Lattice theory: Difference is equal to pseudodifference Consider a distributive lattice $(\mathfrak{A},\cup,\cap$) with least element $\bot$.
Difference $a\setminus b$ is defined by the formulas $(a\setminus b)\cup b=a\cup b$ and $(a\setminus b)\cap b=\bot$.
Pseudodifference $a\setminus^* b$ is defined by the formula $a\setminus^* b = \min \{ z\in\mathfrak{A} \mid a\leq b\cup z \}$.
Prove that if difference exists then pseudodifference also exists and is equal to difference ($a\setminus^* b = a\setminus b$).
I tried to write a formula for an element $z\in \{ z\in\mathfrak{A} \mid a\leq b\cup z \}$ and tried prove by contrary of $a\setminus b$ being the minimum.
This is not a homework.
 A: \begin{align*}
&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\text{Suppose $a\setminus b$ exists.}\\[6pt]
&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\text{Let $x = a\setminus b$, and let $S = \{y \mid a \le b \cup y\}$.}\\[6pt]
&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\text{We need to show}\\[6pt]
&(1)\;\;x \in S\\[4pt]
&(2)\;\;y \in S \implies x \le y\\[6pt]
&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\text{The proof of $(1)$ is immediate:}\\[5pt]
&a \le a \cup b\\[4pt]
\implies\;&a \le x \cup b&&\text{[since $x \cup b = a \cup b$]}\\[4pt]
\implies\;&a \le b \cup x\\[4pt]
\implies\;&x \in S&&\text{[by definition of $S$]}\\[6pt]
&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\text{Next the proof of $(2)$:}\\[6pt]
&y \in S\\[4pt]
\implies\;&a \le b\cup y&&\text{[by definition of $S$]}\\[4pt]
\implies\;&a \cup b \le b\cup y\\[4pt]
\implies\;&x \cup b \le  b\cup y&&\text{[since $x \cup b = a \cup b$]}\\[4pt]
\implies\;&x\cap (x \cup b) \le x  \cap (b\cup  y)\\[4pt]
\implies\;&(x \cap x) \cup (x \cap b) \le  (x \cap b) \cup (x \cap y)
&&\text{[by the distributive law]}\\[4pt]
\implies\;&x \cup \bot \le  \bot \cup (x \cap y)
&&\text{[since $x\cap b = \bot$]}\\[4pt]
\implies\;&x \le  x \cap y\\[4pt]
\implies\;&x \le y\\[6pt]
&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\text{as required.}\\[6pt]
&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\text{This completes the proof.}\\[6pt]
\end{align*}
