Metamath Proof Explorer


Theorem difxp1

Description: Difference law for Cartesian product. (Contributed by Scott Fenton, 18-Feb-2013) (Revised by Mario Carneiro, 26-Jun-2014)

Ref Expression
Assertion difxp1 ( ( 𝐴 ∖ 𝐵 ) × 𝐶 ) = ( ( 𝐴 × 𝐶 ) ∖ ( 𝐵 × 𝐶 ) )

Proof

Step Hyp Ref Expression
1 difxp ⊢ ( ( 𝐴 × 𝐶 ) ∖ ( 𝐵 × 𝐶 ) ) = ( ( ( 𝐴 ∖ 𝐵 ) × 𝐶 ) ∪ ( 𝐴 × ( 𝐶 ∖ 𝐶 ) ) )
2 difid ⊢ ( 𝐶 ∖ 𝐶 ) = ∅
3 2 xpeq2i ⊢ ( 𝐴 × ( 𝐶 ∖ 𝐶 ) ) = ( 𝐴 × ∅ )
4 xp0 ⊢ ( 𝐴 × ∅ ) = ∅
5 3 4 eqtri ⊢ ( 𝐴 × ( 𝐶 ∖ 𝐶 ) ) = ∅
6 5 uneq2i ⊢ ( ( ( 𝐴 ∖ 𝐵 ) × 𝐶 ) ∪ ( 𝐴 × ( 𝐶 ∖ 𝐶 ) ) ) = ( ( ( 𝐴 ∖ 𝐵 ) × 𝐶 ) ∪ ∅ )
7 un0 ⊢ ( ( ( 𝐴 ∖ 𝐵 ) × 𝐶 ) ∪ ∅ ) = ( ( 𝐴 ∖ 𝐵 ) × 𝐶 )
8 1 6 7 3eqtrri ⊢ ( ( 𝐴 ∖ 𝐵 ) × 𝐶 ) = ( ( 𝐴 × 𝐶 ) ∖ ( 𝐵 × 𝐶 ) )