Metamath Proof Explorer


Theorem abv1z

Description: The absolute value of one is one in a non-trivial ring. (Contributed by Mario Carneiro, 8-Sep-2014)

Ref Expression
Hypotheses abv0.a A = AbsVal R
abv1.p 1 ˙ = 1 R
abv1z.z 0 ˙ = 0 R
Assertion abv1z F A 1 ˙ 0 ˙ F 1 ˙ = 1

Proof

Step Hyp Ref Expression
1 abv0.a A = AbsVal R
2 abv1.p 1 ˙ = 1 R
3 abv1z.z 0 ˙ = 0 R
4 1 abvrcl F A R Ring
5 eqid Base R = Base R
6 5 2 ringidcl R Ring 1 ˙ Base R
7 4 6 syl F A 1 ˙ Base R
8 1 5 abvcl F A 1 ˙ Base R F 1 ˙
9 7 8 mpdan F A F 1 ˙
10 9 adantr F A 1 ˙ 0 ˙ F 1 ˙
11 10 recnd F A 1 ˙ 0 ˙ F 1 ˙
12 simpl F A 1 ˙ 0 ˙ F A
13 7 adantr F A 1 ˙ 0 ˙ 1 ˙ Base R
14 simpr F A 1 ˙ 0 ˙ 1 ˙ 0 ˙
15 1 5 3 abvne0 F A 1 ˙ Base R 1 ˙ 0 ˙ F 1 ˙ 0
16 12 13 14 15 syl3anc F A 1 ˙ 0 ˙ F 1 ˙ 0
17 11 11 16 divcan3d F A 1 ˙ 0 ˙ F 1 ˙ F 1 ˙ F 1 ˙ = F 1 ˙
18 eqid R = R
19 5 18 2 ringlidm R Ring 1 ˙ Base R 1 ˙ R 1 ˙ = 1 ˙
20 4 13 19 syl2an2r F A 1 ˙ 0 ˙ 1 ˙ R 1 ˙ = 1 ˙
21 20 fveq2d F A 1 ˙ 0 ˙ F 1 ˙ R 1 ˙ = F 1 ˙
22 1 5 18 abvmul F A 1 ˙ Base R 1 ˙ Base R F 1 ˙ R 1 ˙ = F 1 ˙ F 1 ˙
23 12 13 13 22 syl3anc F A 1 ˙ 0 ˙ F 1 ˙ R 1 ˙ = F 1 ˙ F 1 ˙
24 21 23 eqtr3d F A 1 ˙ 0 ˙ F 1 ˙ = F 1 ˙ F 1 ˙
25 24 oveq1d F A 1 ˙ 0 ˙ F 1 ˙ F 1 ˙ = F 1 ˙ F 1 ˙ F 1 ˙
26 11 16 dividd F A 1 ˙ 0 ˙ F 1 ˙ F 1 ˙ = 1
27 25 26 eqtr3d F A 1 ˙ 0 ˙ F 1 ˙ F 1 ˙ F 1 ˙ = 1
28 17 27 eqtr3d F A 1 ˙ 0 ˙ F 1 ˙ = 1