rewrite my proof: with this feedback: The justification "Applying the distributive property of multiplication over addition in mod arithmetic" is unclear and appears to incorrectly assume was is trying to be proved. \\ Using substitution, we are proving \\ $[x]_{68}*([w]_{68}+[t]_{68})$ 2. Now, before we begin, defining the two operations at hand is crucial:\\ Addition: For the two elements in our set $[w]_{68}$ and $[t]_{68}$, when added together, $[w]_{68}+[t]_{68}$ is stated to be $[w+t]_{68}$\\ Multiplication: For two elements in our set $[x]_{68}$ and an equivalence class $[d]_{68}$, when multiplied together, $[x]_{68}*[d]_{68}$ is stated to be $[x*d]_{68}$.\\ \\ Let's look again at our equation, but in our set $Z_{68}$:\\ $[x]_{68}*([w]_{68} + [t]_{68}) = [x]_{68} * [w]_{68} + [x]_{68} * [t]_{68}$\\ \\ 3. Let's begin proving this by looking at the left side of the equation:\\ $[x]_{68}*([w]_{68} + [t]_{68})$ with the definition of addition, results in $[x]_{68}*[w+t]_{68}$. The definition of multiplication now being used, it results in \\ $[x*(w+t)_{68}]$\\ 4. Relying on the distributive property of multiplication over addition in $Z$ is crucial. It states $a*(b+c)=a*b+a*c$ for any integers $a$,$b$