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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 ‘ ( 𝐴 − 𝐵 ) ) = ( ( 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 ) = ( 𝐴 − 𝐵 ) )
9 8 fveq2d ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ℜ ‘ ( 𝐴 + - 𝐵 ) ) = ( ℜ ‘ ( 𝐴 − 𝐵 ) ) )
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 ‘ ( 𝐴 − 𝐵 ) ) = ( ( ℜ ‘ 𝐴 ) − ( ℜ ‘ 𝐵 ) ) )