# In a lattice, $x \leq y$ and $x \leq z \iff x \leq y ∧ z$

Added later: Proof now in progress.

Let $(L, \leq)$ be a lattice, $x, y, z \in L$. Prove that $x ≤ y$ and $x ≤ z \iff x ≤ y ∧ z$.

I proceed to prove as follows:

The statement can be split into two implications :

(i) $x ≤ y$ and $x ≤ z \implies x ≤ y ∧ z$

(ii) $x ≤ y ∧ z \implies x ≤ y$ and $x ≤ z$

(i) Approach 1 : Proceeding by use of definitions:

$x ≤ y$ and $x ≤ z \implies x$ is a LB in $L$

Let $w = y ∧ z$

Since w is the GLB, by defn. $x ≤ w$, i.e. $x ≤ y ∧ z$

Approach 2 : Proof by contradiction:

Let $x ≤ y$ and $x ≤ z \implies x$ not $≤ y ∧ z$

$x ≤ y$ and $x ≤ z \implies x$ is a LB in $L$

Let $w = y ∧ z$

Since w is the GLB, by defn. $x ≤ w$, i.e. $x ≤ y ∧ z$.

As a contradiction has been reached, the original assertion is true.

(ii) Approach 1 : Proceeding by use of definitions:

$x ≤ y ∧ z \implies x$ is a LB in $L$

By definition of LB, $x ≤$ for all $w ∈ L$, i.e. $x ≤ y$ and $x ≤ z$

Approach 2 : Proof by contradiction:

Let $x ≤ y ∧ z \implies$ not $x ≤ y$ or not $x ≤ z$.

$x ≤ y ∧ z \implies x$ is a LB in $L$

By definition of LB, $x ≤$ for all $w ∈ L$, i.e. $x ≤ y$ and $x ≤ z$.

As a contradiction has been reached, the original assertion is true.

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What is your question? –  Rasmus Nov 21 '10 at 11:10
Your proofs seem basically fine. (One quibble: your writing "$x\leq$ for all $w\in L$" in (ii) doesn't really make sense, but the conclusion is sound.) –  JDH Nov 21 '10 at 11:24
@JDH : I am just beginning this. Thanks for catching the mistake, and I would like to know why the statement doesn't make sense. Is there a proper way/notation of expressing what I want to say through that line? –  user3740 Nov 21 '10 at 20:23
This is not a questio, and there is nothing we can do. Voting to close... If at some later time, user3740 has a question, then he can ask it. –  Mariano Suárez-Alvarez Nov 23 '10 at 20:10
@user3740: I've rolled back to what you had at first. By deleting all the material in the question, not only did you make the question nonsensical (and subject to downvotes), you also made all comments that had been posted ipso-facto wrong. Don't delete information from a question, just add more; at worse, you can use <strike> and <strike/> to "cross out" what you no longer want considered. This was very ill-done. –  Arturo Magidin Nov 23 '10 at 20:33
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## 1 Answer

With the identity, $x \leq y \Leftrightarrow x = x \wedge y$, the implication is straightforward to prove. That is,

\begin{align} x \leq y \text{ and } x \leq z & \Leftrightarrow x = x \wedge y \text{ and } x = x \wedge z \\ & \Leftrightarrow x \wedge x = (x\wedge y) \wedge (x\wedge z)\\ & \Leftrightarrow x = x \wedge (y \wedge z)\\ & \Leftrightarrow x \leq y \wedge z. \end{align}

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I am afraid that the second $<=>$ is not clear to me. $=>$ direction is obvious, but the other direction... –  hkju Apr 27 '12 at 10:26
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