I am reading "The basic concepts of Algebraic Logic" by P. Halmos. He defines a filter over a Boolean Algebra $A$ as a non empty set $P$ such that:
-For all $x,y \in P$, $x \land y \in P$
-For all $x \in P$ , $t \in A$, $x \lor t \in P$.
Then the following remark is made: "A necessary and suficcient condition for $P$ to be a filter is that $1 \in P$ and that if $x \in P$ and $x \rightarrow y \in P$, then $y \in P$ (for any $y \in A$)."
I have managed to prove some things : for one direction, we assume the given definition of filter, and easily we see that if $x \in P$, then $1 = x \lor \neg x \in P$. Then if $x , x \rightarrow y \in P$, we can join them and get $x \land (\neg x \lor y ) \in P$. After reducing, $x \land y \in P$. Then $y = (y \land x) \lor (y \land \neg x) \in P$.
For the other direction, since $1 = (x \rightarrow x \lor y)$ for any $y$, the second condition is satisfied. I am unable to conclude that $P$ is closed under $\land$. Any hint or comment is appreciated.
Assume $x,y\in P$ and consider $$1=x\to(y\to (x\land y))$$ Since $1\in P$ and $x\in P$, we have $y\to (x\land y)\,\in P$, and thus also $x\land y\in P$.