Metamath Proof Explorer
Description: The set of atoms is a subset of the base set. ( atssch analog.)
(Contributed by NM, 21-Oct-2011)
|
|
Ref |
Expression |
|
Hypotheses |
atombase.b |
⊢ 𝐵 = ( Base ‘ 𝐾 ) |
|
|
atombase.a |
⊢ 𝐴 = ( Atoms ‘ 𝐾 ) |
|
Assertion |
atssbase |
⊢ 𝐴 ⊆ 𝐵 |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
atombase.b |
⊢ 𝐵 = ( Base ‘ 𝐾 ) |
2 |
|
atombase.a |
⊢ 𝐴 = ( Atoms ‘ 𝐾 ) |
3 |
1 2
|
atbase |
⊢ ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵 ) |
4 |
3
|
ssriv |
⊢ 𝐴 ⊆ 𝐵 |