Metamath Proof Explorer
Description: Subtraction in a ring, in terms of addition and negation. (Contributed by Jeff Madsen, 19-Jun-2010)
|
|
Ref |
Expression |
|
Hypotheses |
ringnegcl.1 |
⊢ 𝐺 = ( 1st ‘ 𝑅 ) |
|
|
ringnegcl.2 |
⊢ 𝑋 = ran 𝐺 |
|
|
ringnegcl.3 |
⊢ 𝑁 = ( inv ‘ 𝐺 ) |
|
|
ringsub.4 |
⊢ 𝐷 = ( /𝑔 ‘ 𝐺 ) |
|
Assertion |
rngosub |
⊢ ( ( 𝑅 ∈ RingOps ∧ 𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋 ) → ( 𝐴 𝐷 𝐵 ) = ( 𝐴 𝐺 ( 𝑁 ‘ 𝐵 ) ) ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
ringnegcl.1 |
⊢ 𝐺 = ( 1st ‘ 𝑅 ) |
2 |
|
ringnegcl.2 |
⊢ 𝑋 = ran 𝐺 |
3 |
|
ringnegcl.3 |
⊢ 𝑁 = ( inv ‘ 𝐺 ) |
4 |
|
ringsub.4 |
⊢ 𝐷 = ( /𝑔 ‘ 𝐺 ) |
5 |
1
|
rngogrpo |
⊢ ( 𝑅 ∈ RingOps → 𝐺 ∈ GrpOp ) |
6 |
2 3 4
|
grpodivval |
⊢ ( ( 𝐺 ∈ GrpOp ∧ 𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋 ) → ( 𝐴 𝐷 𝐵 ) = ( 𝐴 𝐺 ( 𝑁 ‘ 𝐵 ) ) ) |
7 |
5 6
|
syl3an1 |
⊢ ( ( 𝑅 ∈ RingOps ∧ 𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋 ) → ( 𝐴 𝐷 𝐵 ) = ( 𝐴 𝐺 ( 𝑁 ‘ 𝐵 ) ) ) |