Metamath Proof Explorer

Theorem divneg

Description: Move negative sign inside of a division. (Contributed by NM, 17-Sep-2004)

Ref Expression
Assertion divneg ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → - ( 𝐴 / 𝐵 ) = ( - 𝐴 / 𝐵 ) )

Proof

Step Hyp Ref Expression
1 reccl ⊢ ( ( 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( 1 / 𝐵 ) ∈ ℂ )
2 mulneg1 ⊢ ( ( 𝐴 ∈ ℂ ∧ ( 1 / 𝐵 ) ∈ ℂ ) → ( - 𝐴 · ( 1 / 𝐵 ) ) = - ( 𝐴 · ( 1 / 𝐵 ) ) )
3 1 2 sylan2 ⊢ ( ( 𝐴 ∈ ℂ ∧ ( 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) ) → ( - 𝐴 · ( 1 / 𝐵 ) ) = - ( 𝐴 · ( 1 / 𝐵 ) ) )
4 3 3impb ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( - 𝐴 · ( 1 / 𝐵 ) ) = - ( 𝐴 · ( 1 / 𝐵 ) ) )
5 negcl ⊢ ( 𝐴 ∈ ℂ → - 𝐴 ∈ ℂ )
6 divrec ⊢ ( ( - 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( - 𝐴 / 𝐵 ) = ( - 𝐴 · ( 1 / 𝐵 ) ) )
7 5 6 syl3an1 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( - 𝐴 / 𝐵 ) = ( - 𝐴 · ( 1 / 𝐵 ) ) )
8 divrec ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( 𝐴 / 𝐵 ) = ( 𝐴 · ( 1 / 𝐵 ) ) )
9 8 negeqd ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → - ( 𝐴 / 𝐵 ) = - ( 𝐴 · ( 1 / 𝐵 ) ) )
10 4 7 9 3eqtr4rd ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → - ( 𝐴 / 𝐵 ) = ( - 𝐴 / 𝐵 ) )