Distributive lattice and embedding Good evening everybody, 
I am working on the notion of lattice, which is still very difficult for me to grasp.
I have this question:  
Suppose that a lattice T can be embedded in a distributive lattice D. Show that T is also distributive. 
It is my understanding that an embedding conserves the operations, so if T has ∧,∨, so does D. Then I could verify the distributivity of T by seeing if this is true: 
x ∧ (y ∨ z) = (x ∧ y) ∨ (x ∧ z) 
But then, how am I suppose to do so without any value to verify if it works? Or maybe I'm having a completely wrong angle for this problem? (It's very possible, I'm new to this and completely outside of my field of study.)
Any help would be appreciated,
Thank you
 A: The main idea is: Distributivity can be (theoretically) destroyed by a single counter example. Embedding means injectivity: if $x\neq y$ then $e(x)\neq e(y)$. So if an equation has a counterexample in $T$ it also has a counterexample in $D$.
This can be used formally to describe a direct proof. To avoid unnecessary spoilers, suppose I want to prove that every groupoid (set with a binary operation) that embeds into a semigroup is already a semigroup. So we have two operations $*$ in $T$ and $\cdot$ in $D$. And we know that $D$ is associative. In a first step I write down the associative law in $D$: $$x_D\cdot(y_D \cdot z_D) = (x_D\cdot y_D)\cdot z_D.$$
(This line does not need to occur in the proof, it is only for the construction of the proof).
Now I replace every variable by the embedding of an element of $T$:
$$e(x)\cdot\bigl(e(y)\cdot e(z)\bigr)=\bigl(e(x)\cdot e(y)\bigr)\cdot e(z).$$
The embedding $e$ supports the equation $e(x*y) = e(x)\cdot e(y)$. We apply this to the inner parentheses (the outer operation does not support it, yet):
$$e(x)\cdot e\bigl(y *  z\bigr)=e(x)\cdot\bigl(e(y)\cdot e(z)\bigr)=\bigl(e(x)\cdot e(y)\bigr)\cdot e(z)=e\bigl(x * y\bigr)\cdot e(z)$$
Now, also the outer operations have the structure that allows us to apply the law again:
$$e\bigl(x * (y *  z)\bigr)=e(x)\cdot e\bigl(y *  z\bigr)=e\bigl(x * y\bigr)\cdot e(z)=e\bigl((x * y) *z\bigr)$$
Finally, we apply that $e$ is injective, which implies
$$x * (y *  z)=(x * y) *z$$
q.e.d.
As you can see it is a very formal process. In the same way you can prove your distributivity law. I chose a different example, here as you obviously need some exercise with this method.
