# Metamath Proof Explorer

## Theorem opposet

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

Ref Expression
Assertion opposet
`|- ( K e. OP -> K e. 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 e. OP <-> ( ( K e. Poset /\ ( Base ` K ) e. dom ( lub ` K ) /\ ( Base ` K ) e. dom ( glb ` K ) ) /\ A. x e. ( Base ` K ) A. y e. ( Base ` K ) ( ( ( ( oc ` K ) ` x ) e. ( 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 e. Poset /\ ( Base ` K ) e. dom ( lub ` K ) /\ ( Base ` K ) e. dom ( glb ` K ) ) /\ A. x e. ( Base ` K ) A. y e. ( Base ` K ) ( ( ( ( oc ` K ) ` x ) e. ( 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 e. Poset )`
12 10 11 sylbi
` |-  ( K e. OP -> K e. Poset )`