Metamath Proof Explorer


Theorem cnvss

Description: Subset theorem for converse. (Contributed by NM, 22-Mar-1998) (Proof shortened by Kyle Wyonch, 27-Apr-2021)

Ref Expression
Assertion cnvss
|- ( A C_ B -> `' A C_ `' B )

Proof

Step Hyp Ref Expression
1 ssbr
 |-  ( A C_ B -> ( y A x -> y B x ) )
2 1 ssopab2dv
 |-  ( A C_ B -> { <. x , y >. | y A x } C_ { <. x , y >. | y B x } )
3 df-cnv
 |-  `' A = { <. x , y >. | y A x }
4 df-cnv
 |-  `' B = { <. x , y >. | y B x }
5 2 3 4 3sstr4g
 |-  ( A C_ B -> `' A C_ `' B )