Metamath Proof Explorer
Description: A group element's inverse is a group element. (Contributed by NM, 24-Aug-2011) (Revised by Mario Carneiro, 4-May-2015)
|
|
Ref |
Expression |
|
Hypotheses |
grpinvcl.b |
⊢ 𝐵 = ( Base ‘ 𝐺 ) |
|
|
grpinvcl.n |
⊢ 𝑁 = ( invg ‘ 𝐺 ) |
|
Assertion |
grpinvcl |
⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ) → ( 𝑁 ‘ 𝑋 ) ∈ 𝐵 ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
grpinvcl.b |
⊢ 𝐵 = ( Base ‘ 𝐺 ) |
2 |
|
grpinvcl.n |
⊢ 𝑁 = ( invg ‘ 𝐺 ) |
3 |
1 2
|
grpinvf |
⊢ ( 𝐺 ∈ Grp → 𝑁 : 𝐵 ⟶ 𝐵 ) |
4 |
3
|
ffvelrnda |
⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ) → ( 𝑁 ‘ 𝑋 ) ∈ 𝐵 ) |