Metamath Proof Explorer
Description: The LUB of the set of two comparable elements in a poset is the
greater one of the two. (Contributed by Zhi Wang, 26-Sep-2024)
|
|
Ref |
Expression |
|
Hypotheses |
lubpr.k |
⊢ ( 𝜑 → 𝐾 ∈ Poset ) |
|
|
lubpr.b |
⊢ 𝐵 = ( Base ‘ 𝐾 ) |
|
|
lubpr.x |
⊢ ( 𝜑 → 𝑋 ∈ 𝐵 ) |
|
|
lubpr.y |
⊢ ( 𝜑 → 𝑌 ∈ 𝐵 ) |
|
|
lubpr.l |
⊢ ≤ = ( le ‘ 𝐾 ) |
|
|
lubpr.c |
⊢ ( 𝜑 → 𝑋 ≤ 𝑌 ) |
|
|
lubpr.s |
⊢ ( 𝜑 → 𝑆 = { 𝑋 , 𝑌 } ) |
|
|
lubpr.u |
⊢ 𝑈 = ( lub ‘ 𝐾 ) |
|
Assertion |
lubpr |
⊢ ( 𝜑 → ( 𝑈 ‘ 𝑆 ) = 𝑌 ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
lubpr.k |
⊢ ( 𝜑 → 𝐾 ∈ Poset ) |
2 |
|
lubpr.b |
⊢ 𝐵 = ( Base ‘ 𝐾 ) |
3 |
|
lubpr.x |
⊢ ( 𝜑 → 𝑋 ∈ 𝐵 ) |
4 |
|
lubpr.y |
⊢ ( 𝜑 → 𝑌 ∈ 𝐵 ) |
5 |
|
lubpr.l |
⊢ ≤ = ( le ‘ 𝐾 ) |
6 |
|
lubpr.c |
⊢ ( 𝜑 → 𝑋 ≤ 𝑌 ) |
7 |
|
lubpr.s |
⊢ ( 𝜑 → 𝑆 = { 𝑋 , 𝑌 } ) |
8 |
|
lubpr.u |
⊢ 𝑈 = ( lub ‘ 𝐾 ) |
9 |
1 2 3 4 5 6 7 8
|
lubprlem |
⊢ ( 𝜑 → ( 𝑆 ∈ dom 𝑈 ∧ ( 𝑈 ‘ 𝑆 ) = 𝑌 ) ) |
10 |
9
|
simprd |
⊢ ( 𝜑 → ( 𝑈 ‘ 𝑆 ) = 𝑌 ) |