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 ABA+B=AB

Proof

Step Hyp Ref Expression
1 df-neg B=0B
2 1 oveq2i A+B=A+0-B
3 2 a1i ABA+B=A+0-B
4 0cn 0
5 addsubass A0BA+0-B=A+0-B
6 4 5 mp3an2 ABA+0-B=A+0-B
7 simpl ABA
8 7 addridd ABA+0=A
9 8 oveq1d ABA+0-B=AB
10 3 6 9 3eqtr2d ABA+B=AB