Metamath Proof Explorer

Theorem op01dm

Description: Conditions necessary for zero and unit 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 e. OP -> ( B e. dom U /\ B e. 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 e. OP <-> ( ( K e. Poset /\ B e. dom U /\ B e. dom G ) /\ A. x e. B A. y e. B ( ( ( ( oc ` K ) ` x ) e. 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 e. dom U /\ B e. dom G ) /\ A. x e. B A. y e. B ( ( ( ( oc ` K ) ` x ) e. 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 e. dom U /\ B e. dom G ) )
12 11 3adantl1 
 |-  ( ( ( K e. Poset /\ B e. dom U /\ B e. dom G ) /\ A. x e. B A. y e. B ( ( ( ( oc ` K ) ` x ) e. 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 e. dom U /\ B e. dom G ) )
13 10 12 sylbi 
 |-  ( K e. OP -> ( B e. dom U /\ B e. dom G ) )