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$(X, \leq)$ is a lattice (order).

For all $ a,b,c \in X$, can you prove $$a \leq c \Leftrightarrow [a\lor(b\land c)] \leq [(a\lor b)\land c]?$$

As far as I have managed to do this:

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Let: $[a\lor (b\land c)]\leq [(a\lor b)\land c]$

$a\leq [a\lor(b\land c)]\leq [(a\lor b)\land c] \leq c$

So: $a\leq c$

something that I found on my notes: for $(a,b,c)\in X$

$a\leq a\lor b$, and $b\leq a\lor b$

$a\leq b, c\leq b \Rightarrow a\lor c\leq b$

$(a\land b)\leq a,a\land b\leq b$

$(a\leq b, a\leq c)\Rightarrow a\leq b\land c$ by definition.

De Costa
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    What have you tried. If you show where you are having trouble, people will be in a better position to help you. We are not here to do the problem for you. Also, are there any examples in your notes or your textbook that resemble this problem? – MasB May 28 '20 at 12:59
  • I agree with @BernardMassé. In particular, which direction of the $\iff$ is giving you trouble? One direction looks (to me) easier than the other. – Andreas Blass May 28 '20 at 13:54
  • @AndreasBlass I don't even know how to start, I haven't seen such examples before. It is the first time. So, I'm having trouble both. – De Costa May 28 '20 at 14:07
  • I'd recommend starting with the definitions of $\lor$ and $\land$. They should give you information like (in the case of $\lor$) $x\leq x\lor y$ and $$ x\lor y\leq z\iff(x\leq z\text{ and }y\leq z)$$. Try to find some values for $x,y,z$ that will make this information (and the analogous information about $\land$) relevant to your problem. – Andreas Blass May 28 '20 at 14:15
  • I just added one side, I guess. I don't know if it is correct. What about the other? – De Costa May 28 '20 at 14:22
  • By the way, a lattice for which the inequality on the right-hand side is always an equality whenever the inequality on the left-hand side is true is known as a modular lattice. So, have you seen modular lattices before, BugMeNot? – Geoffrey Trang May 28 '20 at 14:44
  • We have seen "lattices" only... I don't think we have seen it. There is a note that I added to my question, which I have found. – De Costa May 28 '20 at 14:54

1 Answers1

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Suppose that $a \le c$. Then, to show that $a \lor (b \land c) \le (a \lor b) \land c$, one must show four things:

  1. $a \le a \lor b$
  2. $a \le c$
  3. $b \land c \le a \lor b$
  4. $b \land c \le c$

The first inequality holds by the definition of $\lor$.

The second inequality holds by assumption.

The third inequality holds because $b \land c \le b \le a \lor b$ and $\le$ is transitive.

Finally, the third inequality holds by the definition of $\land$.

Since all four inequalities hold, one can conclude that $a \lor (b \land c) \le (a \lor b) \land c$.

Geoffrey Trang
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