Metamath Proof Explorer


Theorem rediv11d

Description: One-to-one relationship for division. (Contributed by SN, 9-Apr-2026)

Ref Expression
Hypotheses rediv23d.a
|- ( ph -> A e. RR )
rediv23d.b
|- ( ph -> B e. RR )
rediv23d.c
|- ( ph -> C e. RR )
rediv23d.z
|- ( ph -> C =/= 0 )
Assertion rediv11d
|- ( ph -> ( ( A /R C ) = ( B /R C ) <-> A = B ) )

Proof

Step Hyp Ref Expression
1 rediv23d.a
 |-  ( ph -> A e. RR )
2 rediv23d.b
 |-  ( ph -> B e. RR )
3 rediv23d.c
 |-  ( ph -> C e. RR )
4 rediv23d.z
 |-  ( ph -> C =/= 0 )
5 2 3 4 sn-redivcld
 |-  ( ph -> ( B /R C ) e. RR )
6 1 5 3 4 redivmul2d
 |-  ( ph -> ( ( A /R C ) = ( B /R C ) <-> A = ( C x. ( B /R C ) ) ) )
7 2 3 4 redivcan2d
 |-  ( ph -> ( C x. ( B /R C ) ) = B )
8 7 eqeq2d
 |-  ( ph -> ( A = ( C x. ( B /R C ) ) <-> A = B ) )
9 6 8 bitrd
 |-  ( ph -> ( ( A /R C ) = ( B /R C ) <-> A = B ) )