Metamath Proof Explorer


Theorem eqscut

Description: Condition for equality to a surreal cut. (Contributed by Scott Fenton, 8-Aug-2024)

Ref Expression
Assertion eqscut
|- ( ( L < ( ( L |s R ) = X <-> ( L <

Proof

Step Hyp Ref Expression
1 scutval
 |-  ( L < ( L |s R ) = ( iota_ x e. { y e. No | ( L <
2 1 adantr
 |-  ( ( L < ( L |s R ) = ( iota_ x e. { y e. No | ( L <
3 sneq
 |-  ( x = y -> { x } = { y } )
4 3 breq2d
 |-  ( x = y -> ( L < L <
5 3 breq1d
 |-  ( x = y -> ( { x } < { y } <
6 4 5 anbi12d
 |-  ( x = y -> ( ( L < ( L <
7 6 riotarab
 |-  ( iota_ x e. { y e. No | ( L <
8 2 7 eqtrdi
 |-  ( ( L < ( L |s R ) = ( iota_ x e. No ( ( L <
9 8 eqeq1d
 |-  ( ( L < ( ( L |s R ) = X <-> ( iota_ x e. No ( ( L <
10 conway
 |-  ( L < E! x e. { y e. No | ( L <
11 6 reurab
 |-  ( E! x e. { y e. No | ( L < E! x e. No ( ( L <
12 10 11 sylib
 |-  ( L < E! x e. No ( ( L <
13 df-3an
 |-  ( ( L < ( ( L <
14 sneq
 |-  ( x = X -> { x } = { X } )
15 14 breq2d
 |-  ( x = X -> ( L < L <
16 14 breq1d
 |-  ( x = X -> ( { x } < { X } <
17 fveqeq2
 |-  ( x = X -> ( ( bday ` x ) = |^| ( bday " { y e. No | ( L < ( bday ` X ) = |^| ( bday " { y e. No | ( L <
18 15 16 17 3anbi123d
 |-  ( x = X -> ( ( L < ( L <
19 13 18 bitr3id
 |-  ( x = X -> ( ( ( L < ( L <
20 19 riota2
 |-  ( ( X e. No /\ E! x e. No ( ( L < ( ( L < ( iota_ x e. No ( ( L <
21 20 ancoms
 |-  ( ( E! x e. No ( ( L < ( ( L < ( iota_ x e. No ( ( L <
22 12 21 sylan
 |-  ( ( L < ( ( L < ( iota_ x e. No ( ( L <
23 9 22 bitr4d
 |-  ( ( L < ( ( L |s R ) = X <-> ( L <