Metamath Proof Explorer

Theorem renegadd

Description: Relationship between real negation and addition. (Contributed by Steven Nguyen, 7-Jan-2023)

Ref Expression
Assertion renegadd 
|- ( ( A e. RR /\ B e. RR ) -> ( ( 0 -R A ) = B <-> ( A + B ) = 0 ) )

Proof

Step Hyp Ref Expression
1 elre0re 
 |-  ( A e. RR -> 0 e. RR )
2 resubval 
 |-  ( ( 0 e. RR /\ A e. RR ) -> ( 0 -R A ) = ( iota_ x e. RR ( A + x ) = 0 ) )
3 1 2 mpancom 
 |-  ( A e. RR -> ( 0 -R A ) = ( iota_ x e. RR ( A + x ) = 0 ) )
4 3 eqeq1d 
 |-  ( A e. RR -> ( ( 0 -R A ) = B <-> ( iota_ x e. RR ( A + x ) = 0 ) = B ) )
5 4 adantr 
 |-  ( ( A e. RR /\ B e. RR ) -> ( ( 0 -R A ) = B <-> ( iota_ x e. RR ( A + x ) = 0 ) = B ) )
6 renegeu 
 |-  ( A e. RR -> E! x e. RR ( A + x ) = 0 )
7 oveq2 
 |-  ( x = B -> ( A + x ) = ( A + B ) )
8 7 eqeq1d 
 |-  ( x = B -> ( ( A + x ) = 0 <-> ( A + B ) = 0 ) )
9 8 riota2 
 |-  ( ( B e. RR /\ E! x e. RR ( A + x ) = 0 ) -> ( ( A + B ) = 0 <-> ( iota_ x e. RR ( A + x ) = 0 ) = B ) )
10 6 9 sylan2 
 |-  ( ( B e. RR /\ A e. RR ) -> ( ( A + B ) = 0 <-> ( iota_ x e. RR ( A + x ) = 0 ) = B ) )
11 10 ancoms 
 |-  ( ( A e. RR /\ B e. RR ) -> ( ( A + B ) = 0 <-> ( iota_ x e. RR ( A + x ) = 0 ) = B ) )
12 5 11 bitr4d 
 |-  ( ( A e. RR /\ B e. RR ) -> ( ( 0 -R A ) = B <-> ( A + B ) = 0 ) )