩ 𝐶 ) ∖ 𝐵 )
12	10 11	eqtrdi	⊢ ( 𝐴 ⊆ 𝐶 → ( 𝐶 ∖ ( ( 𝐶 ∖ 𝐴 ) ∪ 𝐵 ) ) = ( ( 𝐴 ∩ 𝐶 ) ∖ 𝐵 ) )
13	5 12	eqtr4d	⊢ ( 𝐴 ⊆ 𝐶 → ( 𝐴 ∖ 𝐵 ) = ( 𝐶 ∖ ( ( 𝐶 ∖ 𝐴 ) ∪ 𝐵 ) ) )