Suppose that (X,≤) is a lattice and x,y ∈ X. Prove that: x ∧ ( x ∨ y ) = x
I'm having trouble understanding how these types of lattice proofs should be structured. I've looked at similar ones and found this approach:
Let a = x ∧ ( x ∨ y ) Then a ≤ x and a ≤ ( x ∨ y )
But, I'm not sure what "a" is exactly? I have a whole set of these types of proofs to do but I don't have an example to go off of, so if someone could walk me through this one, it would be super helpful.
On
Your first strategy could be to unwrap definitions, and for this problem no more is needed.
Join: $a\vee b$ is an element satisfying $a\vee b\leq a$, $a\vee b\leq b$, and if $z$ satisfies $z\leq a$ and $z\leq b$, then $z\leq a\vee b$
$\phantom{}$
Meet: $a\wedge b$ is an element satisfying $a\wedge b\geq a$, $a\wedge b\geq b$, and if $z$ satisfies $z\geq a$ and $z\geq b$, then $z\geq a\wedge b$
$\phantom{}$
$x\vee y$ is an element such that $x\vee y\leq x$, $x\vee y\leq y$ and if $z\leq x$ and $z\leq y$, then $z\leq x\vee y$.
Now, just verity that $x$ satisfies the definition of $x\wedge (x\vee y)$.
You have that $x\geq x$, $x\geq (x\vee y)$, by definition of $x\vee y$ above,
If $z\geq x$ and $z\geq (x\vee y)$, then in particular $z\geq x$.
$a$ is just an abbreviation for $x\land(x\lor y)$ that they introduced so they don't have to keep writing that long expression.
The fact that $a\le x$ and $a\le (x\lor y)$ follows from the defintion of $\land$ as meaning the greatest lower bound (if it is a lower bound, it is less than both of them).
That $a\ge x$ follows from the fact that $x$ is less than or equal to both $x$ and $x\lor y,$ and thus it is a lower bound for them. And thus $a\ge x$ since $a$ is the greatest lower bound.