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 B A + B = A B

Proof

Step Hyp Ref Expression
1 df-neg B = 0 B
2 1 oveq2i A + B = A + 0 - B
3 2 a1i A B A + B = A + 0 - B
4 0cn 0
5 addsubass A 0 B A + 0 - B = A + 0 - B
6 4 5 mp3an2 A B A + 0 - B = A + 0 - B
7 simpl A B A
8 7 addid1d A B A + 0 = A
9 8 oveq1d A B A + 0 - B = A B
10 3 6 9 3eqtr2d A B A + B = A B