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Metamath Proof Explorer

Theorem opposet

Description: Every orthoposet is a poset. (Contributed by NM, 12-Oct-2011)

Ref Expression
Assertion opposet ( 𝐾 ∈ OP → 𝐾 ∈ Poset )

Proof

Step Hyp Ref Expression
1 eqid ⊢ ( Base ‘ 𝐾 ) = ( Base ‘ 𝐾 )
2 eqid ⊢ ( lub ‘ 𝐾 ) = ( lub ‘ 𝐾 )
3 eqid ⊢ ( glb ‘ 𝐾 ) = ( 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 ∧ ( Base ‘ 𝐾 ) ∈ dom ( lub ‘ 𝐾 ) ∧ ( Base ‘ 𝐾 ) ∈ dom ( glb ‘ 𝐾 ) ) ∧ ∀ 𝑥 ∈ ( Base ‘ 𝐾 ) ∀ 𝑦 ∈ ( Base ‘ 𝐾 ) ( ( ( ( oc ‘ 𝐾 ) ‘ 𝑥 ) ∈ ( Base ‘ 𝐾 ) ∧ ( ( 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 simpl1 ⊢ ( ( ( 𝐾 ∈ Poset ∧ ( Base ‘ 𝐾 ) ∈ dom ( lub ‘ 𝐾 ) ∧ ( Base ‘ 𝐾 ) ∈ dom ( glb ‘ 𝐾 ) ) ∧ ∀ 𝑥 ∈ ( Base ‘ 𝐾 ) ∀ 𝑦 ∈ ( Base ‘ 𝐾 ) ( ( ( ( oc ‘ 𝐾 ) ‘ 𝑥 ) ∈ ( Base ‘ 𝐾 ) ∧ ( ( 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 ) ) ) → 𝐾 ∈ Poset )
12 10 11 sylbi ⊢ ( 𝐾 ∈ OP → 𝐾 ∈ Poset )