$\,m = {\rm lcm}(a,b)\iff a,b\mid m\ \, \& \ \gcd(m/a,m/b)=1$ For $a \in\Bbb  N$, $b\in\Bbb  N$, $μ \in\Bbb N^*$,
we have $μ = \operatorname{lcm}(a,b) \iff  μ = αa\text{ and }μ= βb$ and $\gcd(α,β)$ is $1$
Till now I succeeded to prove the left $\Rightarrow$ right implication, now I need to prove the reciprocal ($\Leftarrow$ way) 
Here is my work:

Can someone help me, or give me a hint? Rules : you can't use the expression ($a \wedge b)(a\vee b)=ab$, $(ka)\vee(vkb) = k(a\vee b)$ 
+ You can also help me improve my redaction if possible..
 A: This answer proves the theorem below using  basic divisibility laws (by exploiting innate $\rm\color{#c00}{duality}$).
Theorem $\  $ If $\,\ a,b\mid m\,\ $ then  $\ \displaystyle \frac{m}{{\rm lcm}(a,b)} \,=\,\gcd\left(\frac{m}a,\frac{m}b\right),\ $ i.e. $\ \color{#c00}{{\rm lcm}(a,b)' = \gcd(a',b')}$
$\begin{align}{\rm We\ seek}\ \   m = {\rm lcm}(a,b) &\iff a,b\mid m\ \ \&\  \gcd\left(\dfrac{m}a,\dfrac{m}b\right) = 1,\ \ \text{which by the above is}\\[.3em]
 &\iff a,b\mid m\ \ \&\ \ \dfrac{m}{{\rm lcm}(a,b)}\, =\, 1,\quad\ \ \ \text{which is clearly true.}
\end{align}$
Remark $ $ The linked proof using cofactor duality reveals your equivalence boils down to
${\rm lcm}(a,b) = m \iff {\rm lcm}(a,b)' = m',\ $ via $\ {\rm lcm}(a,b)' = \gcd(a',b') = \gcd(m/a,m/b)$

Alternatively $ $ we show: $ $ a common multiple $m$ of $\,a,b\,$ is least $\iff \gcd(m/a,m/b)=1$
Proof $\ $ If the gcd $= c> 1\,$ then $\,c\mid m/a,m/b\,$ so $\,a,b\mid m/c,\,$ therefore  $m$ is not least. $ $ Conversely, if $m$ is not least then $\,m = c\ell\,$ for $\,\ell\,$ least, so the gcd $= (c\ell/a,c\ell/b) = c(\ell/a,\ell/b) > 1\,$ by $\,c>1$.
