Metamath Proof Explorer


Theorem negsub

Description: Relationship between subtraction and negative. Theorem I.3 of Apostol p. 18. (Contributed by NM, 21-Jan-1997) (Proof shortened by Mario Carneiro, 27-May-2016)

Ref Expression
Assertion negsub ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( 𝐴 + - 𝐵 ) = ( 𝐴𝐵 ) )

Proof

Step Hyp Ref Expression
1 df-neg - 𝐵 = ( 0 − 𝐵 )
2 1 oveq2i ( 𝐴 + - 𝐵 ) = ( 𝐴 + ( 0 − 𝐵 ) )
3 2 a1i ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( 𝐴 + - 𝐵 ) = ( 𝐴 + ( 0 − 𝐵 ) ) )
4 0cn 0 ∈ ℂ
5 addsubass ( ( 𝐴 ∈ ℂ ∧ 0 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ( 𝐴 + 0 ) − 𝐵 ) = ( 𝐴 + ( 0 − 𝐵 ) ) )
6 4 5 mp3an2 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ( 𝐴 + 0 ) − 𝐵 ) = ( 𝐴 + ( 0 − 𝐵 ) ) )
7 simpl ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → 𝐴 ∈ ℂ )
8 7 addid1d ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( 𝐴 + 0 ) = 𝐴 )
9 8 oveq1d ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ( 𝐴 + 0 ) − 𝐵 ) = ( 𝐴𝐵 ) )
10 3 6 9 3eqtr2d ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( 𝐴 + - 𝐵 ) = ( 𝐴𝐵 ) )