Metamath Proof Explorer
Description: A poset element less than or equal to an atom equals either zero or the
atom. (Contributed by NM, 15-Oct-2013)
|
|
Ref |
Expression |
|
Hypotheses |
leatom.b |
⊢ 𝐵 = ( Base ‘ 𝐾 ) |
|
|
leatom.l |
⊢ ≤ = ( le ‘ 𝐾 ) |
|
|
leatom.z |
⊢ 0 = ( 0. ‘ 𝐾 ) |
|
|
leatom.a |
⊢ 𝐴 = ( Atoms ‘ 𝐾 ) |
|
Assertion |
leat |
⊢ ( ( ( 𝐾 ∈ OP ∧ 𝑋 ∈ 𝐵 ∧ 𝑃 ∈ 𝐴 ) ∧ 𝑋 ≤ 𝑃 ) → ( 𝑋 = 𝑃 ∨ 𝑋 = 0 ) ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
leatom.b |
⊢ 𝐵 = ( Base ‘ 𝐾 ) |
2 |
|
leatom.l |
⊢ ≤ = ( le ‘ 𝐾 ) |
3 |
|
leatom.z |
⊢ 0 = ( 0. ‘ 𝐾 ) |
4 |
|
leatom.a |
⊢ 𝐴 = ( Atoms ‘ 𝐾 ) |
5 |
1 2 3 4
|
leatb |
⊢ ( ( 𝐾 ∈ OP ∧ 𝑋 ∈ 𝐵 ∧ 𝑃 ∈ 𝐴 ) → ( 𝑋 ≤ 𝑃 ↔ ( 𝑋 = 𝑃 ∨ 𝑋 = 0 ) ) ) |
6 |
5
|
biimpa |
⊢ ( ( ( 𝐾 ∈ OP ∧ 𝑋 ∈ 𝐵 ∧ 𝑃 ∈ 𝐴 ) ∧ 𝑋 ≤ 𝑃 ) → ( 𝑋 = 𝑃 ∨ 𝑋 = 0 ) ) |