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Metamath Proof Explorer

Theorem subrecd

Description: Subtraction of reciprocals. (Contributed by Scott Fenton, 9-Jan-2017)

Ref Expression
Hypotheses subrecd.1 
|- ( ph -> A e. CC )
subrecd.2 
|- ( ph -> B e. CC )
subrecd.3 
|- ( ph -> A =/= 0 )
subrecd.4 
|- ( ph -> B =/= 0 )
Assertion subrecd 
|- ( ph -> ( ( 1 / A ) - ( 1 / B ) ) = ( ( B - A ) / ( A x. B ) ) )

Proof

Step Hyp Ref Expression
1 subrecd.1 
 |-  ( ph -> A e. CC )
2 subrecd.2 
 |-  ( ph -> B e. CC )
3 subrecd.3 
 |-  ( ph -> A =/= 0 )
4 subrecd.4 
 |-  ( ph -> B =/= 0 )
5 1cnd 
 |-  ( ph -> 1 e. CC )
6 5 1 5 2 3 4 divsubdivd 
 |-  ( ph -> ( ( 1 / A ) - ( 1 / B ) ) = ( ( ( 1 x. B ) - ( 1 x. A ) ) / ( A x. B ) ) )
7 2 mullidd 
 |-  ( ph -> ( 1 x. B ) = B )
8 1 mullidd 
 |-  ( ph -> ( 1 x. A ) = A )
9 7 8 oveq12d 
 |-  ( ph -> ( ( 1 x. B ) - ( 1 x. A ) ) = ( B - A ) )
10 9 oveq1d 
 |-  ( ph -> ( ( ( 1 x. B ) - ( 1 x. A ) ) / ( A x. B ) ) = ( ( B - A ) / ( A x. B ) ) )
11 6 10 eqtrd 
 |-  ( ph -> ( ( 1 / A ) - ( 1 / B ) ) = ( ( B - A ) / ( A x. B ) ) )