Proving Properties of Lattices: How to Use DeMorgan's Laws

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This discussion focuses on proving properties of bounded, complemented, distributive lattices using DeMorgan's Laws. The user successfully proved the first two properties: $x \wedge y = \bot \Leftrightarrow x \leq y^c$ and $x = (x^c)^c$. They seek assistance with the third property, $x \wedge y \leq z \Leftrightarrow y \leq x^c \vee z$, and provide a partial proof using distributive and complement laws. The discussion emphasizes the importance of understanding the reversibility of logical steps in lattice theory.

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Aryth1
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My problem is this:

Let $L$ be a bounded, complemented, distributive lattice and let $x,y,z\in L$. Prove the following:

1. $x\wedge y = \bot \Leftrightarrow x\leq y^c$
2. $x = (x^c)^c$
3. $x\wedge y \leq z \Leftrightarrow y\leq x^c \vee z$
4. $(x\vee y)^c = x^c \wedge y^c$
5. $(x\wedge y)^c = x^c \vee y^c$

The first two I have finished, and the last two are basically DeMorgan's laws. I'm having some trouble with #3. Any help is appreciated!
 
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Here is one direction of 3:

$y \leq (x^c \vee z) \iff y = y \wedge (x^c \vee z)$

Thus:

$x \wedge y = x \wedge [y \wedge (x^c \vee z)]$

$= x \wedge [(y \wedge x^c) \vee (y \wedge z)]$ (distributive law)

$= [x \wedge (y \wedge x^c) ] \vee [x \wedge (y \wedge z)]$ (distributive law, again)

$ = [x \wedge (x^c \wedge y)] \vee [x \wedge (y \wedge z)]$ (commutativity)

$ = [(x \wedge x^c) \wedge y] \vee [x \wedge (y \wedge z)]$ (associativity)

$ = [0 \wedge y] \vee [x \wedge (y \wedge z)]$ (complement law)

$ = 0 \vee [x \wedge (y \wedge z)]$ (zero law? forget what this is called)

$= x \wedge (y \wedge z)$ (identity law)

$= (x \wedge y) \wedge z$ (associativity)

which shows that $x \wedge y = (x \wedge y) \wedge z$, that is:

$x \wedge y \leq z$

Ask yourself, are these steps reversible?
 
Deveno said:
Here is one direction of 3:

$y \leq (x^c \vee z) \iff y = y \wedge (x^c \vee z)$

Thus:

$x \wedge y = x \wedge [y \wedge (x^c \vee z)]$

$= x \wedge [(y \wedge x^c) \vee (y \wedge z)]$ (distributive law)

$= [x \wedge (y \wedge x^c) ] \vee [x \wedge (y \wedge z)]$ (distributive law, again)

$ = [x \wedge (x^c \wedge y)] \vee [x \wedge (y \wedge z)]$ (commutativity)

$ = [(x \wedge x^c) \wedge y] \vee [x \wedge (y \wedge z)]$ (associativity)

$ = [0 \wedge y] \vee [x \wedge (y \wedge z)]$ (complement law)

$ = 0 \vee [x \wedge (y \wedge z)]$ (zero law? forget what this is called)

$= x \wedge (y \wedge z)$ (identity law)

$= (x \wedge y) \wedge z$ (associativity)

which shows that $x \wedge y = (x \wedge y) \wedge z$, that is:

$x \wedge y \leq z$

Ask yourself, are these steps reversible?

Thanks a lot for your help! I hadn't thought about your first $\iff$. That was what I was missing. I'm learning this on my own with monthly meetings with a professor, so sometimes I don't manage to see simple things. Thank you!
 

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