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Metamath Proof Explorer

Theorem divdiv1

Description: Division into a fraction. (Contributed by NM, 31-Dec-2007)

Ref Expression
Assertion divdiv1 ( ( 𝐴 ∈ ℂ ∧ ( 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( ( 𝐴 / 𝐵 ) / 𝐶 ) = ( 𝐴 / ( 𝐵 · 𝐶 ) ) )

Proof

Step Hyp Ref Expression
1 ax-1cn ⊢ 1 ∈ ℂ
2 ax-1ne0 ⊢ 1 ≠ 0
3 1 2 pm3.2i ⊢ ( 1 ∈ ℂ ∧ 1 ≠ 0 )
4 divdivdiv ⊢ ( ( ( 𝐴 ∈ ℂ ∧ ( 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) ) ∧ ( ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ∧ ( 1 ∈ ℂ ∧ 1 ≠ 0 ) ) ) → ( ( 𝐴 / 𝐵 ) / ( 𝐶 / 1 ) ) = ( ( 𝐴 · 1 ) / ( 𝐵 · 𝐶 ) ) )
5 3 4 mpanr2 ⊢ ( ( ( 𝐴 ∈ ℂ ∧ ( 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) ) ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( ( 𝐴 / 𝐵 ) / ( 𝐶 / 1 ) ) = ( ( 𝐴 · 1 ) / ( 𝐵 · 𝐶 ) ) )
6 5 3impa ⊢ ( ( 𝐴 ∈ ℂ ∧ ( 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( ( 𝐴 / 𝐵 ) / ( 𝐶 / 1 ) ) = ( ( 𝐴 · 1 ) / ( 𝐵 · 𝐶 ) ) )
7 div1 ⊢ ( 𝐶 ∈ ℂ → ( 𝐶 / 1 ) = 𝐶 )
8 7 oveq2d ⊢ ( 𝐶 ∈ ℂ → ( ( 𝐴 / 𝐵 ) / ( 𝐶 / 1 ) ) = ( ( 𝐴 / 𝐵 ) / 𝐶 ) )
9 8 ad2antrl ⊢ ( ( ( 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( ( 𝐴 / 𝐵 ) / ( 𝐶 / 1 ) ) = ( ( 𝐴 / 𝐵 ) / 𝐶 ) )
10 9 3adant1 ⊢ ( ( 𝐴 ∈ ℂ ∧ ( 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( ( 𝐴 / 𝐵 ) / ( 𝐶 / 1 ) ) = ( ( 𝐴 / 𝐵 ) / 𝐶 ) )
11 mulrid ⊢ ( 𝐴 ∈ ℂ → ( 𝐴 · 1 ) = 𝐴 )
12 11 oveq1d ⊢ ( 𝐴 ∈ ℂ → ( ( 𝐴 · 1 ) / ( 𝐵 · 𝐶 ) ) = ( 𝐴 / ( 𝐵 · 𝐶 ) ) )
13 12 3ad2ant1 ⊢ ( ( 𝐴 ∈ ℂ ∧ ( 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( ( 𝐴 · 1 ) / ( 𝐵 · 𝐶 ) ) = ( 𝐴 / ( 𝐵 · 𝐶 ) ) )
14 6 10 13 3eqtr3d ⊢ ( ( 𝐴 ∈ ℂ ∧ ( 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( ( 𝐴 / 𝐵 ) / 𝐶 ) = ( 𝐴 / ( 𝐵 · 𝐶 ) ) )