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Metamath Proof Explorer

Theorem pleval2i

Description: One direction of pleval2 . (Contributed by Mario Carneiro, 8-Feb-2015)

Ref Expression
Hypotheses pleval2.b 
|- B = ( Base ` K )
pleval2.l 
|- .<_ = ( le ` K )
pleval2.s 
|- .< = ( lt ` K )
Assertion pleval2i 
|- ( ( X e. B /\ Y e. B ) -> ( X .<_ Y -> ( X .< Y \/ X = Y ) ) )

Proof

Step Hyp Ref Expression
1 pleval2.b 
 |-  B = ( Base ` K )
2 pleval2.l 
 |-  .<_ = ( le ` K )
3 pleval2.s 
 |-  .< = ( lt ` K )
4 elfvdm 
 |-  ( X e. ( Base ` K ) -> K e. dom Base )
5 4 1 eleq2s 
 |-  ( X e. B -> K e. dom Base )
6 5 adantr 
 |-  ( ( X e. B /\ Y e. B ) -> K e. dom Base )
7 2 3 pltval 
 |-  ( ( K e. dom Base /\ X e. B /\ Y e. B ) -> ( X .< Y <-> ( X .<_ Y /\ X =/= Y ) ) )
8 7 3expb 
 |-  ( ( K e. dom Base /\ ( X e. B /\ Y e. B ) ) -> ( X .< Y <-> ( X .<_ Y /\ X =/= Y ) ) )
9 6 8 mpancom 
 |-  ( ( X e. B /\ Y e. B ) -> ( X .< Y <-> ( X .<_ Y /\ X =/= Y ) ) )
10 9 biimpar 
 |-  ( ( ( X e. B /\ Y e. B ) /\ ( X .<_ Y /\ X =/= Y ) ) -> X .< Y )
11 10 expr 
 |-  ( ( ( X e. B /\ Y e. B ) /\ X .<_ Y ) -> ( X =/= Y -> X .< Y ) )
12 11 necon1bd 
 |-  ( ( ( X e. B /\ Y e. B ) /\ X .<_ Y ) -> ( -. X .< Y -> X = Y ) )
13 12 orrd 
 |-  ( ( ( X e. B /\ Y e. B ) /\ X .<_ Y ) -> ( X .< Y \/ X = Y ) )
14 13 ex 
 |-  ( ( X e. B /\ Y e. B ) -> ( X .<_ Y -> ( X .< Y \/ X = Y ) ) )