Metamath Proof Explorer


Theorem divmul24d

Description: Swap the numerators in the product of two ratios. (Contributed by Mario Carneiro, 27-May-2016)

Ref Expression
Hypotheses div1d.1 ( 𝜑𝐴 ∈ ℂ )
divcld.2 ( 𝜑𝐵 ∈ ℂ )
divmuld.3 ( 𝜑𝐶 ∈ ℂ )
divmuldivd.4 ( 𝜑𝐷 ∈ ℂ )
divmuldivd.5 ( 𝜑𝐵 ≠ 0 )
divmuldivd.6 ( 𝜑𝐷 ≠ 0 )
Assertion divmul24d ( 𝜑 → ( ( 𝐴 / 𝐵 ) · ( 𝐶 / 𝐷 ) ) = ( ( 𝐴 / 𝐷 ) · ( 𝐶 / 𝐵 ) ) )

Proof

Step Hyp Ref Expression
1 div1d.1 ( 𝜑𝐴 ∈ ℂ )
2 divcld.2 ( 𝜑𝐵 ∈ ℂ )
3 divmuld.3 ( 𝜑𝐶 ∈ ℂ )
4 divmuldivd.4 ( 𝜑𝐷 ∈ ℂ )
5 divmuldivd.5 ( 𝜑𝐵 ≠ 0 )
6 divmuldivd.6 ( 𝜑𝐷 ≠ 0 )
7 2 5 jca ( 𝜑 → ( 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) )
8 4 6 jca ( 𝜑 → ( 𝐷 ∈ ℂ ∧ 𝐷 ≠ 0 ) )
9 divmul24 ( ( ( 𝐴 ∈ ℂ ∧ 𝐶 ∈ ℂ ) ∧ ( ( 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) ∧ ( 𝐷 ∈ ℂ ∧ 𝐷 ≠ 0 ) ) ) → ( ( 𝐴 / 𝐵 ) · ( 𝐶 / 𝐷 ) ) = ( ( 𝐴 / 𝐷 ) · ( 𝐶 / 𝐵 ) ) )
10 1 3 7 8 9 syl22anc ( 𝜑 → ( ( 𝐴 / 𝐵 ) · ( 𝐶 / 𝐷 ) ) = ( ( 𝐴 / 𝐷 ) · ( 𝐶 / 𝐵 ) ) )