Metamath Proof Explorer

Theorem submul2

Description: Convert a subtraction to addition using multiplication by a negative. (Contributed by NM, 2-Feb-2007)

Ref Expression
Assertion submul2 
|- ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( A - ( B x. C ) ) = ( A + ( B x. -u C ) ) )

Proof

Step Hyp Ref Expression
1 mulneg2 
 |-  ( ( B e. CC /\ C e. CC ) -> ( B x. -u C ) = -u ( B x. C ) )
2 1 adantl 
 |-  ( ( A e. CC /\ ( B e. CC /\ C e. CC ) ) -> ( B x. -u C ) = -u ( B x. C ) )
3 2 oveq2d 
 |-  ( ( A e. CC /\ ( B e. CC /\ C e. CC ) ) -> ( A + ( B x. -u C ) ) = ( A + -u ( B x. C ) ) )
4 mulcl 
 |-  ( ( B e. CC /\ C e. CC ) -> ( B x. C ) e. CC )
5 negsub 
 |-  ( ( A e. CC /\ ( B x. C ) e. CC ) -> ( A + -u ( B x. C ) ) = ( A - ( B x. C ) ) )
6 4 5 sylan2 
 |-  ( ( A e. CC /\ ( B e. CC /\ C e. CC ) ) -> ( A + -u ( B x. C ) ) = ( A - ( B x. C ) ) )
7 3 6 eqtr2d 
 |-  ( ( A e. CC /\ ( B e. CC /\ C e. CC ) ) -> ( A - ( B x. C ) ) = ( A + ( B x. -u C ) ) )
8 7 3impb 
 |-  ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( A - ( B x. C ) ) = ( A + ( B x. -u C ) ) )