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Metamath Proof Explorer

Theorem posasymb

Description: A poset ordering is asymmetric. (Contributed by NM, 21-Oct-2011)

Ref Expression
Hypotheses posi.b 
|- B = ( Base ` K )
posi.l 
|- .<_ = ( le ` K )
Assertion posasymb 
|- ( ( K e. Poset /\ X e. B /\ Y e. B ) -> ( ( X .<_ Y /\ Y .<_ X ) <-> X = Y ) )

Proof

Step Hyp Ref Expression
1 posi.b 
 |-  B = ( Base ` K )
2 posi.l 
 |-  .<_ = ( le ` K )
3 simp1 
 |-  ( ( K e. Poset /\ X e. B /\ Y e. B ) -> K e. Poset )
4 simp2 
 |-  ( ( K e. Poset /\ X e. B /\ Y e. B ) -> X e. B )
5 simp3 
 |-  ( ( K e. Poset /\ X e. B /\ Y e. B ) -> Y e. B )
6 1 2 posi 
 |-  ( ( K e. Poset /\ ( X e. B /\ Y e. B /\ Y e. B ) ) -> ( X .<_ X /\ ( ( X .<_ Y /\ Y .<_ X ) -> X = Y ) /\ ( ( X .<_ Y /\ Y .<_ Y ) -> X .<_ Y ) ) )
7 3 4 5 5 6 syl13anc 
 |-  ( ( K e. Poset /\ X e. B /\ Y e. B ) -> ( X .<_ X /\ ( ( X .<_ Y /\ Y .<_ X ) -> X = Y ) /\ ( ( X .<_ Y /\ Y .<_ Y ) -> X .<_ Y ) ) )
8 7 simp2d 
 |-  ( ( K e. Poset /\ X e. B /\ Y e. B ) -> ( ( X .<_ Y /\ Y .<_ X ) -> X = Y ) )
9 1 2 posref 
 |-  ( ( K e. Poset /\ X e. B ) -> X .<_ X )
10 breq2 
 |-  ( X = Y -> ( X .<_ X <-> X .<_ Y ) )
11 9 10 syl5ibcom 
 |-  ( ( K e. Poset /\ X e. B ) -> ( X = Y -> X .<_ Y ) )
12 breq1 
 |-  ( X = Y -> ( X .<_ X <-> Y .<_ X ) )
13 9 12 syl5ibcom 
 |-  ( ( K e. Poset /\ X e. B ) -> ( X = Y -> Y .<_ X ) )
14 11 13 jcad 
 |-  ( ( K e. Poset /\ X e. B ) -> ( X = Y -> ( X .<_ Y /\ Y .<_ X ) ) )
15 14 3adant3 
 |-  ( ( K e. Poset /\ X e. B /\ Y e. B ) -> ( X = Y -> ( X .<_ Y /\ Y .<_ X ) ) )
16 8 15 impbid 
 |-  ( ( K e. Poset /\ X e. B /\ Y e. B ) -> ( ( X .<_ Y /\ Y .<_ X ) <-> X = Y ) )