Metamath Proof Explorer


Theorem resub

Description: Real part distributes over subtraction. (Contributed by NM, 17-Mar-2005)

Ref Expression
Assertion resub ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ℜ ‘ ( 𝐴𝐵 ) ) = ( ( ℜ ‘ 𝐴 ) − ( ℜ ‘ 𝐵 ) ) )

Proof

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