Metamath Proof Explorer


Theorem ispos

Description: The predicate "is a poset". (Contributed by NM, 18-Oct-2012) (Revised by Mario Carneiro, 4-Nov-2013)

Ref Expression
Hypotheses ispos.b B=BaseK
ispos.l ˙=K
Assertion ispos KPosetKVxByBzBx˙xx˙yy˙xx=yx˙yy˙zx˙z

Proof

Step Hyp Ref Expression
1 ispos.b B=BaseK
2 ispos.l ˙=K
3 fveq2 p=KBasep=BaseK
4 3 1 eqtr4di p=KBasep=B
5 4 eqeq2d p=Kb=Basepb=B
6 fveq2 p=Kp=K
7 6 2 eqtr4di p=Kp=˙
8 7 eqeq2d p=Kr=pr=˙
9 5 8 3anbi12d p=Kb=Basepr=pxbybzbxrxxryyrxx=yxryyrzxrzb=Br=˙xbybzbxrxxryyrxx=yxryyrzxrz
10 9 2exbidv p=Kbrb=Basepr=pxbybzbxrxxryyrxx=yxryyrzxrzbrb=Br=˙xbybzbxrxxryyrxx=yxryyrzxrz
11 df-poset Poset=p|brb=Basepr=pxbybzbxrxxryyrxx=yxryyrzxrz
12 10 11 elab4g KPosetKVbrb=Br=˙xbybzbxrxxryyrxx=yxryyrzxrz
13 1 fvexi BV
14 2 fvexi ˙V
15 raleq b=Bzbxrxxryyrxx=yxryyrzxrzzBxrxxryyrxx=yxryyrzxrz
16 15 raleqbi1dv b=Bybzbxrxxryyrxx=yxryyrzxrzyBzBxrxxryyrxx=yxryyrzxrz
17 16 raleqbi1dv b=Bxbybzbxrxxryyrxx=yxryyrzxrzxByBzBxrxxryyrxx=yxryyrzxrz
18 breq r=˙xrxx˙x
19 breq r=˙xryx˙y
20 breq r=˙yrxy˙x
21 19 20 anbi12d r=˙xryyrxx˙yy˙x
22 21 imbi1d r=˙xryyrxx=yx˙yy˙xx=y
23 breq r=˙yrzy˙z
24 19 23 anbi12d r=˙xryyrzx˙yy˙z
25 breq r=˙xrzx˙z
26 24 25 imbi12d r=˙xryyrzxrzx˙yy˙zx˙z
27 18 22 26 3anbi123d r=˙xrxxryyrxx=yxryyrzxrzx˙xx˙yy˙xx=yx˙yy˙zx˙z
28 27 ralbidv r=˙zBxrxxryyrxx=yxryyrzxrzzBx˙xx˙yy˙xx=yx˙yy˙zx˙z
29 28 2ralbidv r=˙xByBzBxrxxryyrxx=yxryyrzxrzxByBzBx˙xx˙yy˙xx=yx˙yy˙zx˙z
30 13 14 17 29 ceqsex2v brb=Br=˙xbybzbxrxxryyrxx=yxryyrzxrzxByBzBx˙xx˙yy˙xx=yx˙yy˙zx˙z
31 30 anbi2i KVbrb=Br=˙xbybzbxrxxryyrxx=yxryyrzxrzKVxByBzBx˙xx˙yy˙xx=yx˙yy˙zx˙z
32 12 31 bitri KPosetKVxByBzBx˙xx˙yy˙xx=yx˙yy˙zx˙z