The homomorphism theorem states that every boolean ideal $I$ of a boolean algebra $A$ is the kernel of a boolean isomorphism. I'm reading a paper where the author presents a short proof of this theorem, saying that the following definition of equivalence relation $a\equiv b \iff \exists i,j\in I: a\lor i = b \lor j$ will do the work.
I'm having trouble to prove that the complement in the quotient $A/\equiv$ (defined as $\neg[a]=[\neg a]$) does not depend on the choice of $a$. That is, if $a \lor i = b \lor j$ for some $i,j \in I$, then there will be $i',j' \in I$ such that $\neg a \lor i' = \neg b \lor j'$.
So my question is: is this equivalence really fit for the job? If so, a hint on how to prove the above will be appreciated.
Let $i'=j'=k=i\vee j\in I$.
Then $a\vee k=b\vee k$.
Proof: $a\vee k=a\vee(i\vee j)=(a\vee i)\vee j=(b\vee j)\vee j=b\vee(j\vee j)=b\vee j$
$=a\vee i=a\vee(i\vee i)=(a\vee i)\vee i=(b\vee j)\vee i=b\vee(i\vee j)=b\vee k.$
Also $\neg a\vee k=k\vee\neg(a\vee k).$
Proof: $\neg a\vee k=(\neg a\vee k)\wedge(k\vee\neg k)=[(\neg a\vee k)\wedge k]\vee[(\neg a\vee k)\wedge\neg k]=k\vee(\neg a\wedge\neg k)$
$=k\vee\neg(a\vee k).$
Likewise $\neg b\vee k=k\vee\neg(b\vee k).$
Proof: same as above, with $a$ replaced by $b.$
Since $a\vee k=b\vee k$, it follows that $\neg a\vee k=k\vee\neg(a\vee k)=k\vee\neg(b\vee k)=\neg b\vee k,$
i.e., $a\equiv b$.