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Metamath Proof Explorer

Theorem divmul24

Description: Swap the numerators in the product of two ratios. (Contributed by NM, 3-May-2005)

Ref Expression
Assertion divmul24 ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) ∧ ( ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ∧ ( 𝐷 ∈ ℂ ∧ 𝐷 ≠ 0 ) ) ) → ( ( 𝐴 / 𝐶 ) · ( 𝐵 / 𝐷 ) ) = ( ( 𝐴 / 𝐷 ) · ( 𝐵 / 𝐶 ) ) )

Proof

Step Hyp Ref Expression
1 mulcom ⊢ ( ( 𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ) → ( 𝐶 · 𝐷 ) = ( 𝐷 · 𝐶 ) )
2 1 ad2ant2r ⊢ ( ( ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ∧ ( 𝐷 ∈ ℂ ∧ 𝐷 ≠ 0 ) ) → ( 𝐶 · 𝐷 ) = ( 𝐷 · 𝐶 ) )
3 2 adantl ⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) ∧ ( ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ∧ ( 𝐷 ∈ ℂ ∧ 𝐷 ≠ 0 ) ) ) → ( 𝐶 · 𝐷 ) = ( 𝐷 · 𝐶 ) )
4 3 oveq2d ⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) ∧ ( ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ∧ ( 𝐷 ∈ ℂ ∧ 𝐷 ≠ 0 ) ) ) → ( ( 𝐴 · 𝐵 ) / ( 𝐶 · 𝐷 ) ) = ( ( 𝐴 · 𝐵 ) / ( 𝐷 · 𝐶 ) ) )
5 divmuldiv ⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) ∧ ( ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ∧ ( 𝐷 ∈ ℂ ∧ 𝐷 ≠ 0 ) ) ) → ( ( 𝐴 / 𝐶 ) · ( 𝐵 / 𝐷 ) ) = ( ( 𝐴 · 𝐵 ) / ( 𝐶 · 𝐷 ) ) )
6 divmuldiv ⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) ∧ ( ( 𝐷 ∈ ℂ ∧ 𝐷 ≠ 0 ) ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) ) → ( ( 𝐴 / 𝐷 ) · ( 𝐵 / 𝐶 ) ) = ( ( 𝐴 · 𝐵 ) / ( 𝐷 · 𝐶 ) ) )
7 6 ancom2s ⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) ∧ ( ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ∧ ( 𝐷 ∈ ℂ ∧ 𝐷 ≠ 0 ) ) ) → ( ( 𝐴 / 𝐷 ) · ( 𝐵 / 𝐶 ) ) = ( ( 𝐴 · 𝐵 ) / ( 𝐷 · 𝐶 ) ) )
8 4 5 7 3eqtr4d ⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) ∧ ( ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ∧ ( 𝐷 ∈ ℂ ∧ 𝐷 ≠ 0 ) ) ) → ( ( 𝐴 / 𝐶 ) · ( 𝐵 / 𝐷 ) ) = ( ( 𝐴 / 𝐷 ) · ( 𝐵 / 𝐶 ) ) )