Define the relation $\leq$ on a boolean algebra $B$ by for all $x,y\in B$, $x\leq y \iff x\lor y=y$, show that $\leq$ is a partial order relation.
First of all what exactly does boolean algebra $B$ look like? can you give me an example of a set $A$ that is a boolean algebra???
I have done alot of examples to prove equivalence relation last week and the idea is straight forward and with this one I first tried to prove reflexivity antisymmetry and transitive
Reflexivity: for any element $x$ that is in $B$, $x\leq x$ that is $x\lor x=x$?? This part doesnt make sense to me nor do I know what does $x\lor x$ mean, $x$ or $x$ as a set operation 'OR' if so could you guys lead me in a bit
I think once I understand what exactly $x\lor y=y$ means and how I can manupulate it I think I will be ok to prove antisymmetry transitivity and reflexivity.
Before trying to read this answer, please have a look at some equivalent definitions of a Boolean algebra.
The canonical example of a Boolean algebra $(B,\lor,\land,\neg)$ is the power set of a set: $(\mathcal P(S), \cup,\cap,\complement)$ where $\complement$ denotes the complement operation $\complement(S') = \{s \in S: s \notin S'\}$.
Note that for $\le$ to be a partial order, it should be reflexive, transitive and antisymmetric: $x\le y$ and $y \le x$ imply $x = y$. A partial order is not the same as an equivalence relation (which replaces antisymmetry with symmetry).
A proof that $x \lor x = x$ is given here.
What may have been given as an axiom for BAs to you is that $\lor$ is associative. Then, if $x \le y$ and $y \le z$, we have: $$x \lor z = x \lor (y \lor z) = (x \lor y) \lor z = y \lor z = z$$ hence $x \le z$.
Finally, if $x \le y$ and $y \le z$, we have $y = x \lor y = y \lor x = x$.
Hence $\le$ is reflexive, transitive, and antisymmetric: a partial ordering.