Theorem nmeq0

Description: The identity is the only element of the group with zero norm. First part of Problem 2 of Kreyszig p. 64. (Contributed by NM, 24-Nov-2006) (Revised by Mario Carneiro, 4-Oct-2015)

	Ref	Expression
Hypotheses	nmf.x	⊢ 𝑋 = ( Base ‘ 𝐺 )
	nmf.n	⊢ 𝑁 = ( norm ‘ 𝐺 )
	nmeq0.z	⊢ 0 = ( 0g ‘ 𝐺 )
Assertion	nmeq0	⊢ ( ( 𝐺 ∈ NrmGrp ∧ 𝐴 ∈ 𝑋 ) → ( ( 𝑁 ‘ 𝐴 ) = 0 ↔ 𝐴 = 0 ) )

Proof

Step	Hyp	Ref	Expression
1		nmf.x	⊢ 𝑋 = ( Base ‘ 𝐺 )
2		nmf.n	⊢ 𝑁 = ( norm ‘ 𝐺 )
3		nmeq0.z	⊢ 0 = ( 0g ‘ 𝐺 )
4		eqid	⊢ ( dist ‘ 𝐺 ) = ( dist ‘ 𝐺 )
5	2 1 3 4	nmval	⊢ ( 𝐴 ∈ 𝑋 → ( 𝑁 ‘ 𝐴 ) = ( 𝐴 ( dist ‘ 𝐺 ) 0 ) )
6	5	adantl	⊢ ( ( 𝐺 ∈ NrmGrp ∧ 𝐴 ∈ 𝑋 ) → ( 𝑁 ‘ 𝐴 ) = ( 𝐴 ( dist ‘ 𝐺 ) 0 ) )
7	6	eqeq1d	⊢ ( ( 𝐺 ∈ NrmGrp ∧ 𝐴 ∈ 𝑋 ) → ( ( 𝑁 ‘ 𝐴 ) = 0 ↔ ( 𝐴 ( dist ‘ 𝐺 ) 0 ) = 0 ) )
8		ngpgrp	⊢ ( 𝐺 ∈ NrmGrp → 𝐺 ∈ Grp )
9	8	adantr	⊢ ( ( 𝐺 ∈ NrmGrp ∧ 𝐴 ∈ 𝑋 ) → 𝐺 ∈ Grp )
10	1 3	grpidcl	⊢ ( 𝐺 ∈ Grp → 0 ∈ 𝑋 )
11	9 10	syl	⊢ ( ( 𝐺 ∈ NrmGrp ∧ 𝐴 ∈ 𝑋 ) → 0 ∈ 𝑋 )
12		ngpxms	⊢ ( 𝐺 ∈ NrmGrp → 𝐺 ∈ ∞MetSp )
13	1 4	xmseq0	⊢ ( ( 𝐺 ∈ ∞MetSp ∧ 𝐴 ∈ 𝑋 ∧ 0 ∈ 𝑋 ) → ( ( 𝐴 ( dist ‘ 𝐺 ) 0 ) = 0 ↔ 𝐴 = 0 ) )
14	12 13	syl3an1	⊢ ( ( 𝐺 ∈ NrmGrp ∧ 𝐴 ∈ 𝑋 ∧ 0 ∈ 𝑋 ) → ( ( 𝐴 ( dist ‘ 𝐺 ) 0 ) = 0 ↔ 𝐴 = 0 ) )
15	11 14	mpd3an3	⊢ ( ( 𝐺 ∈ NrmGrp ∧ 𝐴 ∈ 𝑋 ) → ( ( 𝐴 ( dist ‘ 𝐺 ) 0 ) = 0 ↔ 𝐴 = 0 ) )
16	7 15	bitrd	⊢ ( ( 𝐺 ∈ NrmGrp ∧ 𝐴 ∈ 𝑋 ) → ( ( 𝑁 ‘ 𝐴 ) = 0 ↔ 𝐴 = 0 ) )