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Metamath Proof Explorer

Theorem leat3

Description: A poset element less than or equal to an atom is either an atom or zero. (Contributed by NM, 2-Dec-2012)

Ref Expression
Hypotheses leatom.b ⊢ 𝐵 = ( Base ‘ 𝐾 )
leatom.l ⊢ ≤ = ( le ‘ 𝐾 )
leatom.z ⊢ 0 = ( 0. ‘ 𝐾 )
leatom.a ⊢ 𝐴 = ( Atoms ‘ 𝐾 )
Assertion leat3 ( ( ( 𝐾 ∈ OP ∧ 𝑋 ∈ 𝐵 ∧ 𝑃 ∈ 𝐴 ) ∧ 𝑋 ≤ 𝑃 ) → ( 𝑋 ∈ 𝐴 ∨ 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 leat ⊢ ( ( ( 𝐾 ∈ OP ∧ 𝑋 ∈ 𝐵 ∧ 𝑃 ∈ 𝐴 ) ∧ 𝑋 ≤ 𝑃 ) → ( 𝑋 = 𝑃 ∨ 𝑋 = 0 ) )
6 simpl3 ⊢ ( ( ( 𝐾 ∈ OP ∧ 𝑋 ∈ 𝐵 ∧ 𝑃 ∈ 𝐴 ) ∧ 𝑋 ≤ 𝑃 ) → 𝑃 ∈ 𝐴 )
7 eleq1a ⊢ ( 𝑃 ∈ 𝐴 → ( 𝑋 = 𝑃 → 𝑋 ∈ 𝐴 ) )
8 6 7 syl ⊢ ( ( ( 𝐾 ∈ OP ∧ 𝑋 ∈ 𝐵 ∧ 𝑃 ∈ 𝐴 ) ∧ 𝑋 ≤ 𝑃 ) → ( 𝑋 = 𝑃 → 𝑋 ∈ 𝐴 ) )
9 8 orim1d ⊢ ( ( ( 𝐾 ∈ OP ∧ 𝑋 ∈ 𝐵 ∧ 𝑃 ∈ 𝐴 ) ∧ 𝑋 ≤ 𝑃 ) → ( ( 𝑋 = 𝑃 ∨ 𝑋 = 0 ) → ( 𝑋 ∈ 𝐴 ∨ 𝑋 = 0 ) ) )
10 5 9 mpd ⊢ ( ( ( 𝐾 ∈ OP ∧ 𝑋 ∈ 𝐵 ∧ 𝑃 ∈ 𝐴 ) ∧ 𝑋 ≤ 𝑃 ) → ( 𝑋 ∈ 𝐴 ∨ 𝑋 = 0 ) )