Metamath Proof Explorer


Theorem f1rhm0to0ALT

Description: Alternate proof for f1ghm0to0 . Using ghmf1 does not make the proof shorter and requires disjoint variable restrictions! (Contributed by AV, 24-Oct-2019) (New usage is discouraged.) (Proof modification is discouraged.)

Ref Expression
Hypotheses gim0to0ALT.a A = Base R
gim0to0ALT.b B = Base S
gim0to0ALT.n N = 0 S
gim0to0ALT.0 0 ˙ = 0 R
Assertion f1rhm0to0ALT F R RingHom S F : A 1-1 B X A F X = N X = 0 ˙

Proof

Step Hyp Ref Expression
1 gim0to0ALT.a A = Base R
2 gim0to0ALT.b B = Base S
3 gim0to0ALT.n N = 0 S
4 gim0to0ALT.0 0 ˙ = 0 R
5 rhmghm F R RingHom S F R GrpHom S
6 5 adantr F R RingHom S X A F R GrpHom S
7 1 2 4 3 ghmf1 F R GrpHom S F : A 1-1 B x A F x = N x = 0 ˙
8 6 7 syl F R RingHom S X A F : A 1-1 B x A F x = N x = 0 ˙
9 fveq2 x = X F x = F X
10 9 eqeq1d x = X F x = N F X = N
11 eqeq1 x = X x = 0 ˙ X = 0 ˙
12 10 11 imbi12d x = X F x = N x = 0 ˙ F X = N X = 0 ˙
13 12 rspcv X A x A F x = N x = 0 ˙ F X = N X = 0 ˙
14 13 adantl F R RingHom S X A x A F x = N x = 0 ˙ F X = N X = 0 ˙
15 8 14 sylbid F R RingHom S X A F : A 1-1 B F X = N X = 0 ˙
16 15 ex F R RingHom S X A F : A 1-1 B F X = N X = 0 ˙
17 16 com23 F R RingHom S F : A 1-1 B X A F X = N X = 0 ˙
18 17 3imp F R RingHom S F : A 1-1 B X A F X = N X = 0 ˙
19 fveq2 X = 0 ˙ F X = F 0 ˙
20 4 3 ghmid F R GrpHom S F 0 ˙ = N
21 5 20 syl F R RingHom S F 0 ˙ = N
22 21 3ad2ant1 F R RingHom S F : A 1-1 B X A F 0 ˙ = N
23 19 22 sylan9eqr F R RingHom S F : A 1-1 B X A X = 0 ˙ F X = N
24 23 ex F R RingHom S F : A 1-1 B X A X = 0 ˙ F X = N
25 18 24 impbid F R RingHom S F : A 1-1 B X A F X = N X = 0 ˙