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 ∈ ℂ → ( ( ( 𝑋 + 1 ) / 2 ) − 1 ) = ( ( 𝑋 − 1 ) / 2 ) )

Proof

Step Hyp Ref Expression
1 peano2cn ⊢ ( 𝑋 ∈ ℂ → ( 𝑋 + 1 ) ∈ ℂ )
2 1 halfcld ⊢ ( 𝑋 ∈ ℂ → ( ( 𝑋 + 1 ) / 2 ) ∈ ℂ )
3 peano2cnm ⊢ ( ( ( 𝑋 + 1 ) / 2 ) ∈ ℂ → ( ( ( 𝑋 + 1 ) / 2 ) − 1 ) ∈ ℂ )
4 2 3 syl ⊢ ( 𝑋 ∈ ℂ → ( ( ( 𝑋 + 1 ) / 2 ) − 1 ) ∈ ℂ )
5 peano2cnm ⊢ ( 𝑋 ∈ ℂ → ( 𝑋 − 1 ) ∈ ℂ )
6 5 halfcld ⊢ ( 𝑋 ∈ ℂ → ( ( 𝑋 − 1 ) / 2 ) ∈ ℂ )
7 2cnd ⊢ ( 𝑋 ∈ ℂ → 2 ∈ ℂ )
8 2ne0 ⊢ 2 ≠ 0
9 8 a1i ⊢ ( 𝑋 ∈ ℂ → 2 ≠ 0 )
10 1cnd ⊢ ( 𝑋 ∈ ℂ → 1 ∈ ℂ )
11 2 10 7 subdird ⊢ ( 𝑋 ∈ ℂ → ( ( ( ( 𝑋 + 1 ) / 2 ) − 1 ) · 2 ) = ( ( ( ( 𝑋 + 1 ) / 2 ) · 2 ) − ( 1 · 2 ) ) )
12 1 7 9 divcan1d ⊢ ( 𝑋 ∈ ℂ → ( ( ( 𝑋 + 1 ) / 2 ) · 2 ) = ( 𝑋 + 1 ) )
13 7 mulid2d ⊢ ( 𝑋 ∈ ℂ → ( 1 · 2 ) = 2 )
14 12 13 oveq12d ⊢ ( 𝑋 ∈ ℂ → ( ( ( ( 𝑋 + 1 ) / 2 ) · 2 ) − ( 1 · 2 ) ) = ( ( 𝑋 + 1 ) − 2 ) )
15 5 7 9 divcan1d ⊢ ( 𝑋 ∈ ℂ → ( ( ( 𝑋 − 1 ) / 2 ) · 2 ) = ( 𝑋 − 1 ) )
16 2m1e1 ⊢ ( 2 − 1 ) = 1
17 16 a1i ⊢ ( 𝑋 ∈ ℂ → ( 2 − 1 ) = 1 )
18 17 oveq2d ⊢ ( 𝑋 ∈ ℂ → ( 𝑋 − ( 2 − 1 ) ) = ( 𝑋 − 1 ) )
19 id ⊢ ( 𝑋 ∈ ℂ → 𝑋 ∈ ℂ )
20 19 7 10 subsub3d ⊢ ( 𝑋 ∈ ℂ → ( 𝑋 − ( 2 − 1 ) ) = ( ( 𝑋 + 1 ) − 2 ) )
21 15 18 20 3eqtr2rd ⊢ ( 𝑋 ∈ ℂ → ( ( 𝑋 + 1 ) − 2 ) = ( ( ( 𝑋 − 1 ) / 2 ) · 2 ) )
22 11 14 21 3eqtrd ⊢ ( 𝑋 ∈ ℂ → ( ( ( ( 𝑋 + 1 ) / 2 ) − 1 ) · 2 ) = ( ( ( 𝑋 − 1 ) / 2 ) · 2 ) )
23 4 6 7 9 22 mulcan2ad ⊢ ( 𝑋 ∈ ℂ → ( ( ( 𝑋 + 1 ) / 2 ) − 1 ) = ( ( 𝑋 − 1 ) / 2 ) )