Metamath Proof Explorer

Theorem sn-00idlem2

Description: Lemma for sn-00id . (Contributed by SN, 25-Dec-2023)

Ref Expression
Assertion sn-00idlem2 
|- ( ( 0 -R 0 ) =/= 0 -> ( 0 -R 0 ) = 1 )

Proof

Step Hyp Ref Expression
1 0re 
 |-  0 e. RR
2 rennncan2 
 |-  ( ( 0 e. RR /\ 0 e. RR /\ 0 e. RR ) -> ( ( 0 -R 0 ) -R ( 0 -R 0 ) ) = ( 0 -R 0 ) )
3 1 1 1 2 mp3an 
 |-  ( ( 0 -R 0 ) -R ( 0 -R 0 ) ) = ( 0 -R 0 )
4 re1m1e0m0 
 |-  ( 1 -R 1 ) = ( 0 -R 0 )
5 3 4 eqtr4i 
 |-  ( ( 0 -R 0 ) -R ( 0 -R 0 ) ) = ( 1 -R 1 )
6 rernegcl 
 |-  ( 0 e. RR -> ( 0 -R 0 ) e. RR )
7 1 6 ax-mp 
 |-  ( 0 -R 0 ) e. RR
8 sn-00idlem1 
 |-  ( ( 0 -R 0 ) e. RR -> ( ( 0 -R 0 ) x. ( 0 -R 0 ) ) = ( ( 0 -R 0 ) -R ( 0 -R 0 ) ) )
9 7 8 ax-mp 
 |-  ( ( 0 -R 0 ) x. ( 0 -R 0 ) ) = ( ( 0 -R 0 ) -R ( 0 -R 0 ) )
10 1re 
 |-  1 e. RR
11 sn-00idlem1 
 |-  ( 1 e. RR -> ( 1 x. ( 0 -R 0 ) ) = ( 1 -R 1 ) )
12 10 11 ax-mp 
 |-  ( 1 x. ( 0 -R 0 ) ) = ( 1 -R 1 )
13 5 9 12 3eqtr4i 
 |-  ( ( 0 -R 0 ) x. ( 0 -R 0 ) ) = ( 1 x. ( 0 -R 0 ) )
14 7 a1i 
 |-  ( ( 0 -R 0 ) =/= 0 -> ( 0 -R 0 ) e. RR )
15 1red 
 |-  ( ( 0 -R 0 ) =/= 0 -> 1 e. RR )
16 id 
 |-  ( ( 0 -R 0 ) =/= 0 -> ( 0 -R 0 ) =/= 0 )
17 14 15 14 16 remulcan2d 
 |-  ( ( 0 -R 0 ) =/= 0 -> ( ( ( 0 -R 0 ) x. ( 0 -R 0 ) ) = ( 1 x. ( 0 -R 0 ) ) <-> ( 0 -R 0 ) = 1 ) )
18 13 17 mpbii 
 |-  ( ( 0 -R 0 ) =/= 0 -> ( 0 -R 0 ) = 1 )