Lattice theory: Difference is equal to pseudodifference

Question

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.

Answer 1

\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*}