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Metamath Proof Explorer

Theorem leat3

Description: A poset element less than or equal to an atom is either an atom or zero. (Contributed by NM, 2-Dec-2012)

Ref Expression
Hypotheses leatom.b 
|- B = ( Base ` K )
leatom.l 
|- .<_ = ( le ` K )
leatom.z 
|- .0. = ( 0. ` K )
leatom.a 
|- A = ( Atoms ` K )
Assertion leat3 
|- ( ( ( K e. OP /\ X e. B /\ P e. A ) /\ X .<_ P ) -> ( X e. A \/ X = .0. ) )

Proof

Step Hyp Ref Expression
1 leatom.b 
 |-  B = ( Base ` K )
2 leatom.l 
 |-  .<_ = ( le ` K )
3 leatom.z 
 |-  .0. = ( 0. ` K )
4 leatom.a 
 |-  A = ( Atoms ` K )
5 1 2 3 4 leat 
 |-  ( ( ( K e. OP /\ X e. B /\ P e. A ) /\ X .<_ P ) -> ( X = P \/ X = .0. ) )
6 simpl3 
 |-  ( ( ( K e. OP /\ X e. B /\ P e. A ) /\ X .<_ P ) -> P e. A )
7 eleq1a 
 |-  ( P e. A -> ( X = P -> X e. A ) )
8 6 7 syl 
 |-  ( ( ( K e. OP /\ X e. B /\ P e. A ) /\ X .<_ P ) -> ( X = P -> X e. A ) )
9 8 orim1d 
 |-  ( ( ( K e. OP /\ X e. B /\ P e. A ) /\ X .<_ P ) -> ( ( X = P \/ X = .0. ) -> ( X e. A \/ X = .0. ) ) )
10 5 9 mpd 
 |-  ( ( ( K e. OP /\ X e. B /\ P e. A ) /\ X .<_ P ) -> ( X e. A \/ X = .0. ) )