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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 ) )