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.
` and `– 2010-11-23` to "cross out" what you no longer want considered. This was very ill-done.