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Metamath Proof Explorer

Theorem drnginvrr

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

Ref Expression
Hypotheses drnginvrl.b ⊢ 𝐵 = ( Base ‘ 𝑅 )
drnginvrl.z ⊢ 0 = ( 0g ‘ 𝑅 )
drnginvrl.t ⊢ · = ( .r ‘ 𝑅 )
drnginvrl.u ⊢ 1 = ( 1r ‘ 𝑅 )
drnginvrl.i ⊢ 𝐼 = ( invr ‘ 𝑅 )
Assertion drnginvrr ( ( 𝑅 ∈ DivRing ∧ 𝑋 ∈ 𝐵 ∧ 𝑋 ≠ 0 ) → ( 𝑋 · ( 𝐼 ‘ 𝑋 ) ) = 1 )

Proof

Step Hyp Ref Expression
1 drnginvrl.b ⊢ 𝐵 = ( Base ‘ 𝑅 )
2 drnginvrl.z ⊢ 0 = ( 0g ‘ 𝑅 )
3 drnginvrl.t ⊢ · = ( .r ‘ 𝑅 )
4 drnginvrl.u ⊢ 1 = ( 1r ‘ 𝑅 )
5 drnginvrl.i ⊢ 𝐼 = ( invr ‘ 𝑅 )
6 eqid ⊢ ( Unit ‘ 𝑅 ) = ( Unit ‘ 𝑅 )
7 1 6 2 drngunit ⊢ ( 𝑅 ∈ DivRing → ( 𝑋 ∈ ( Unit ‘ 𝑅 ) ↔ ( 𝑋 ∈ 𝐵 ∧ 𝑋 ≠ 0 ) ) )
8 drngring ⊢ ( 𝑅 ∈ DivRing → 𝑅 ∈ Ring )
9 6 5 3 4 unitrinv ⊢ ( ( 𝑅 ∈ Ring ∧ 𝑋 ∈ ( Unit ‘ 𝑅 ) ) → ( 𝑋 · ( 𝐼 ‘ 𝑋 ) ) = 1 )
10 9 ex ⊢ ( 𝑅 ∈ Ring → ( 𝑋 ∈ ( Unit ‘ 𝑅 ) → ( 𝑋 · ( 𝐼 ‘ 𝑋 ) ) = 1 ) )
11 8 10 syl ⊢ ( 𝑅 ∈ DivRing → ( 𝑋 ∈ ( Unit ‘ 𝑅 ) → ( 𝑋 · ( 𝐼 ‘ 𝑋 ) ) = 1 ) )
12 7 11 sylbird ⊢ ( 𝑅 ∈ DivRing → ( ( 𝑋 ∈ 𝐵 ∧ 𝑋 ≠ 0 ) → ( 𝑋 · ( 𝐼 ‘ 𝑋 ) ) = 1 ) )
13 12 3impib ⊢ ( ( 𝑅 ∈ DivRing ∧ 𝑋 ∈ 𝐵 ∧ 𝑋 ≠ 0 ) → ( 𝑋 · ( 𝐼 ‘ 𝑋 ) ) = 1 )