Metamath Proof Explorer


Theorem halfpm6th

Description: One half plus or minus one sixth. (Contributed by Paul Chapman, 17-Jan-2008) (Proof shortened by SN, 22-Oct-2025)

Ref Expression
Assertion halfpm6th ( ( ( 1 / 2 ) − ( 1 / 6 ) ) = ( 1 / 3 ) ∧ ( ( 1 / 2 ) + ( 1 / 6 ) ) = ( 2 / 3 ) )

Proof

Step Hyp Ref Expression
1 3cn 3 ∈ ℂ
2 3ne0 3 ≠ 0
3 1 2 reccli ( 1 / 3 ) ∈ ℂ
4 6cn 6 ∈ ℂ
5 6re 6 ∈ ℝ
6 6pos 0 < 6
7 5 6 gt0ne0ii 6 ≠ 0
8 4 7 reccli ( 1 / 6 ) ∈ ℂ
9 halfcn ( 1 / 2 ) ∈ ℂ
10 3 9 pncan3i ( ( 1 / 3 ) + ( ( 1 / 2 ) − ( 1 / 3 ) ) ) = ( 1 / 2 )
11 halfthird ( ( 1 / 2 ) − ( 1 / 3 ) ) = ( 1 / 6 )
12 11 oveq2i ( ( 1 / 3 ) + ( ( 1 / 2 ) − ( 1 / 3 ) ) ) = ( ( 1 / 3 ) + ( 1 / 6 ) )
13 10 12 eqtr3i ( 1 / 2 ) = ( ( 1 / 3 ) + ( 1 / 6 ) )
14 3 8 13 mvrraddi ( ( 1 / 2 ) − ( 1 / 6 ) ) = ( 1 / 3 )
15 11 oveq2i ( ( 1 / 2 ) + ( ( 1 / 2 ) − ( 1 / 3 ) ) ) = ( ( 1 / 2 ) + ( 1 / 6 ) )
16 9 9 3 addsubassi ( ( ( 1 / 2 ) + ( 1 / 2 ) ) − ( 1 / 3 ) ) = ( ( 1 / 2 ) + ( ( 1 / 2 ) − ( 1 / 3 ) ) )
17 2cn 2 ∈ ℂ
18 17 1 2 divcli ( 2 / 3 ) ∈ ℂ
19 ax-1cn 1 ∈ ℂ
20 2halves ( 1 ∈ ℂ → ( ( 1 / 2 ) + ( 1 / 2 ) ) = 1 )
21 19 20 ax-mp ( ( 1 / 2 ) + ( 1 / 2 ) ) = 1
22 2p1e3 ( 2 + 1 ) = 3
23 22 oveq1i ( ( 2 + 1 ) / 3 ) = ( 3 / 3 )
24 1 2 dividi ( 3 / 3 ) = 1
25 23 24 eqtri ( ( 2 + 1 ) / 3 ) = 1
26 17 19 1 2 divdiri ( ( 2 + 1 ) / 3 ) = ( ( 2 / 3 ) + ( 1 / 3 ) )
27 21 25 26 3eqtr2i ( ( 1 / 2 ) + ( 1 / 2 ) ) = ( ( 2 / 3 ) + ( 1 / 3 ) )
28 18 3 27 mvrraddi ( ( ( 1 / 2 ) + ( 1 / 2 ) ) − ( 1 / 3 ) ) = ( 2 / 3 )
29 16 28 eqtr3i ( ( 1 / 2 ) + ( ( 1 / 2 ) − ( 1 / 3 ) ) ) = ( 2 / 3 )
30 15 29 eqtr3i ( ( 1 / 2 ) + ( 1 / 6 ) ) = ( 2 / 3 )
31 14 30 pm3.2i ( ( ( 1 / 2 ) − ( 1 / 6 ) ) = ( 1 / 3 ) ∧ ( ( 1 / 2 ) + ( 1 / 6 ) ) = ( 2 / 3 ) )