Metamath Proof Explorer


Theorem divdiv23zi

Description: Swap denominators in a division. (Contributed by NM, 15-Sep-1999)

Ref Expression
Hypotheses divclz.1 𝐴 ∈ ℂ
divclz.2 𝐵 ∈ ℂ
divmulz.3 𝐶 ∈ ℂ
Assertion divdiv23zi ( ( 𝐵 ≠ 0 ∧ 𝐶 ≠ 0 ) → ( ( 𝐴 / 𝐵 ) / 𝐶 ) = ( ( 𝐴 / 𝐶 ) / 𝐵 ) )

Proof

Step Hyp Ref Expression
1 divclz.1 𝐴 ∈ ℂ
2 divclz.2 𝐵 ∈ ℂ
3 divmulz.3 𝐶 ∈ ℂ
4 divdiv32 ( ( 𝐴 ∈ ℂ ∧ ( 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( ( 𝐴 / 𝐵 ) / 𝐶 ) = ( ( 𝐴 / 𝐶 ) / 𝐵 ) )
5 1 4 mp3an1 ( ( ( 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( ( 𝐴 / 𝐵 ) / 𝐶 ) = ( ( 𝐴 / 𝐶 ) / 𝐵 ) )
6 3 5 mpanr1 ( ( ( 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) ∧ 𝐶 ≠ 0 ) → ( ( 𝐴 / 𝐵 ) / 𝐶 ) = ( ( 𝐴 / 𝐶 ) / 𝐵 ) )
7 2 6 mpanl1 ( ( 𝐵 ≠ 0 ∧ 𝐶 ≠ 0 ) → ( ( 𝐴 / 𝐵 ) / 𝐶 ) = ( ( 𝐴 / 𝐶 ) / 𝐵 ) )