Metamath Proof Explorer


Theorem div2negd

Description: Quotient of two negatives. (Contributed by Mario Carneiro, 27-May-2016)

Ref Expression
Hypotheses div1d.1 ( 𝜑𝐴 ∈ ℂ )
divcld.2 ( 𝜑𝐵 ∈ ℂ )
divcld.3 ( 𝜑𝐵 ≠ 0 )
Assertion div2negd ( 𝜑 → ( - 𝐴 / - 𝐵 ) = ( 𝐴 / 𝐵 ) )

Proof

Step Hyp Ref Expression
1 div1d.1 ( 𝜑𝐴 ∈ ℂ )
2 divcld.2 ( 𝜑𝐵 ∈ ℂ )
3 divcld.3 ( 𝜑𝐵 ≠ 0 )
4 div2neg ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( - 𝐴 / - 𝐵 ) = ( 𝐴 / 𝐵 ) )
5 1 2 3 4 syl3anc ( 𝜑 → ( - 𝐴 / - 𝐵 ) = ( 𝐴 / 𝐵 ) )