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 )