Metamath Proof Explorer


Theorem mulcanenq

Description: Lemma for distributive law: cancellation of common factor. (Contributed by NM, 2-Sep-1995) (Revised by Mario Carneiro, 8-May-2013) (New usage is discouraged.)

Ref Expression
Assertion mulcanenq A 𝑵 B 𝑵 C 𝑵 A 𝑵 B A 𝑵 C ~ 𝑸 B C

Proof

Step Hyp Ref Expression
1 oveq2 b = B A 𝑵 b = A 𝑵 B
2 1 opeq1d b = B A 𝑵 b A 𝑵 c = A 𝑵 B A 𝑵 c
3 opeq1 b = B b c = B c
4 2 3 breq12d b = B A 𝑵 b A 𝑵 c ~ 𝑸 b c A 𝑵 B A 𝑵 c ~ 𝑸 B c
5 4 imbi2d b = B A 𝑵 A 𝑵 b A 𝑵 c ~ 𝑸 b c A 𝑵 A 𝑵 B A 𝑵 c ~ 𝑸 B c
6 oveq2 c = C A 𝑵 c = A 𝑵 C
7 6 opeq2d c = C A 𝑵 B A 𝑵 c = A 𝑵 B A 𝑵 C
8 opeq2 c = C B c = B C
9 7 8 breq12d c = C A 𝑵 B A 𝑵 c ~ 𝑸 B c A 𝑵 B A 𝑵 C ~ 𝑸 B C
10 9 imbi2d c = C A 𝑵 A 𝑵 B A 𝑵 c ~ 𝑸 B c A 𝑵 A 𝑵 B A 𝑵 C ~ 𝑸 B C
11 mulcompi b 𝑵 c = c 𝑵 b
12 11 oveq2i A 𝑵 b 𝑵 c = A 𝑵 c 𝑵 b
13 mulasspi A 𝑵 b 𝑵 c = A 𝑵 b 𝑵 c
14 mulasspi A 𝑵 c 𝑵 b = A 𝑵 c 𝑵 b
15 12 13 14 3eqtr4i A 𝑵 b 𝑵 c = A 𝑵 c 𝑵 b
16 mulclpi A 𝑵 b 𝑵 A 𝑵 b 𝑵
17 16 3adant3 A 𝑵 b 𝑵 c 𝑵 A 𝑵 b 𝑵
18 mulclpi A 𝑵 c 𝑵 A 𝑵 c 𝑵
19 18 3adant2 A 𝑵 b 𝑵 c 𝑵 A 𝑵 c 𝑵
20 3simpc A 𝑵 b 𝑵 c 𝑵 b 𝑵 c 𝑵
21 enqbreq A 𝑵 b 𝑵 A 𝑵 c 𝑵 b 𝑵 c 𝑵 A 𝑵 b A 𝑵 c ~ 𝑸 b c A 𝑵 b 𝑵 c = A 𝑵 c 𝑵 b
22 17 19 20 21 syl21anc A 𝑵 b 𝑵 c 𝑵 A 𝑵 b A 𝑵 c ~ 𝑸 b c A 𝑵 b 𝑵 c = A 𝑵 c 𝑵 b
23 15 22 mpbiri A 𝑵 b 𝑵 c 𝑵 A 𝑵 b A 𝑵 c ~ 𝑸 b c
24 23 3expb A 𝑵 b 𝑵 c 𝑵 A 𝑵 b A 𝑵 c ~ 𝑸 b c
25 24 expcom b 𝑵 c 𝑵 A 𝑵 A 𝑵 b A 𝑵 c ~ 𝑸 b c
26 5 10 25 vtocl2ga B 𝑵 C 𝑵 A 𝑵 A 𝑵 B A 𝑵 C ~ 𝑸 B C
27 26 impcom A 𝑵 B 𝑵 C 𝑵 A 𝑵 B A 𝑵 C ~ 𝑸 B C
28 27 3impb A 𝑵 B 𝑵 C 𝑵 A 𝑵 B A 𝑵 C ~ 𝑸 B C