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 simpll 
 |-  ( ( ( A e. CC /\ A =/= 0 ) /\ ( B e. CC /\ B =/= 0 ) ) -> A e. CC )
2 simprl 
 |-  ( ( ( A e. CC /\ A =/= 0 ) /\ ( B e. CC /\ B =/= 0 ) ) -> B e. CC )
3 simplr 
 |-  ( ( ( A e. CC /\ A =/= 0 ) /\ ( B e. CC /\ B =/= 0 ) ) -> A =/= 0 )
4 simprr 
 |-  ( ( ( A e. CC /\ A =/= 0 ) /\ ( B e. CC /\ B =/= 0 ) ) -> B =/= 0 )
5 1 2 3 4 subrecd 
 |-  ( ( ( A e. CC /\ A =/= 0 ) /\ ( B e. CC /\ B =/= 0 ) ) -> ( ( 1 / A ) - ( 1 / B ) ) = ( ( B - A ) / ( A x. B ) ) )