Metamath Proof Explorer


Theorem nmmul

Description: The norm of a product in a normed ring. (Contributed by Mario Carneiro, 5-Oct-2015)

Ref Expression
Hypotheses nmmul.x 𝑋 = ( Base ‘ 𝑅 )
nmmul.n 𝑁 = ( norm ‘ 𝑅 )
nmmul.t · = ( .r𝑅 )
Assertion nmmul ( ( 𝑅 ∈ NrmRing ∧ 𝐴𝑋𝐵𝑋 ) → ( 𝑁 ‘ ( 𝐴 · 𝐵 ) ) = ( ( 𝑁𝐴 ) · ( 𝑁𝐵 ) ) )

Proof

Step Hyp Ref Expression
1 nmmul.x 𝑋 = ( Base ‘ 𝑅 )
2 nmmul.n 𝑁 = ( norm ‘ 𝑅 )
3 nmmul.t · = ( .r𝑅 )
4 eqid ( AbsVal ‘ 𝑅 ) = ( AbsVal ‘ 𝑅 )
5 2 4 nrgabv ( 𝑅 ∈ NrmRing → 𝑁 ∈ ( AbsVal ‘ 𝑅 ) )
6 4 1 3 abvmul ( ( 𝑁 ∈ ( AbsVal ‘ 𝑅 ) ∧ 𝐴𝑋𝐵𝑋 ) → ( 𝑁 ‘ ( 𝐴 · 𝐵 ) ) = ( ( 𝑁𝐴 ) · ( 𝑁𝐵 ) ) )
7 5 6 syl3an1 ( ( 𝑅 ∈ NrmRing ∧ 𝐴𝑋𝐵𝑋 ) → ( 𝑁 ‘ ( 𝐴 · 𝐵 ) ) = ( ( 𝑁𝐴 ) · ( 𝑁𝐵 ) ) )