Metamath Proof Explorer


Theorem difprsn2

Description: Removal of a singleton from an unordered pair. (Contributed by Alexander van der Vekens, 5-Oct-2017)

Ref Expression
Assertion difprsn2 ( 𝐴 ≠ 𝐵 → ( { 𝐴 , 𝐵 } ∖ { 𝐵 } ) = { 𝐴 } )

Proof

Step Hyp Ref Expression
1 prcom ⊢ { 𝐴 , 𝐵 } = { 𝐵 , 𝐴 }
2 1 difeq1i ⊢ ( { 𝐴 , 𝐵 } ∖ { 𝐵 } ) = ( { 𝐵 , 𝐴 } ∖ { 𝐵 } )
3 necom ⊢ ( 𝐴 ≠ 𝐵 ↔ 𝐵 ≠ 𝐴 )
4 difprsn1 ⊢ ( 𝐵 ≠ 𝐴 → ( { 𝐵 , 𝐴 } ∖ { 𝐵 } ) = { 𝐴 } )
5 3 4 sylbi ⊢ ( 𝐴 ≠ 𝐵 → ( { 𝐵 , 𝐴 } ∖ { 𝐵 } ) = { 𝐴 } )
6 2 5 eqtrid ⊢ ( 𝐴 ≠ 𝐵 → ( { 𝐴 , 𝐵 } ∖ { 𝐵 } ) = { 𝐴 } )