Metamath Proof Explorer


Theorem subdir

Description: Distribution of multiplication over subtraction. Theorem I.5 of Apostol p. 18. (Contributed by NM, 30-Dec-2005)

Ref Expression
Assertion subdir ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴𝐵 ) · 𝐶 ) = ( ( 𝐴 · 𝐶 ) − ( 𝐵 · 𝐶 ) ) )

Proof

Step Hyp Ref Expression
1 subdi ( ( 𝐶 ∈ ℂ ∧ 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( 𝐶 · ( 𝐴𝐵 ) ) = ( ( 𝐶 · 𝐴 ) − ( 𝐶 · 𝐵 ) ) )
2 1 3coml ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐶 · ( 𝐴𝐵 ) ) = ( ( 𝐶 · 𝐴 ) − ( 𝐶 · 𝐵 ) ) )
3 subcl ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( 𝐴𝐵 ) ∈ ℂ )
4 mulcom ( ( ( 𝐴𝐵 ) ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴𝐵 ) · 𝐶 ) = ( 𝐶 · ( 𝐴𝐵 ) ) )
5 3 4 stoic3 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴𝐵 ) · 𝐶 ) = ( 𝐶 · ( 𝐴𝐵 ) ) )
6 mulcom ( ( 𝐴 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 · 𝐶 ) = ( 𝐶 · 𝐴 ) )
7 6 3adant2 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 · 𝐶 ) = ( 𝐶 · 𝐴 ) )
8 mulcom ( ( 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐵 · 𝐶 ) = ( 𝐶 · 𝐵 ) )
9 8 3adant1 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐵 · 𝐶 ) = ( 𝐶 · 𝐵 ) )
10 7 9 oveq12d ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 · 𝐶 ) − ( 𝐵 · 𝐶 ) ) = ( ( 𝐶 · 𝐴 ) − ( 𝐶 · 𝐵 ) ) )
11 2 5 10 3eqtr4d ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴𝐵 ) · 𝐶 ) = ( ( 𝐴 · 𝐶 ) − ( 𝐵 · 𝐶 ) ) )