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 )