Metamath Proof Explorer


Theorem abv1

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

Ref Expression
Hypotheses abv0.a 𝐴 = ( AbsVal ‘ 𝑅 )
abv1.p 1 = ( 1r𝑅 )
Assertion abv1 ( ( 𝑅 ∈ DivRing ∧ 𝐹𝐴 ) → ( 𝐹1 ) = 1 )

Proof

Step Hyp Ref Expression
1 abv0.a 𝐴 = ( AbsVal ‘ 𝑅 )
2 abv1.p 1 = ( 1r𝑅 )
3 id ( 𝐹𝐴𝐹𝐴 )
4 eqid ( 0g𝑅 ) = ( 0g𝑅 )
5 4 2 drngunz ( 𝑅 ∈ DivRing → 1 ≠ ( 0g𝑅 ) )
6 1 2 4 abv1z ( ( 𝐹𝐴1 ≠ ( 0g𝑅 ) ) → ( 𝐹1 ) = 1 )
7 3 5 6 syl2anr ( ( 𝑅 ∈ DivRing ∧ 𝐹𝐴 ) → ( 𝐹1 ) = 1 )