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$L$ is a distributive lattice with top and bottom element($1$ and $0$ respectively). Show that if an element has a complement, the complement must be unique.

This is what I have so far, but am getting stuck on where to go, or if this is even the right route to take.

Let $a\in L$ and $b_1,b_2$ be two elements in $L$ s.t. $a\vee b_1=1, a\wedge b_1=0, a\vee b_2=1,a\wedge b_2=0$ Then by distribution $b_1\vee (a\wedge b_2)=b_1\vee 0=b_1$ and $(b_1\vee a)\wedge (b_1\vee b_2)=1\vee (b_1\vee b_2)=b_1\vee b_2$ So I have $b_1=b_1 \vee b_2$ but this does not help.

Any ideas would be great, it seems like I am close I just cant see the missing part.

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You have $b_1=b_1\vee b_2$, equivalently, $b_2\leq b_1,$ and similarly $b_2=b_1\vee b_2$ so that $b_1\leq b_2$.

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  • $\begingroup$ ok great, this is my first time with this stuff so its good to see the little tricks:) $\endgroup$ – tmpys Oct 13 '14 at 22:04
  • $\begingroup$ Glad that was helpful. $\endgroup$ – Kevin Arlin Oct 14 '14 at 3:05
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You're on the right track.

By symmetry (exchanging $b_1$ and $b_2$), we also have $b_2=b_1\lor b_2$.

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  • $\begingroup$ Great ty. I wish I could give two answers:0 $\endgroup$ – tmpys Oct 13 '14 at 22:04
  • $\begingroup$ Am I correct that you posted this first? $\endgroup$ – tmpys Oct 13 '14 at 22:05
  • $\begingroup$ ok I gave it to the other guy because you have 36k and he showed me that $b_1=b_1 \vee b_2 \implies b_2\le b_1$ I know it's obvious but I did not think about it. $\endgroup$ – tmpys Oct 13 '14 at 22:07

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