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Metamath Proof Explorer

Theorem sn-00idlem1

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

Ref Expression
Assertion sn-00idlem1 
|- ( A e. RR -> ( A x. ( 0 -R 0 ) ) = ( A -R A ) )

Proof

Step Hyp Ref Expression
1 1re 
 |-  1 e. RR
2 resubdi 
 |-  ( ( A e. RR /\ 1 e. RR /\ 1 e. RR ) -> ( A x. ( 1 -R 1 ) ) = ( ( A x. 1 ) -R ( A x. 1 ) ) )
3 1 1 2 mp3an23 
 |-  ( A e. RR -> ( A x. ( 1 -R 1 ) ) = ( ( A x. 1 ) -R ( A x. 1 ) ) )
4 re1m1e0m0 
 |-  ( 1 -R 1 ) = ( 0 -R 0 )
5 4 oveq2i 
 |-  ( A x. ( 1 -R 1 ) ) = ( A x. ( 0 -R 0 ) )
6 5 a1i 
 |-  ( A e. RR -> ( A x. ( 1 -R 1 ) ) = ( A x. ( 0 -R 0 ) ) )
7 ax-1rid 
 |-  ( A e. RR -> ( A x. 1 ) = A )
8 7 7 oveq12d 
 |-  ( A e. RR -> ( ( A x. 1 ) -R ( A x. 1 ) ) = ( A -R A ) )
9 3 6 8 3eqtr3d 
 |-  ( A e. RR -> ( A x. ( 0 -R 0 ) ) = ( A -R A ) )