Metamath Proof Explorer

Theorem reneg

Description: Real part of negative. (Contributed by NM, 17-Mar-2005) (Revised by Mario Carneiro, 14-Jul-2014)

Ref Expression
Assertion reneg ( 𝐴 ∈ ℂ → ( ℜ ‘ - 𝐴 ) = - ( ℜ ‘ 𝐴 ) )

Proof

Step Hyp Ref Expression
1 recl ⊢ ( 𝐴 ∈ ℂ → ( ℜ ‘ 𝐴 ) ∈ ℝ )
2 1 recnd ⊢ ( 𝐴 ∈ ℂ → ( ℜ ‘ 𝐴 ) ∈ ℂ )
3 ax-icn ⊢ i ∈ ℂ
4 imcl ⊢ ( 𝐴 ∈ ℂ → ( ℑ ‘ 𝐴 ) ∈ ℝ )
5 4 recnd ⊢ ( 𝐴 ∈ ℂ → ( ℑ ‘ 𝐴 ) ∈ ℂ )
6 mulcl ⊢ ( ( i ∈ ℂ ∧ ( ℑ ‘ 𝐴 ) ∈ ℂ ) → ( i · ( ℑ ‘ 𝐴 ) ) ∈ ℂ )
7 3 5 6 sylancr ⊢ ( 𝐴 ∈ ℂ → ( i · ( ℑ ‘ 𝐴 ) ) ∈ ℂ )
8 2 7 negdid ⊢ ( 𝐴 ∈ ℂ → - ( ( ℜ ‘ 𝐴 ) + ( i · ( ℑ ‘ 𝐴 ) ) ) = ( - ( ℜ ‘ 𝐴 ) + - ( i · ( ℑ ‘ 𝐴 ) ) ) )
9 replim ⊢ ( 𝐴 ∈ ℂ → 𝐴 = ( ( ℜ ‘ 𝐴 ) + ( i · ( ℑ ‘ 𝐴 ) ) ) )
10 9 negeqd ⊢ ( 𝐴 ∈ ℂ → - 𝐴 = - ( ( ℜ ‘ 𝐴 ) + ( i · ( ℑ ‘ 𝐴 ) ) ) )
11 mulneg2 ⊢ ( ( i ∈ ℂ ∧ ( ℑ ‘ 𝐴 ) ∈ ℂ ) → ( i · - ( ℑ ‘ 𝐴 ) ) = - ( i · ( ℑ ‘ 𝐴 ) ) )
12 3 5 11 sylancr ⊢ ( 𝐴 ∈ ℂ → ( i · - ( ℑ ‘ 𝐴 ) ) = - ( i · ( ℑ ‘ 𝐴 ) ) )
13 12 oveq2d ⊢ ( 𝐴 ∈ ℂ → ( - ( ℜ ‘ 𝐴 ) + ( i · - ( ℑ ‘ 𝐴 ) ) ) = ( - ( ℜ ‘ 𝐴 ) + - ( i · ( ℑ ‘ 𝐴 ) ) ) )
14 8 10 13 3eqtr4d ⊢ ( 𝐴 ∈ ℂ → - 𝐴 = ( - ( ℜ ‘ 𝐴 ) + ( i · - ( ℑ ‘ 𝐴 ) ) ) )
15 14 fveq2d ⊢ ( 𝐴 ∈ ℂ → ( ℜ ‘ - 𝐴 ) = ( ℜ ‘ ( - ( ℜ ‘ 𝐴 ) + ( i · - ( ℑ ‘ 𝐴 ) ) ) ) )
16 1 renegcld ⊢ ( 𝐴 ∈ ℂ → - ( ℜ ‘ 𝐴 ) ∈ ℝ )
17 4 renegcld ⊢ ( 𝐴 ∈ ℂ → - ( ℑ ‘ 𝐴 ) ∈ ℝ )
18 crre ⊢ ( ( - ( ℜ ‘ 𝐴 ) ∈ ℝ ∧ - ( ℑ ‘ 𝐴 ) ∈ ℝ ) → ( ℜ ‘ ( - ( ℜ ‘ 𝐴 ) + ( i · - ( ℑ ‘ 𝐴 ) ) ) ) = - ( ℜ ‘ 𝐴 ) )
19 16 17 18 syl2anc ⊢ ( 𝐴 ∈ ℂ → ( ℜ ‘ ( - ( ℜ ‘ 𝐴 ) + ( i · - ( ℑ ‘ 𝐴 ) ) ) ) = - ( ℜ ‘ 𝐴 ) )
20 15 19 eqtrd ⊢ ( 𝐴 ∈ ℂ → ( ℜ ‘ - 𝐴 ) = - ( ℜ ‘ 𝐴 ) )