Metamath Proof Explorer


Theorem leat

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 ) )