Metamath Proof Explorer


Theorem rediv23d

Description: A "commutative"/associative law 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 rediv23d
|- ( ph -> ( ( A x. B ) /R C ) = ( ( A /R C ) x. 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 3 4 sn-rereccld
 |-  ( ph -> ( 1 /R C ) e. RR )
6 5 recnd
 |-  ( ph -> ( 1 /R C ) e. CC )
7 1 recnd
 |-  ( ph -> A e. CC )
8 2 recnd
 |-  ( ph -> B e. CC )
9 6 7 8 mulassd
 |-  ( ph -> ( ( ( 1 /R C ) x. A ) x. B ) = ( ( 1 /R C ) x. ( A x. B ) ) )
10 1 3 4 redivrec2d
 |-  ( ph -> ( A /R C ) = ( ( 1 /R C ) x. A ) )
11 10 oveq1d
 |-  ( ph -> ( ( A /R C ) x. B ) = ( ( ( 1 /R C ) x. A ) x. B ) )
12 1 2 remulcld
 |-  ( ph -> ( A x. B ) e. RR )
13 12 3 4 redivrec2d
 |-  ( ph -> ( ( A x. B ) /R C ) = ( ( 1 /R C ) x. ( A x. B ) ) )
14 9 11 13 3eqtr4rd
 |-  ( ph -> ( ( A x. B ) /R C ) = ( ( A /R C ) x. B ) )