Metamath Proof Explorer


Theorem redivcan2d

Description: A cancellation law for division. (Contributed by SN, 25-Nov-2025)

Ref Expression
Hypotheses redivcan2d.a
|- ( ph -> A e. RR )
redivcan2d.b
|- ( ph -> B e. RR )
redivcan2d.z
|- ( ph -> B =/= 0 )
Assertion redivcan2d
|- ( ph -> ( B x. ( A /R B ) ) = A )

Proof

Step Hyp Ref Expression
1 redivcan2d.a
 |-  ( ph -> A e. RR )
2 redivcan2d.b
 |-  ( ph -> B e. RR )
3 redivcan2d.z
 |-  ( ph -> B =/= 0 )
4 eqidd
 |-  ( ph -> ( A /R B ) = ( A /R B ) )
5 1 2 3 sn-redivcld
 |-  ( ph -> ( A /R B ) e. RR )
6 1 5 2 3 redivmuld
 |-  ( ph -> ( ( A /R B ) = ( A /R B ) <-> ( B x. ( A /R B ) ) = A ) )
7 4 6 mpbid
 |-  ( ph -> ( B x. ( A /R B ) ) = A )