Metamath Proof Explorer


Theorem opcon1b

Description: Orthocomplement contraposition law. ( negcon1 analog.) (Contributed by NM, 24-Jan-2012)

Ref Expression
Hypotheses opoccl.b
|- B = ( Base ` K )
opoccl.o
|- ._|_ = ( oc ` K )
Assertion opcon1b
|- ( ( K e. OP /\ X e. B /\ Y e. B ) -> ( ( ._|_ ` X ) = Y <-> ( ._|_ ` Y ) = X ) )

Proof

Step Hyp Ref Expression
1 opoccl.b
 |-  B = ( Base ` K )
2 opoccl.o
 |-  ._|_ = ( oc ` K )
3 1 2 opcon2b
 |-  ( ( K e. OP /\ X e. B /\ Y e. B ) -> ( X = ( ._|_ ` Y ) <-> Y = ( ._|_ ` X ) ) )
4 eqcom
 |-  ( ( ._|_ ` Y ) = X <-> X = ( ._|_ ` Y ) )
5 eqcom
 |-  ( ( ._|_ ` X ) = Y <-> Y = ( ._|_ ` X ) )
6 3 4 5 3bitr4g
 |-  ( ( K e. OP /\ X e. B /\ Y e. B ) -> ( ( ._|_ ` Y ) = X <-> ( ._|_ ` X ) = Y ) )
7 6 bicomd
 |-  ( ( K e. OP /\ X e. B /\ Y e. B ) -> ( ( ._|_ ` X ) = Y <-> ( ._|_ ` Y ) = X ) )