Metamath Proof Explorer


Theorem latasymd

Description: Deduce equality from lattice ordering. ( eqssd analog.) (Contributed by NM, 18-Nov-2011)

Ref Expression
Hypotheses latasymd.b
|- B = ( Base ` K )
latasymd.l
|- .<_ = ( le ` K )
latasymd.3
|- ( ph -> K e. Lat )
latasymd.4
|- ( ph -> X e. B )
latasymd.5
|- ( ph -> Y e. B )
latasymd.6
|- ( ph -> X .<_ Y )
latasymd.7
|- ( ph -> Y .<_ X )
Assertion latasymd
|- ( ph -> X = Y )

Proof

Step Hyp Ref Expression
1 latasymd.b
 |-  B = ( Base ` K )
2 latasymd.l
 |-  .<_ = ( le ` K )
3 latasymd.3
 |-  ( ph -> K e. Lat )
4 latasymd.4
 |-  ( ph -> X e. B )
5 latasymd.5
 |-  ( ph -> Y e. B )
6 latasymd.6
 |-  ( ph -> X .<_ Y )
7 latasymd.7
 |-  ( ph -> Y .<_ X )
8 1 2 latasymb
 |-  ( ( K e. Lat /\ X e. B /\ Y e. B ) -> ( ( X .<_ Y /\ Y .<_ X ) <-> X = Y ) )
9 3 4 5 8 syl3anc
 |-  ( ph -> ( ( X .<_ Y /\ Y .<_ X ) <-> X = Y ) )
10 6 7 9 mpbi2and
 |-  ( ph -> X = Y )