Metamath Proof Explorer
Description: Base set of an opposite group. (Contributed by Stefan O'Rear, 26-Aug-2015)
|
|
Ref |
Expression |
|
Hypotheses |
oppgbas.1 |
⊢ 𝑂 = ( oppg ‘ 𝑅 ) |
|
|
oppgbas.2 |
⊢ 𝐵 = ( Base ‘ 𝑅 ) |
|
Assertion |
oppgbas |
⊢ 𝐵 = ( Base ‘ 𝑂 ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
oppgbas.1 |
⊢ 𝑂 = ( oppg ‘ 𝑅 ) |
2 |
|
oppgbas.2 |
⊢ 𝐵 = ( Base ‘ 𝑅 ) |
3 |
|
df-base |
⊢ Base = Slot 1 |
4 |
|
1nn |
⊢ 1 ∈ ℕ |
5 |
|
1ne2 |
⊢ 1 ≠ 2 |
6 |
1 3 4 5
|
oppglem |
⊢ ( Base ‘ 𝑅 ) = ( Base ‘ 𝑂 ) |
7 |
2 6
|
eqtri |
⊢ 𝐵 = ( Base ‘ 𝑂 ) |