Metamath Proof Explorer


Theorem mul4r

Description: Rearrangement of 4 factors: swap the right factors in the factors of a product of two products. (Contributed by AV, 4-Mar-2023)

Ref Expression
Assertion mul4r ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) ∧ ( 𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ) ) → ( ( 𝐴 · 𝐵 ) · ( 𝐶 · 𝐷 ) ) = ( ( 𝐴 · 𝐷 ) · ( 𝐶 · 𝐵 ) ) )

Proof

Step Hyp Ref Expression
1 mulcom ( ( 𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ) → ( 𝐶 · 𝐷 ) = ( 𝐷 · 𝐶 ) )
2 1 adantl ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) ∧ ( 𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ) ) → ( 𝐶 · 𝐷 ) = ( 𝐷 · 𝐶 ) )
3 2 oveq2d ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) ∧ ( 𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ) ) → ( ( 𝐴 · 𝐵 ) · ( 𝐶 · 𝐷 ) ) = ( ( 𝐴 · 𝐵 ) · ( 𝐷 · 𝐶 ) ) )
4 mul4 ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) ∧ ( 𝐷 ∈ ℂ ∧ 𝐶 ∈ ℂ ) ) → ( ( 𝐴 · 𝐵 ) · ( 𝐷 · 𝐶 ) ) = ( ( 𝐴 · 𝐷 ) · ( 𝐵 · 𝐶 ) ) )
5 4 ancom2s ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) ∧ ( 𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ) ) → ( ( 𝐴 · 𝐵 ) · ( 𝐷 · 𝐶 ) ) = ( ( 𝐴 · 𝐷 ) · ( 𝐵 · 𝐶 ) ) )
6 simplr ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) ∧ ( 𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ) ) → 𝐵 ∈ ℂ )
7 simprl ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) ∧ ( 𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ) ) → 𝐶 ∈ ℂ )
8 6 7 mulcomd ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) ∧ ( 𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ) ) → ( 𝐵 · 𝐶 ) = ( 𝐶 · 𝐵 ) )
9 8 oveq2d ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) ∧ ( 𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ) ) → ( ( 𝐴 · 𝐷 ) · ( 𝐵 · 𝐶 ) ) = ( ( 𝐴 · 𝐷 ) · ( 𝐶 · 𝐵 ) ) )
10 3 5 9 3eqtrd ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) ∧ ( 𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ) ) → ( ( 𝐴 · 𝐵 ) · ( 𝐶 · 𝐷 ) ) = ( ( 𝐴 · 𝐷 ) · ( 𝐶 · 𝐵 ) ) )