Metamath Proof Explorer


Theorem div2subd

Description: Swap subtrahend and minuend inside the numerator and denominator of a fraction. Deduction form of div2sub . (Contributed by David Moews, 28-Feb-2017)

Ref Expression
Hypotheses div2subd.1 ( 𝜑𝐴 ∈ ℂ )
div2subd.2 ( 𝜑𝐵 ∈ ℂ )
div2subd.3 ( 𝜑𝐶 ∈ ℂ )
div2subd.4 ( 𝜑𝐷 ∈ ℂ )
div2subd.5 ( 𝜑𝐶𝐷 )
Assertion div2subd ( 𝜑 → ( ( 𝐴𝐵 ) / ( 𝐶𝐷 ) ) = ( ( 𝐵𝐴 ) / ( 𝐷𝐶 ) ) )

Proof

Step Hyp Ref Expression
1 div2subd.1 ( 𝜑𝐴 ∈ ℂ )
2 div2subd.2 ( 𝜑𝐵 ∈ ℂ )
3 div2subd.3 ( 𝜑𝐶 ∈ ℂ )
4 div2subd.4 ( 𝜑𝐷 ∈ ℂ )
5 div2subd.5 ( 𝜑𝐶𝐷 )
6 div2sub ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) ∧ ( 𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ∧ 𝐶𝐷 ) ) → ( ( 𝐴𝐵 ) / ( 𝐶𝐷 ) ) = ( ( 𝐵𝐴 ) / ( 𝐷𝐶 ) ) )
7 1 2 3 4 5 6 syl23anc ( 𝜑 → ( ( 𝐴𝐵 ) / ( 𝐶𝐷 ) ) = ( ( 𝐵𝐴 ) / ( 𝐷𝐶 ) ) )