Metamath Proof Explorer


Theorem difcom

Description: Swap the arguments of a class difference. (Contributed by NM, 29-Mar-2007)

Ref Expression
Assertion difcom ( ( 𝐴𝐵 ) ⊆ 𝐶 ↔ ( 𝐴𝐶 ) ⊆ 𝐵 )

Proof

Step Hyp Ref Expression
1 uncom ( 𝐵𝐶 ) = ( 𝐶𝐵 )
2 1 sseq2i ( 𝐴 ⊆ ( 𝐵𝐶 ) ↔ 𝐴 ⊆ ( 𝐶𝐵 ) )
3 ssundif ( 𝐴 ⊆ ( 𝐵𝐶 ) ↔ ( 𝐴𝐵 ) ⊆ 𝐶 )
4 ssundif ( 𝐴 ⊆ ( 𝐶𝐵 ) ↔ ( 𝐴𝐶 ) ⊆ 𝐵 )
5 2 3 4 3bitr3i ( ( 𝐴𝐵 ) ⊆ 𝐶 ↔ ( 𝐴𝐶 ) ⊆ 𝐵 )