Metamath Proof Explorer


Theorem lrrecval2

Description: Next, we establish an alternate expression for R . (Contributed by Scott Fenton, 19-Aug-2024)

Ref Expression
Hypothesis lrrec.1
|- R = { <. x , y >. | x e. ( ( _L ` y ) u. ( _R ` y ) ) }
Assertion lrrecval2
|- ( ( A e. No /\ B e. No ) -> ( A R B <-> ( bday ` A ) e. ( bday ` B ) ) )

Proof

Step Hyp Ref Expression
1 lrrec.1
 |-  R = { <. x , y >. | x e. ( ( _L ` y ) u. ( _R ` y ) ) }
2 1 lrrecval
 |-  ( ( A e. No /\ B e. No ) -> ( A R B <-> A e. ( ( _L ` B ) u. ( _R ` B ) ) ) )
3 lrold
 |-  ( ( _L ` B ) u. ( _R ` B ) ) = ( _Old ` ( bday ` B ) )
4 3 a1i
 |-  ( ( A e. No /\ B e. No ) -> ( ( _L ` B ) u. ( _R ` B ) ) = ( _Old ` ( bday ` B ) ) )
5 4 eleq2d
 |-  ( ( A e. No /\ B e. No ) -> ( A e. ( ( _L ` B ) u. ( _R ` B ) ) <-> A e. ( _Old ` ( bday ` B ) ) ) )
6 bdayelon
 |-  ( bday ` B ) e. On
7 oldbday
 |-  ( ( ( bday ` B ) e. On /\ A e. No ) -> ( A e. ( _Old ` ( bday ` B ) ) <-> ( bday ` A ) e. ( bday ` B ) ) )
8 6 7 mpan
 |-  ( A e. No -> ( A e. ( _Old ` ( bday ` B ) ) <-> ( bday ` A ) e. ( bday ` B ) ) )
9 8 adantr
 |-  ( ( A e. No /\ B e. No ) -> ( A e. ( _Old ` ( bday ` B ) ) <-> ( bday ` A ) e. ( bday ` B ) ) )
10 2 5 9 3bitrd
 |-  ( ( A e. No /\ B e. No ) -> ( A R B <-> ( bday ` A ) e. ( bday ` B ) ) )