Metamath Proof Explorer


Theorem 5recm6rec

Description: One fifth minus one sixth. (Contributed by Scott Fenton, 9-Jan-2017)

Ref Expression
Assertion 5recm6rec
|- ( ( 1 / 5 ) - ( 1 / 6 ) ) = ( 1 / ; 3 0 )

Proof

Step Hyp Ref Expression
1 5cn
 |-  5 e. CC
2 6cn
 |-  6 e. CC
3 5re
 |-  5 e. RR
4 5pos
 |-  0 < 5
5 3 4 gt0ne0ii
 |-  5 =/= 0
6 6re
 |-  6 e. RR
7 6pos
 |-  0 < 6
8 6 7 gt0ne0ii
 |-  6 =/= 0
9 1 2 5 8 subreci
 |-  ( ( 1 / 5 ) - ( 1 / 6 ) ) = ( ( 6 - 5 ) / ( 5 x. 6 ) )
10 ax-1cn
 |-  1 e. CC
11 5p1e6
 |-  ( 5 + 1 ) = 6
12 2 1 10 11 subaddrii
 |-  ( 6 - 5 ) = 1
13 6t5e30
 |-  ( 6 x. 5 ) = ; 3 0
14 2 1 13 mulcomli
 |-  ( 5 x. 6 ) = ; 3 0
15 12 14 oveq12i
 |-  ( ( 6 - 5 ) / ( 5 x. 6 ) ) = ( 1 / ; 3 0 )
16 9 15 eqtri
 |-  ( ( 1 / 5 ) - ( 1 / 6 ) ) = ( 1 / ; 3 0 )