Metamath Proof Explorer


Theorem inxpssidinxp

Description: Two ways to say that intersections with Cartesian products are in a subclass relation, special case of inxpss2 . (Contributed by Peter Mazsa, 4-Jul-2019)

Ref Expression
Assertion inxpssidinxp
|- ( ( R i^i ( A X. B ) ) C_ ( _I i^i ( A X. B ) ) <-> A. x e. A A. y e. B ( x R y -> x = y ) )

Proof

Step Hyp Ref Expression
1 inxpss2
 |-  ( ( R i^i ( A X. B ) ) C_ ( _I i^i ( A X. B ) ) <-> A. x e. A A. y e. B ( x R y -> x _I y ) )
2 ideqg
 |-  ( y e. _V -> ( x _I y <-> x = y ) )
3 2 elv
 |-  ( x _I y <-> x = y )
4 3 imbi2i
 |-  ( ( x R y -> x _I y ) <-> ( x R y -> x = y ) )
5 4 2ralbii
 |-  ( A. x e. A A. y e. B ( x R y -> x _I y ) <-> A. x e. A A. y e. B ( x R y -> x = y ) )
6 1 5 bitri
 |-  ( ( R i^i ( A X. B ) ) C_ ( _I i^i ( A X. B ) ) <-> A. x e. A A. y e. B ( x R y -> x = y ) )