Metamath Proof Explorer

Theorem subrec

Description: Subtraction of reciprocals. (Contributed by Scott Fenton, 9-Jul-2015)

Ref Expression
Assertion subrec 
|- ( ( ( A e. CC /\ A =/= 0 ) /\ ( B e. CC /\ B =/= 0 ) ) -> ( ( 1 / A ) - ( 1 / B ) ) = ( ( B - A ) / ( A x. B ) ) )

Proof

Step Hyp Ref Expression
1 1cnd 
 |-  ( ( ( A e. CC /\ A =/= 0 ) /\ ( B e. CC /\ B =/= 0 ) ) -> 1 e. CC )
2 simpll 
 |-  ( ( ( A e. CC /\ A =/= 0 ) /\ ( B e. CC /\ B =/= 0 ) ) -> A e. CC )
3 simprl 
 |-  ( ( ( A e. CC /\ A =/= 0 ) /\ ( B e. CC /\ B =/= 0 ) ) -> B e. CC )
4 simplr 
 |-  ( ( ( A e. CC /\ A =/= 0 ) /\ ( B e. CC /\ B =/= 0 ) ) -> A =/= 0 )
5 simprr 
 |-  ( ( ( A e. CC /\ A =/= 0 ) /\ ( B e. CC /\ B =/= 0 ) ) -> B =/= 0 )
6 1 2 1 3 4 5 divsubdivd 
 |-  ( ( ( A e. CC /\ A =/= 0 ) /\ ( B e. CC /\ B =/= 0 ) ) -> ( ( 1 / A ) - ( 1 / B ) ) = ( ( ( 1 x. B ) - ( 1 x. A ) ) / ( A x. B ) ) )
7 3 mulid2d 
 |-  ( ( ( A e. CC /\ A =/= 0 ) /\ ( B e. CC /\ B =/= 0 ) ) -> ( 1 x. B ) = B )
8 2 mulid2d 
 |-  ( ( ( A e. CC /\ A =/= 0 ) /\ ( B e. CC /\ B =/= 0 ) ) -> ( 1 x. A ) = A )
9 7 8 oveq12d 
 |-  ( ( ( A e. CC /\ A =/= 0 ) /\ ( B e. CC /\ B =/= 0 ) ) -> ( ( 1 x. B ) - ( 1 x. A ) ) = ( B - A ) )
10 9 oveq1d 
 |-  ( ( ( A e. CC /\ A =/= 0 ) /\ ( B e. CC /\ B =/= 0 ) ) -> ( ( ( 1 x. B ) - ( 1 x. A ) ) / ( A x. B ) ) = ( ( B - A ) / ( A x. B ) ) )
11 6 10 eqtrd 
 |-  ( ( ( A e. CC /\ A =/= 0 ) /\ ( B e. CC /\ B =/= 0 ) ) -> ( ( 1 / A ) - ( 1 / B ) ) = ( ( B - A ) / ( A x. B ) ) )