Answer
$\forall x \text{ }\forall y\text{ } \forall z\text{ }[(x+y)\cdot z=x\cdot z+y\cdot z]$
$\forall x \text{ }\forall y\text{ } \forall z\text{ }[x\cdot (y+z)=x\cdot y+x\cdot z]$
Work Step by Step
Our goal is to express the distributive laws of multiplication over addition using quantifiers,
Let's consider the real numbers $x$, $y$, $z$. The two distributive laws of multiplication over addition are:
$$\begin{align*}
(x+y)\cdot z&=x\cdot z+y\cdot z\\
x\cdot (y+z)&=x\cdot y+x\cdot z.
\end{align*}$$
Since these statements are true for any real numbers $x$, $y$, $z$, we will use the universal quantifier $\forall$ in front of each variable:
$$\begin{align*}
\forall x \text{ }\forall y\text{ } \forall z\text{ }[(x+y)\cdot z&=x\cdot z+y\cdot z]\\
\forall x \text{ }\forall y\text{ } \forall z\text{ }[x\cdot (y+z)&=x\cdot y+x\cdot z].
\end{align*}$$
