Metamath Proof Explorer


Theorem divdir

Description: Distribution of division over addition. (Contributed by NM, 31-Jul-2004) (Proof shortened by Mario Carneiro, 27-May-2016)

Ref Expression
Assertion divdir ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( ( 𝐴 + 𝐵 ) / 𝐶 ) = ( ( 𝐴 / 𝐶 ) + ( 𝐵 / 𝐶 ) ) )

Proof

Step Hyp Ref Expression
1 simp1 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → 𝐴 ∈ ℂ )
2 simp2 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → 𝐵 ∈ ℂ )
3 reccl ( ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) → ( 1 / 𝐶 ) ∈ ℂ )
4 3 3ad2ant3 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( 1 / 𝐶 ) ∈ ℂ )
5 1 2 4 adddird ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( ( 𝐴 + 𝐵 ) · ( 1 / 𝐶 ) ) = ( ( 𝐴 · ( 1 / 𝐶 ) ) + ( 𝐵 · ( 1 / 𝐶 ) ) ) )
6 1 2 addcld ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( 𝐴 + 𝐵 ) ∈ ℂ )
7 simp3l ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → 𝐶 ∈ ℂ )
8 simp3r ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → 𝐶 ≠ 0 )
9 divrec ( ( ( 𝐴 + 𝐵 ) ∈ ℂ ∧ 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) → ( ( 𝐴 + 𝐵 ) / 𝐶 ) = ( ( 𝐴 + 𝐵 ) · ( 1 / 𝐶 ) ) )
10 6 7 8 9 syl3anc ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( ( 𝐴 + 𝐵 ) / 𝐶 ) = ( ( 𝐴 + 𝐵 ) · ( 1 / 𝐶 ) ) )
11 divrec ( ( 𝐴 ∈ ℂ ∧ 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) → ( 𝐴 / 𝐶 ) = ( 𝐴 · ( 1 / 𝐶 ) ) )
12 1 7 8 11 syl3anc ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( 𝐴 / 𝐶 ) = ( 𝐴 · ( 1 / 𝐶 ) ) )
13 divrec ( ( 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) → ( 𝐵 / 𝐶 ) = ( 𝐵 · ( 1 / 𝐶 ) ) )
14 2 7 8 13 syl3anc ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( 𝐵 / 𝐶 ) = ( 𝐵 · ( 1 / 𝐶 ) ) )
15 12 14 oveq12d ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( ( 𝐴 / 𝐶 ) + ( 𝐵 / 𝐶 ) ) = ( ( 𝐴 · ( 1 / 𝐶 ) ) + ( 𝐵 · ( 1 / 𝐶 ) ) ) )
16 5 10 15 3eqtr4d ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( ( 𝐴 + 𝐵 ) / 𝐶 ) = ( ( 𝐴 / 𝐶 ) + ( 𝐵 / 𝐶 ) ) )