Metamath Proof Explorer


Theorem difres

Description: Case when class difference in unaffected by restriction. (Contributed by Thierry Arnoux, 1-Jan-2020)

Ref Expression
Assertion difres ( 𝐴 ⊆ ( 𝐵 × V ) → ( 𝐴 ∖ ( 𝐶𝐵 ) ) = ( 𝐴𝐶 ) )

Proof

Step Hyp Ref Expression
1 df-res ( 𝐶𝐵 ) = ( 𝐶 ∩ ( 𝐵 × V ) )
2 1 difeq2i ( 𝐴 ∖ ( 𝐶𝐵 ) ) = ( 𝐴 ∖ ( 𝐶 ∩ ( 𝐵 × V ) ) )
3 difindi ( 𝐴 ∖ ( 𝐶 ∩ ( 𝐵 × V ) ) ) = ( ( 𝐴𝐶 ) ∪ ( 𝐴 ∖ ( 𝐵 × V ) ) )
4 ssdif ( 𝐴 ⊆ ( 𝐵 × V ) → ( 𝐴 ∖ ( 𝐵 × V ) ) ⊆ ( ( 𝐵 × V ) ∖ ( 𝐵 × V ) ) )
5 difid ( ( 𝐵 × V ) ∖ ( 𝐵 × V ) ) = ∅
6 4 5 sseqtrdi ( 𝐴 ⊆ ( 𝐵 × V ) → ( 𝐴 ∖ ( 𝐵 × V ) ) ⊆ ∅ )
7 ss0 ( ( 𝐴 ∖ ( 𝐵 × V ) ) ⊆ ∅ → ( 𝐴 ∖ ( 𝐵 × V ) ) = ∅ )
8 6 7 syl ( 𝐴 ⊆ ( 𝐵 × V ) → ( 𝐴 ∖ ( 𝐵 × V ) ) = ∅ )
9 8 uneq2d ( 𝐴 ⊆ ( 𝐵 × V ) → ( ( 𝐴𝐶 ) ∪ ( 𝐴 ∖ ( 𝐵 × V ) ) ) = ( ( 𝐴𝐶 ) ∪ ∅ ) )
10 3 9 syl5eq ( 𝐴 ⊆ ( 𝐵 × V ) → ( 𝐴 ∖ ( 𝐶 ∩ ( 𝐵 × V ) ) ) = ( ( 𝐴𝐶 ) ∪ ∅ ) )
11 un0 ( ( 𝐴𝐶 ) ∪ ∅ ) = ( 𝐴𝐶 )
12 10 11 eqtrdi ( 𝐴 ⊆ ( 𝐵 × V ) → ( 𝐴 ∖ ( 𝐶 ∩ ( 𝐵 × V ) ) ) = ( 𝐴𝐶 ) )
13 2 12 syl5eq ( 𝐴 ⊆ ( 𝐵 × V ) → ( 𝐴 ∖ ( 𝐶𝐵 ) ) = ( 𝐴𝐶 ) )