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Metamath Proof Explorer

Theorem op01dm

Description: Conditions necessary for zero and unity elements to exist. (Contributed by NM, 14-Sep-2018)

Ref Expression
Hypotheses op01dm.b ⊢ 𝐵 = ( Base ‘ 𝐾 )
op01dm.u ⊢ 𝑈 = ( lub ‘ 𝐾 )
op01dm.g ⊢ 𝐺 = ( glb ‘ 𝐾 )
Assertion op01dm ( 𝐾 ∈ OP → ( 𝐵 ∈ dom 𝑈 ∧ 𝐵 ∈ dom 𝐺 ) )

Proof

Step Hyp Ref Expression
1 op01dm.b ⊢ 𝐵 = ( Base ‘ 𝐾 )
2 op01dm.u ⊢ 𝑈 = ( lub ‘ 𝐾 )
3 op01dm.g ⊢ 𝐺 = ( glb ‘ 𝐾 )
4 eqid ⊢ ( le ‘ 𝐾 ) = ( le ‘ 𝐾 )
5 eqid ⊢ ( oc ‘ 𝐾 ) = ( oc ‘ 𝐾 )
6 eqid ⊢ ( join ‘ 𝐾 ) = ( join ‘ 𝐾 )
7 eqid ⊢ ( meet ‘ 𝐾 ) = ( meet ‘ 𝐾 )
8 eqid ⊢ ( 0. ‘ 𝐾 ) = ( 0. ‘ 𝐾 )
9 eqid ⊢ ( 1. ‘ 𝐾 ) = ( 1. ‘ 𝐾 )
10 1 2 3 4 5 6 7 8 9 isopos ⊢ ( 𝐾 ∈ OP ↔ ( ( 𝐾 ∈ Poset ∧ 𝐵 ∈ dom 𝑈 ∧ 𝐵 ∈ dom 𝐺 ) ∧ ∀ 𝑥 ∈ 𝐵 ∀ 𝑦 ∈ 𝐵 ( ( ( ( oc ‘ 𝐾 ) ‘ 𝑥 ) ∈ 𝐵 ∧ ( ( oc ‘ 𝐾 ) ‘ ( ( oc ‘ 𝐾 ) ‘ 𝑥 ) ) = 𝑥 ∧ ( 𝑥 ( le ‘ 𝐾 ) 𝑦 → ( ( oc ‘ 𝐾 ) ‘ 𝑦 ) ( le ‘ 𝐾 ) ( ( oc ‘ 𝐾 ) ‘ 𝑥 ) ) ) ∧ ( 𝑥 ( join ‘ 𝐾 ) ( ( oc ‘ 𝐾 ) ‘ 𝑥 ) ) = ( 1. ‘ 𝐾 ) ∧ ( 𝑥 ( meet ‘ 𝐾 ) ( ( oc ‘ 𝐾 ) ‘ 𝑥 ) ) = ( 0. ‘ 𝐾 ) ) ) )
11 simpl ⊢ ( ( ( 𝐵 ∈ dom 𝑈 ∧ 𝐵 ∈ dom 𝐺 ) ∧ ∀ 𝑥 ∈ 𝐵 ∀ 𝑦 ∈ 𝐵 ( ( ( ( oc ‘ 𝐾 ) ‘ 𝑥 ) ∈ 𝐵 ∧ ( ( oc ‘ 𝐾 ) ‘ ( ( oc ‘ 𝐾 ) ‘ 𝑥 ) ) = 𝑥 ∧ ( 𝑥 ( le ‘ 𝐾 ) 𝑦 → ( ( oc ‘ 𝐾 ) ‘ 𝑦 ) ( le ‘ 𝐾 ) ( ( oc ‘ 𝐾 ) ‘ 𝑥 ) ) ) ∧ ( 𝑥 ( join ‘ 𝐾 ) ( ( oc ‘ 𝐾 ) ‘ 𝑥 ) ) = ( 1. ‘ 𝐾 ) ∧ ( 𝑥 ( meet ‘ 𝐾 ) ( ( oc ‘ 𝐾 ) ‘ 𝑥 ) ) = ( 0. ‘ 𝐾 ) ) ) → ( 𝐵 ∈ dom 𝑈 ∧ 𝐵 ∈ dom 𝐺 ) )
12 11 3adantl1 ⊢ ( ( ( 𝐾 ∈ Poset ∧ 𝐵 ∈ dom 𝑈 ∧ 𝐵 ∈ dom 𝐺 ) ∧ ∀ 𝑥 ∈ 𝐵 ∀ 𝑦 ∈ 𝐵 ( ( ( ( oc ‘ 𝐾 ) ‘ 𝑥 ) ∈ 𝐵 ∧ ( ( oc ‘ 𝐾 ) ‘ ( ( oc ‘ 𝐾 ) ‘ 𝑥 ) ) = 𝑥 ∧ ( 𝑥 ( le ‘ 𝐾 ) 𝑦 → ( ( oc ‘ 𝐾 ) ‘ 𝑦 ) ( le ‘ 𝐾 ) ( ( oc ‘ 𝐾 ) ‘ 𝑥 ) ) ) ∧ ( 𝑥 ( join ‘ 𝐾 ) ( ( oc ‘ 𝐾 ) ‘ 𝑥 ) ) = ( 1. ‘ 𝐾 ) ∧ ( 𝑥 ( meet ‘ 𝐾 ) ( ( oc ‘ 𝐾 ) ‘ 𝑥 ) ) = ( 0. ‘ 𝐾 ) ) ) → ( 𝐵 ∈ dom 𝑈 ∧ 𝐵 ∈ dom 𝐺 ) )
13 10 12 sylbi ⊢ ( 𝐾 ∈ OP → ( 𝐵 ∈ dom 𝑈 ∧ 𝐵 ∈ dom 𝐺 ) )