Metamath Proof Explorer
Description: A group element's inverse is a group element. (Contributed by SN, 29-Jan-2025)
|
|
Ref |
Expression |
|
Hypotheses |
grpinvcld.b |
⊢ 𝐵 = ( Base ‘ 𝐺 ) |
|
|
grpinvcld.n |
⊢ 𝑁 = ( invg ‘ 𝐺 ) |
|
|
grpinvcld.g |
⊢ ( 𝜑 → 𝐺 ∈ Grp ) |
|
|
grpinvcld.1 |
⊢ ( 𝜑 → 𝑋 ∈ 𝐵 ) |
|
Assertion |
grpinvcld |
⊢ ( 𝜑 → ( 𝑁 ‘ 𝑋 ) ∈ 𝐵 ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
grpinvcld.b |
⊢ 𝐵 = ( Base ‘ 𝐺 ) |
2 |
|
grpinvcld.n |
⊢ 𝑁 = ( invg ‘ 𝐺 ) |
3 |
|
grpinvcld.g |
⊢ ( 𝜑 → 𝐺 ∈ Grp ) |
4 |
|
grpinvcld.1 |
⊢ ( 𝜑 → 𝑋 ∈ 𝐵 ) |
5 |
1 2
|
grpinvcl |
⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ) → ( 𝑁 ‘ 𝑋 ) ∈ 𝐵 ) |
6 |
3 4 5
|
syl2anc |
⊢ ( 𝜑 → ( 𝑁 ‘ 𝑋 ) ∈ 𝐵 ) |