@vv I put the corrected version up in a ME chat room, i will just put the latex here for you to copy and paste.

make not of the part below outline the part that still requires explicit proof.

We seek to prove:

$a | b {\rm \: \Rightarrow \:}\phi(a) | \phi(b)$ (1)

$\phi(a) | \phi(b) {\rm \: \not\Rightarrow \:}a | b$ (2)

for (1) we have:

$a | b{\rm \: \Rightarrow \:} {rad(a)} |\, {rad(b)}$

${rad(a)} |\, {rad(b)}{\rm \: \Rightarrow \:}\prod_{p|a} p \, | \prod_{p|b} p$

$\prod_{p|a} p \, | \prod_{p|b} p{\rm \: \Rightarrow \:}\prod_{p|a} \Bigl(1-\frac{1}{p}\Bigr) | \,b\cdot \prod_{p|b} \Bigl(1-\frac{1}{p}\Bigr)$

$\prod_{p|a} \Bigl(1-\frac{1}{p}\Bigr) | \,b\cdot \prod_{p|b} \Bigl(1-\frac{1}{p}\Bigr){\rm \: \Rightarrow \:}{\Biggl\{\frac{b \cdot \prod_{p|b} (1-\frac{1}{p})}{\prod_{p|a} (1-\frac{1}{p})}}\Biggr\}=0$

${\Biggl\{\frac{b \cdot\prod_{p|b} (1-\frac{1}{p})}{\prod_{p|a} (1-\frac{1}{p})}}\Biggr\}=0 \land {\Bigl\{\frac{b}{a}}\Bigr\}=0{\rm \: \Rightarrow \:}{\Biggl\{\frac{b\prod_{p|b} (1-\frac{1}{p})}{a\prod_{p|a} (1-\frac{1}{p})}}\Biggr\}=0$

${\Biggl\{\frac{b\prod_{p|b} (1-\frac{1}{p})}{a\prod_{p|a} (1-\frac{1}{p})}}\Biggr\}=0{\rm \: \Rightarrow \:}{\Bigl\{\frac{\phi(b)}{\phi(a)}}\Bigr\}=0$

${\Bigl\{\frac{\phi(b)}{\phi(a)}}\Bigr\}=0{\rm \: \Rightarrow \:}\phi(a) | \phi(b)$

Thus establishing $a | b {\rm \: \Rightarrow \:}\phi(a) | \phi(b)$ is true.

We now look to establishing that (2) is true:

$\phi(a) | \phi(b) {\rm \: \Rightarrow \:}{\Bigl\{\frac{\phi(b)}{\phi(a)}}\Bigr\}=0$

Consider $z_1=x_1 y_1 \land z_2=x_2 y_2$ and $x_1,x_2,y_1,y_2 \in \mathbb Q$

And suppose we can assert ${\Biggl\{\frac{z_1}{z_2}}\Biggr\}=0$ knowing $z_2$ to divide $z_1$.

We then consider the individual divisibility criteria between the four integer variables in such a ratio with no remainder, as to allow us to see that the reason for the converse of (1) having a false truth value is elementary in nature, by finally substituting these generic variables for the four arithmetic functions appropriate in the division of the totients of $a$ and $b$, allowing us to see that either $a | b$ or it's negation can be true when only knowing $\phi(a) | \phi(b)$.

$\frac{x_2}{x_1} | \frac{ y_2}{ y_1}{\rm \: \Rightarrow \:}{\Biggl\{\frac{x_2}{x_1}}\Biggr\}=0 \land {\Biggl\{\frac{ y_2}{ y_1}}\Biggr\}=0{\rm \: \Rightarrow \:}{\Biggl\{\frac{x_1 y_1}{x_2 y_2}}\Biggr\}=0$

$\frac{x_2}{y_1} | \frac{ y_2}{ x_1}{\rm \: \Rightarrow \:}{\Biggl\{\frac{x_2}{y_1}}\Biggr\}=0 \land {\Biggl\{\frac{ y_2}{ x_1}}\Biggr\}=0{\rm \: \Rightarrow \:}{\Biggl\{\frac{x_1 y_1}{x_2 y_2}}\Biggr\}=0$

$\frac{x_2 y_2}{y_1} | {x_1}{\rm \: \Rightarrow \:}{\Biggl\{\frac{x_2 y_2}{y_1}}\Biggr\}=0 \land {\Biggl\{{ x_1}}\Biggr\}=0{\rm \: \Rightarrow \:}{\Biggl\{\frac{x_1 y_1}{x_2 y_2}}\Biggr\}=0$

