# 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 ) ) )`