Metamath Proof Explorer


Theorem dfss7

Description: Alternate definition of subclass relationship. (Contributed by AV, 1-Aug-2022)

Ref Expression
Assertion dfss7 ( 𝐵𝐴 ↔ { 𝑥𝐴𝑥𝐵 } = 𝐵 )

Proof

Step Hyp Ref Expression
1 df-ss ( 𝐵𝐴 ↔ ( 𝐵𝐴 ) = 𝐵 )
2 incom ( 𝐵𝐴 ) = ( 𝐴𝐵 )
3 dfin5 ( 𝐴𝐵 ) = { 𝑥𝐴𝑥𝐵 }
4 2 3 eqtri ( 𝐵𝐴 ) = { 𝑥𝐴𝑥𝐵 }
5 4 eqeq1i ( ( 𝐵𝐴 ) = 𝐵 ↔ { 𝑥𝐴𝑥𝐵 } = 𝐵 )
6 1 5 bitri ( 𝐵𝐴 ↔ { 𝑥𝐴𝑥𝐵 } = 𝐵 )