Metamath Proof Explorer
Description: Closure of a dividing element. (Contributed by Mario Carneiro, 5-Dec-2014)
|
|
Ref |
Expression |
|
Hypotheses |
dvdsr.1 |
⊢ 𝐵 = ( Base ‘ 𝑅 ) |
|
|
dvdsr.2 |
⊢ ∥ = ( ∥r ‘ 𝑅 ) |
|
Assertion |
dvdsrcl |
⊢ ( 𝑋 ∥ 𝑌 → 𝑋 ∈ 𝐵 ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
dvdsr.1 |
⊢ 𝐵 = ( Base ‘ 𝑅 ) |
2 |
|
dvdsr.2 |
⊢ ∥ = ( ∥r ‘ 𝑅 ) |
3 |
|
eqid |
⊢ ( .r ‘ 𝑅 ) = ( .r ‘ 𝑅 ) |
4 |
1 2 3
|
dvdsr |
⊢ ( 𝑋 ∥ 𝑌 ↔ ( 𝑋 ∈ 𝐵 ∧ ∃ 𝑥 ∈ 𝐵 ( 𝑥 ( .r ‘ 𝑅 ) 𝑋 ) = 𝑌 ) ) |
5 |
4
|
simplbi |
⊢ ( 𝑋 ∥ 𝑌 → 𝑋 ∈ 𝐵 ) |