Metamath Proof Explorer

Theorem xp1d2m1eqxm1d2

Description: A complex number increased by 1, then divided by 2, then decreased by 1 equals the complex number decreased by 1 and then divided by 2. (Contributed by AV, 24-May-2020)

Ref Expression
Assertion xp1d2m1eqxm1d2 
|- ( X e. CC -> ( ( ( X + 1 ) / 2 ) - 1 ) = ( ( X - 1 ) / 2 ) )

Proof

Step Hyp Ref Expression
1 peano2cn 
 |-  ( X e. CC -> ( X + 1 ) e. CC )
2 1 halfcld 
 |-  ( X e. CC -> ( ( X + 1 ) / 2 ) e. CC )
3 peano2cnm 
 |-  ( ( ( X + 1 ) / 2 ) e. CC -> ( ( ( X + 1 ) / 2 ) - 1 ) e. CC )
4 2 3 syl 
 |-  ( X e. CC -> ( ( ( X + 1 ) / 2 ) - 1 ) e. CC )
5 peano2cnm 
 |-  ( X e. CC -> ( X - 1 ) e. CC )
6 5 halfcld 
 |-  ( X e. CC -> ( ( X - 1 ) / 2 ) e. CC )
7 2cnd 
 |-  ( X e. CC -> 2 e. CC )
8 2ne0 
 |-  2 =/= 0
9 8 a1i 
 |-  ( X e. CC -> 2 =/= 0 )
10 1cnd 
 |-  ( X e. CC -> 1 e. CC )
11 2 10 7 subdird 
 |-  ( X e. CC -> ( ( ( ( X + 1 ) / 2 ) - 1 ) x. 2 ) = ( ( ( ( X + 1 ) / 2 ) x. 2 ) - ( 1 x. 2 ) ) )
12 1 7 9 divcan1d 
 |-  ( X e. CC -> ( ( ( X + 1 ) / 2 ) x. 2 ) = ( X + 1 ) )
13 7 mulid2d 
 |-  ( X e. CC -> ( 1 x. 2 ) = 2 )
14 12 13 oveq12d 
 |-  ( X e. CC -> ( ( ( ( X + 1 ) / 2 ) x. 2 ) - ( 1 x. 2 ) ) = ( ( X + 1 ) - 2 ) )
15 5 7 9 divcan1d 
 |-  ( X e. CC -> ( ( ( X - 1 ) / 2 ) x. 2 ) = ( X - 1 ) )
16 2m1e1 
 |-  ( 2 - 1 ) = 1
17 16 a1i 
 |-  ( X e. CC -> ( 2 - 1 ) = 1 )
18 17 oveq2d 
 |-  ( X e. CC -> ( X - ( 2 - 1 ) ) = ( X - 1 ) )
19 id 
 |-  ( X e. CC -> X e. CC )
20 19 7 10 subsub3d 
 |-  ( X e. CC -> ( X - ( 2 - 1 ) ) = ( ( X + 1 ) - 2 ) )
21 15 18 20 3eqtr2rd 
 |-  ( X e. CC -> ( ( X + 1 ) - 2 ) = ( ( ( X - 1 ) / 2 ) x. 2 ) )
22 11 14 21 3eqtrd 
 |-  ( X e. CC -> ( ( ( ( X + 1 ) / 2 ) - 1 ) x. 2 ) = ( ( ( X - 1 ) / 2 ) x. 2 ) )
23 4 6 7 9 22 mulcan2ad 
 |-  ( X e. CC -> ( ( ( X + 1 ) / 2 ) - 1 ) = ( ( X - 1 ) / 2 ) )