$\frac{x_2 y_2}{x_1} | {y_1}{\rm \: \Rightarrow \:}{\Biggl\{\frac{x_2 y_2}{x_1}}\Biggr\}=0 \land {\Biggl\{{ y_1}}\Biggr\}=0 {\rm \: \Rightarrow \:}{\Biggl\{\frac{x_1 y_1}{x_2 y_2}}\Biggr\}=0$

Therefore:

${\Biggl\{\frac{z_1}{z_2}}\Biggr\}=0{\rm \: \Rightarrow \:}\frac{x_2}{x_1} | \frac{ y_2}{ y_1} \lor \frac{x_2}{y_1} | \frac{ y_2}{ x_1} \lor \frac{x_2 y_2}{y_1} | {x_1} \lor \frac{x_2 y_2}{x_1} | {y_1}$

Thus:

${\Bigl\{\frac{z_1}{z_2}}\Bigr\}=0{\rm \: \Rightarrow \:}S$

$S:({\Bigl\{\frac{ y_2}{ y_1}}\Bigr\}=0 \land {\Bigl\{\frac{x_1}{x_2}}\Bigr\}=0)\lor({\Bigl\{\frac{x_2}{y_1}}\Bigr\}=0 \land {\Bigl\{\frac{ y_2}{ x_1}}\Bigr\}=0)\lor({\Bigl\{\frac{x_2 y_2}{y_1}}\Bigr\}=0)\lor({\Bigl\{\frac{x_2 y_2}{x_1}}\Bigr\}=0)$

Then finally we make the substitutions as follows:

$$x_1=b$$

$$x_2=a$$

$$y_1=\prod_{p|b} (1-\frac{1}{p})$$

$$y_2=\prod_{p|a} (1-\frac{1}{p})$$

$\phi(a) | \phi(b){\rm \: \Rightarrow \:}{\Bigl\{\frac{\phi(b)}{\phi(a)}}\Bigr\}=0{\rm \: \Rightarrow \:}S_{\phi(a),\phi(b)}$

$S_{\phi(a),\phi(b)}:S_{\phi(a),\phi(b),1}\lor S_{\phi(a),\phi(b),2}\lor S_{\phi(a),\phi(b),3} \lor S_{\phi(a),\phi(b),4}$

Where:

$S_{\phi(a),\phi(b),1}:{\Bigl\{\frac{ \prod_{p|a} (1-\frac{1}{p})}{\prod_{p|b} (1-\frac{1}{p})}}\Bigr\}=0 \land {\Bigl\{\frac{b}{a}}\Bigr\}=0$

$S_{\phi(a),\phi(b),2}:{\Bigl\{\frac{a}{\prod_{p|b} (1-\frac{1}{p})}}\Bigr\}=0 \land {\Bigl\{\frac{ \prod_{p|a} (1-\frac{1}{p})}{ b}}\Bigr\}=0$

$S_{\phi(a),\phi(b),3}:{\Bigl\{\frac{a \prod_{p|a} (1-\frac{1}{p})}{\prod_{p|b} (1-\frac{1}{p})}}\Bigr\}=0$

$S_{\phi(a),\phi(b),4}:{\Bigl\{\frac{a \prod_{p|a} (1-\frac{1}{p})}{b}}\Bigr\}=0$

And finally it is clear that:

$S_{\phi(a),\phi(b)}{\rm \: \Rightarrow \:} a | b \lor \lnot(a | b)$

Therefore:

$\phi(a) | \phi(b) {\rm \: \not\Rightarrow \:}a | b$

MISSING PART:

Show that:

$\Biggl(\prod_{p|a} \Bigl(1-\frac{1}{p}\Bigr) | \,b\cdot \prod_{p|b} \Bigl(1-\frac{1}{p}\Bigr) \Biggr)$

is true $\forall a,b \in \mathbb N$ such that $a |b$

Case 1:

$\lnot \Biggl(\prod_{p|a} \Bigl(1-\frac{1}{p}\Bigr) | \,b \Biggr)\land \lnot\Biggl(\prod_{p|a} \Bigl(1-\frac{1}{p}\Bigr) | \prod_{p|b} \Bigl(1-\frac{1}{p}\Bigr) \Biggr)$

Case 2:

$\Biggl(\prod_{p|a} \Bigl(1-\frac{1}{p}\Bigr) | \,b \Biggr)\land \lnot\Biggl(\prod_{p|a} \Bigl(1-\frac{1}{p}\Bigr) | \prod_{p|b} \Bigl(1-\frac{1}{p}\Bigr) \Biggr)$

Case 3:

$\lnot \Biggl(\prod_{p|a} \Bigl(1-\frac{1}{p}\Bigr) | \,b \Biggr)\land \Biggl(\prod_{p|a} \Bigl(1-\frac{1}{p}\Bigr) | \prod_{p|b} \Bigl(1-\frac{1}{p}\Bigr) \Biggr)$