Metamath Proof Explorer


Theorem zmulcom

Description: Multiplication is commutative for integers. Proven without ax-mulcom . From this result and grpcominv1 , we can show that rationals commute under multiplication without using ax-mulcom . (Contributed by SN, 25-Jan-2025)

Ref Expression
Assertion zmulcom A B A B = B A

Proof

Step Hyp Ref Expression
1 reelznn0nn A A 0 A 0 - A
2 reelznn0nn B B 0 B 0 - B
3 nn0mulcom A 0 B 0 A B = B A
4 zmulcomlem A 0 - A B 0 A B = B A
5 zmulcomlem B 0 - B A 0 B A = A B
6 5 eqcomd B 0 - B A 0 A B = B A
7 6 ancoms A 0 B 0 - B A B = B A
8 nnmulcom 0 - A 0 - B 0 - A 0 - B = 0 - B 0 - A
9 8 ad2ant2l A 0 - A B 0 - B 0 - A 0 - B = 0 - B 0 - A
10 9 oveq2d A 0 - A B 0 - B 0 - 0 - A 0 - B = 0 - 0 - B 0 - A
11 rernegcl A 0 - A
12 11 ad2antrr A 0 - A B 0 - B 0 - A
13 simprr A 0 - A B 0 - B 0 - B
14 12 13 renegmulnnass A 0 - A B 0 - B 0 - 0 - A 0 - B = 0 - 0 - A 0 - B
15 rernegcl B 0 - B
16 15 ad2antrl A 0 - A B 0 - B 0 - B
17 simplr A 0 - A B 0 - B 0 - A
18 16 17 renegmulnnass A 0 - A B 0 - B 0 - 0 - B 0 - A = 0 - 0 - B 0 - A
19 10 14 18 3eqtr4d A 0 - A B 0 - B 0 - 0 - A 0 - B = 0 - 0 - B 0 - A
20 19 oveq2d A 0 - A B 0 - B 0 - 0 - 0 - A 0 - B = 0 - 0 - 0 - B 0 - A
21 rernegcl 0 - A 0 - 0 - A
22 11 21 syl A 0 - 0 - A
23 22 ad2antrr A 0 - A B 0 - B 0 - 0 - A
24 23 16 remulneg2d A 0 - A B 0 - B 0 - 0 - A 0 - 0 - B = 0 - 0 - 0 - A 0 - B
25 rernegcl 0 - B 0 - 0 - B
26 15 25 syl B 0 - 0 - B
27 26 ad2antrl A 0 - A B 0 - B 0 - 0 - B
28 27 12 remulneg2d A 0 - A B 0 - B 0 - 0 - B 0 - 0 - A = 0 - 0 - 0 - B 0 - A
29 20 24 28 3eqtr4d A 0 - A B 0 - B 0 - 0 - A 0 - 0 - B = 0 - 0 - B 0 - 0 - A
30 renegneg A 0 - 0 - A = A
31 30 ad2antrr A 0 - A B 0 - B 0 - 0 - A = A
32 renegneg B 0 - 0 - B = B
33 32 ad2antrl A 0 - A B 0 - B 0 - 0 - B = B
34 31 33 oveq12d A 0 - A B 0 - B 0 - 0 - A 0 - 0 - B = A B
35 33 31 oveq12d A 0 - A B 0 - B 0 - 0 - B 0 - 0 - A = B A
36 29 34 35 3eqtr3d A 0 - A B 0 - B A B = B A
37 3 4 7 36 ccase A 0 A 0 - A B 0 B 0 - B A B = B A
38 1 2 37 syl2anb A B A B = B A