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Metamath Proof Explorer

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 eqtrid ⊢ ( 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 ) ) )