Metamath Proof Explorer


Theorem drnginvrcl

Description: Closure of the multiplicative inverse in a division ring. ( reccl analog.) (Contributed by NM, 19-Apr-2014)

Ref Expression
Hypotheses invrcl.b 𝐵 = ( Base ‘ 𝑅 )
invrcl.z 0 = ( 0g𝑅 )
invrcl.i 𝐼 = ( invr𝑅 )
Assertion drnginvrcl ( ( 𝑅 ∈ DivRing ∧ 𝑋𝐵𝑋0 ) → ( 𝐼𝑋 ) ∈ 𝐵 )

Proof

Step Hyp Ref Expression
1 invrcl.b 𝐵 = ( Base ‘ 𝑅 )
2 invrcl.z 0 = ( 0g𝑅 )
3 invrcl.i 𝐼 = ( invr𝑅 )
4 eqid ( Unit ‘ 𝑅 ) = ( Unit ‘ 𝑅 )
5 1 4 2 drngunit ( 𝑅 ∈ DivRing → ( 𝑋 ∈ ( Unit ‘ 𝑅 ) ↔ ( 𝑋𝐵𝑋0 ) ) )
6 drngring ( 𝑅 ∈ DivRing → 𝑅 ∈ Ring )
7 4 3 1 ringinvcl ( ( 𝑅 ∈ Ring ∧ 𝑋 ∈ ( Unit ‘ 𝑅 ) ) → ( 𝐼𝑋 ) ∈ 𝐵 )
8 7 ex ( 𝑅 ∈ Ring → ( 𝑋 ∈ ( Unit ‘ 𝑅 ) → ( 𝐼𝑋 ) ∈ 𝐵 ) )
9 6 8 syl ( 𝑅 ∈ DivRing → ( 𝑋 ∈ ( Unit ‘ 𝑅 ) → ( 𝐼𝑋 ) ∈ 𝐵 ) )
10 5 9 sylbird ( 𝑅 ∈ DivRing → ( ( 𝑋𝐵𝑋0 ) → ( 𝐼𝑋 ) ∈ 𝐵 ) )
11 10 3impib ( ( 𝑅 ∈ DivRing ∧ 𝑋𝐵𝑋0 ) → ( 𝐼𝑋 ) ∈ 𝐵 )