Metamath Proof Explorer


Theorem divsubdir

Description: Distribution of division over subtraction. (Contributed by NM, 4-Mar-2005)

Ref Expression
Assertion divsubdir ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( ( 𝐴𝐵 ) / 𝐶 ) = ( ( 𝐴 / 𝐶 ) − ( 𝐵 / 𝐶 ) ) )

Proof

Step Hyp Ref Expression
1 negcl ( 𝐵 ∈ ℂ → - 𝐵 ∈ ℂ )
2 divdir ( ( 𝐴 ∈ ℂ ∧ - 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( ( 𝐴 + - 𝐵 ) / 𝐶 ) = ( ( 𝐴 / 𝐶 ) + ( - 𝐵 / 𝐶 ) ) )
3 1 2 syl3an2 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( ( 𝐴 + - 𝐵 ) / 𝐶 ) = ( ( 𝐴 / 𝐶 ) + ( - 𝐵 / 𝐶 ) ) )
4 negsub ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( 𝐴 + - 𝐵 ) = ( 𝐴𝐵 ) )
5 4 oveq1d ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ( 𝐴 + - 𝐵 ) / 𝐶 ) = ( ( 𝐴𝐵 ) / 𝐶 ) )
6 5 3adant3 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( ( 𝐴 + - 𝐵 ) / 𝐶 ) = ( ( 𝐴𝐵 ) / 𝐶 ) )
7 3 6 eqtr3d ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( ( 𝐴 / 𝐶 ) + ( - 𝐵 / 𝐶 ) ) = ( ( 𝐴𝐵 ) / 𝐶 ) )
8 divneg ( ( 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) → - ( 𝐵 / 𝐶 ) = ( - 𝐵 / 𝐶 ) )
9 8 3expb ( ( 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → - ( 𝐵 / 𝐶 ) = ( - 𝐵 / 𝐶 ) )
10 9 3adant1 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → - ( 𝐵 / 𝐶 ) = ( - 𝐵 / 𝐶 ) )
11 10 oveq2d ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( ( 𝐴 / 𝐶 ) + - ( 𝐵 / 𝐶 ) ) = ( ( 𝐴 / 𝐶 ) + ( - 𝐵 / 𝐶 ) ) )
12 divcl ( ( 𝐴 ∈ ℂ ∧ 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) → ( 𝐴 / 𝐶 ) ∈ ℂ )
13 12 3expb ( ( 𝐴 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( 𝐴 / 𝐶 ) ∈ ℂ )
14 13 3adant2 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( 𝐴 / 𝐶 ) ∈ ℂ )
15 divcl ( ( 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) → ( 𝐵 / 𝐶 ) ∈ ℂ )
16 15 3expb ( ( 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( 𝐵 / 𝐶 ) ∈ ℂ )
17 16 3adant1 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( 𝐵 / 𝐶 ) ∈ ℂ )
18 14 17 negsubd ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( ( 𝐴 / 𝐶 ) + - ( 𝐵 / 𝐶 ) ) = ( ( 𝐴 / 𝐶 ) − ( 𝐵 / 𝐶 ) ) )
19 11 18 eqtr3d ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( ( 𝐴 / 𝐶 ) + ( - 𝐵 / 𝐶 ) ) = ( ( 𝐴 / 𝐶 ) − ( 𝐵 / 𝐶 ) ) )
20 7 19 eqtr3d ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( ( 𝐴𝐵 ) / 𝐶 ) = ( ( 𝐴 / 𝐶 ) − ( 𝐵 / 𝐶 ) ) )