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Metamath Proof Explorer

Theorem divdiv1d

Description: Division into a fraction. (Contributed by Mario Carneiro, 27-May-2016)

Ref Expression
Hypotheses div1d.1 ⊢ ( 𝜑 → 𝐴 ∈ ℂ )
divcld.2 ⊢ ( 𝜑 → 𝐵 ∈ ℂ )
divmuld.3 ⊢ ( 𝜑 → 𝐶 ∈ ℂ )
divmuld.4 ⊢ ( 𝜑 → 𝐵 ≠ 0 )
divdiv23d.5 ⊢ ( 𝜑 → 𝐶 ≠ 0 )
Assertion divdiv1d ( 𝜑 → ( ( 𝐴 / 𝐵 ) / 𝐶 ) = ( 𝐴 / ( 𝐵 · 𝐶 ) ) )

Proof

Step Hyp Ref Expression
1 div1d.1 ⊢ ( 𝜑 → 𝐴 ∈ ℂ )
2 divcld.2 ⊢ ( 𝜑 → 𝐵 ∈ ℂ )
3 divmuld.3 ⊢ ( 𝜑 → 𝐶 ∈ ℂ )
4 divmuld.4 ⊢ ( 𝜑 → 𝐵 ≠ 0 )
5 divdiv23d.5 ⊢ ( 𝜑 → 𝐶 ≠ 0 )
6 divdiv1 ⊢ ( ( 𝐴 ∈ ℂ ∧ ( 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( ( 𝐴 / 𝐵 ) / 𝐶 ) = ( 𝐴 / ( 𝐵 · 𝐶 ) ) )
7 1 2 4 3 5 6 syl122anc ⊢ ( 𝜑 → ( ( 𝐴 / 𝐵 ) / 𝐶 ) = ( 𝐴 / ( 𝐵 · 𝐶 ) ) )