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Metamath Proof Explorer

Theorem imsub

Description: Imaginary part distributes over subtraction. (Contributed by NM, 18-Mar-2005)

Ref Expression
Assertion imsub ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ℑ ‘ ( 𝐴 − 𝐵 ) ) = ( ( ℑ ‘ 𝐴 ) − ( ℑ ‘ 𝐵 ) ) )

Proof

Step Hyp Ref Expression
1 negcl ⊢ ( 𝐵 ∈ ℂ → - 𝐵 ∈ ℂ )
2 imadd ⊢ ( ( 𝐴 ∈ ℂ ∧ - 𝐵 ∈ ℂ ) → ( ℑ ‘ ( 𝐴 + - 𝐵 ) ) = ( ( ℑ ‘ 𝐴 ) + ( ℑ ‘ - 𝐵 ) ) )
3 1 2 sylan2 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ℑ ‘ ( 𝐴 + - 𝐵 ) ) = ( ( ℑ ‘ 𝐴 ) + ( ℑ ‘ - 𝐵 ) ) )
4 imneg ⊢ ( 𝐵 ∈ ℂ → ( ℑ ‘ - 𝐵 ) = - ( ℑ ‘ 𝐵 ) )
5 4 adantl ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ℑ ‘ - 𝐵 ) = - ( ℑ ‘ 𝐵 ) )
6 5 oveq2d ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ( ℑ ‘ 𝐴 ) + ( ℑ ‘ - 𝐵 ) ) = ( ( ℑ ‘ 𝐴 ) + - ( ℑ ‘ 𝐵 ) ) )
7 3 6 eqtrd ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ℑ ‘ ( 𝐴 + - 𝐵 ) ) = ( ( ℑ ‘ 𝐴 ) + - ( ℑ ‘ 𝐵 ) ) )
8 negsub ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( 𝐴 + - 𝐵 ) = ( 𝐴 − 𝐵 ) )
9 8 fveq2d ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ℑ ‘ ( 𝐴 + - 𝐵 ) ) = ( ℑ ‘ ( 𝐴 − 𝐵 ) ) )
10 imcl ⊢ ( 𝐴 ∈ ℂ → ( ℑ ‘ 𝐴 ) ∈ ℝ )
11 10 recnd ⊢ ( 𝐴 ∈ ℂ → ( ℑ ‘ 𝐴 ) ∈ ℂ )
12 imcl ⊢ ( 𝐵 ∈ ℂ → ( ℑ ‘ 𝐵 ) ∈ ℝ )
13 12 recnd ⊢ ( 𝐵 ∈ ℂ → ( ℑ ‘ 𝐵 ) ∈ ℂ )
14 negsub ⊢ ( ( ( ℑ ‘ 𝐴 ) ∈ ℂ ∧ ( ℑ ‘ 𝐵 ) ∈ ℂ ) → ( ( ℑ ‘ 𝐴 ) + - ( ℑ ‘ 𝐵 ) ) = ( ( ℑ ‘ 𝐴 ) − ( ℑ ‘ 𝐵 ) ) )
15 11 13 14 syl2an ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ( ℑ ‘ 𝐴 ) + - ( ℑ ‘ 𝐵 ) ) = ( ( ℑ ‘ 𝐴 ) − ( ℑ ‘ 𝐵 ) ) )
16 7 9 15 3eqtr3d ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ℑ ‘ ( 𝐴 − 𝐵 ) ) = ( ( ℑ ‘ 𝐴 ) − ( ℑ ‘ 𝐵 ) ) )