Distributive property is as follows:
$a \wedge (b \vee c) = (a \wedge b) \vee (a \wedge c)$
For "divides by" lattice the meet $(\wedge)$ operation is $\gcd$ and the join $(\vee)$ operation is $lcm.$ So, we have to prove:$$ \gcd (a, lcm(b, c) )= lcm (\gcd(a,b), \gcd(a,c))$$ Forward Direction Proof:
Let $p$ be any prime factor of $ \gcd (a, lcm(b, c) )$ and $\alpha$ be the largest integer such that $p^\alpha$ divides $ \gcd (a, lcm(b, c) )$
$\implies p^\alpha$ divides $a$ and $p^\alpha$ divides $lcm(b, c).$
$\implies p^\alpha$ divides $a$ and $(p^\alpha$ divides $b$ OR $p^\alpha$ divides $c)$
$\implies (p^\alpha$ divides $a$ and $p^\alpha$ divides $b)$ OR $(p^\alpha$ divides $a$ AND $p^\alpha$ divides $c)$
$\implies (p^\alpha$ divides $\gcd(a,b))$ OR $(p^\alpha$ divides $\gcd(a,c))$
$\implies p^\alpha $ divides $ lcm (\gcd(a,b), \gcd(a,c))$
Reverse Direction Proof:
Let $p$ be any prime factor of $ lcm (\gcd(a,b), \gcd(a,c))$ and $\alpha$ be the largest integer such that $p^\alpha$ divides $\gcd(a,b)$ or $\gcd(a,c).$ So, $p^{\alpha}$ divides $a$ and either $b$ or $c$
$\implies p^\alpha$ divides both $a$ and $lcm(b,c).$
$\implies p^\alpha$ divides $\gcd(a, lcm(b,c))$
Hence, $ \gcd (a, lcm(b, c) )= lcm (\gcd(a,b), \gcd(a,c))$