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Metamath Proof Explorer

Theorem resubadd

Description: Relation between real subtraction and addition. Based on subadd . (Contributed by Steven Nguyen, 7-Jan-2023)

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

Proof

Step Hyp Ref Expression
1 resubval 
 |-  ( ( A e. RR /\ B e. RR ) -> ( A -R B ) = ( iota_ x e. RR ( B + x ) = A ) )
2 1 eqeq1d 
 |-  ( ( A e. RR /\ B e. RR ) -> ( ( A -R B ) = C <-> ( iota_ x e. RR ( B + x ) = A ) = C ) )
3 2 3adant3 
 |-  ( ( A e. RR /\ B e. RR /\ C e. RR ) -> ( ( A -R B ) = C <-> ( iota_ x e. RR ( B + x ) = A ) = C ) )
4 resubeu 
 |-  ( ( B e. RR /\ A e. RR ) -> E! x e. RR ( B + x ) = A )
5 oveq2 
 |-  ( x = C -> ( B + x ) = ( B + C ) )
6 5 eqeq1d 
 |-  ( x = C -> ( ( B + x ) = A <-> ( B + C ) = A ) )
7 6 riota2 
 |-  ( ( C e. RR /\ E! x e. RR ( B + x ) = A ) -> ( ( B + C ) = A <-> ( iota_ x e. RR ( B + x ) = A ) = C ) )
8 4 7 sylan2 
 |-  ( ( C e. RR /\ ( B e. RR /\ A e. RR ) ) -> ( ( B + C ) = A <-> ( iota_ x e. RR ( B + x ) = A ) = C ) )
9 8 3impb 
 |-  ( ( C e. RR /\ B e. RR /\ A e. RR ) -> ( ( B + C ) = A <-> ( iota_ x e. RR ( B + x ) = A ) = C ) )
10 9 3com13 
 |-  ( ( A e. RR /\ B e. RR /\ C e. RR ) -> ( ( B + C ) = A <-> ( iota_ x e. RR ( B + x ) = A ) = C ) )
11 3 10 bitr4d 
 |-  ( ( A e. RR /\ B e. RR /\ C e. RR ) -> ( ( A -R B ) = C <-> ( B + C ) = A ) )