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Metamath Proof Explorer

Theorem ldiv

Description: Left-division. (Contributed by BJ, 6-Jun-2019)

Ref Expression
Hypotheses ldiv.a 
|- ( ph -> A e. CC )
ldiv.b 
|- ( ph -> B e. CC )
ldiv.c 
|- ( ph -> C e. CC )
ldiv.bn0 
|- ( ph -> B =/= 0 )
Assertion ldiv 
|- ( ph -> ( ( A x. B ) = C <-> A = ( C / B ) ) )

Proof

Step Hyp Ref Expression
1 ldiv.a 
 |-  ( ph -> A e. CC )
2 ldiv.b 
 |-  ( ph -> B e. CC )
3 ldiv.c 
 |-  ( ph -> C e. CC )
4 ldiv.bn0 
 |-  ( ph -> B =/= 0 )
5 oveq1 
 |-  ( ( A x. B ) = C -> ( ( A x. B ) / B ) = ( C / B ) )
6 1 2 4 divcan4d 
 |-  ( ph -> ( ( A x. B ) / B ) = A )
7 6 eqeq1d 
 |-  ( ph -> ( ( ( A x. B ) / B ) = ( C / B ) <-> A = ( C / B ) ) )
8 5 7 imbitrid 
 |-  ( ph -> ( ( A x. B ) = C -> A = ( C / B ) ) )
9 oveq1 
 |-  ( A = ( C / B ) -> ( A x. B ) = ( ( C / B ) x. B ) )
10 3 2 4 divcan1d 
 |-  ( ph -> ( ( C / B ) x. B ) = C )
11 10 eqeq2d 
 |-  ( ph -> ( ( A x. B ) = ( ( C / B ) x. B ) <-> ( A x. B ) = C ) )
12 9 11 imbitrid 
 |-  ( ph -> ( A = ( C / B ) -> ( A x. B ) = C ) )
13 8 12 impbid 
 |-  ( ph -> ( ( A x. B ) = C <-> A = ( C / B ) ) )