Metamath Proof Explorer


Theorem difsnid

Description: If we remove a single element from a class then put it back in, we end up with the original class. (Contributed by NM, 2-Oct-2006)

Ref Expression
Assertion difsnid ( 𝐵𝐴 → ( ( 𝐴 ∖ { 𝐵 } ) ∪ { 𝐵 } ) = 𝐴 )

Proof

Step Hyp Ref Expression
1 snssi ( 𝐵𝐴 → { 𝐵 } ⊆ 𝐴 )
2 undifr ( { 𝐵 } ⊆ 𝐴 ↔ ( ( 𝐴 ∖ { 𝐵 } ) ∪ { 𝐵 } ) = 𝐴 )
3 1 2 sylib ( 𝐵𝐴 → ( ( 𝐴 ∖ { 𝐵 } ) ∪ { 𝐵 } ) = 𝐴 )