这是一份manchester曼切斯特大学 MATH20101作业代写的成功案例
Every complete lattice $L$ in which the infinite distributive law
$$
x \wedge \bigvee_{i \in I} y_{i}=\bigvee_{i \in I}\left(x \wedge y_{i}\right)
$$
holds is pseudocomplemented. Clearly, for every $x \in L$ we have
$$
x^{\star}=\sup {y \in L \mid x \wedge y=0} .
$$
Let $B$ be a boolean algebra and define
$$
B^{[2]}={(a, b) \in B \times B \mid a \leqslant b} .
$$
Then if $a \leqslant b$ and $x \leqslant y$ we have
$$
(a, b) \wedge(x, y)=(0,0) \Longleftrightarrow a \wedge x=0=b \wedge y \Longleftrightarrow x \leqslant y \leqslant b^{\prime} .
$$
It follows that $(a, b)^{\star}$ exists and is $\left(b^{\prime}, b^{\prime}\right)$. Thus $B^{[2]}$ is pseudocomplemented.
MATH20201 COURSE NOTES ：
$\Leftarrow$ : Suppose conversely that $(1),(2),(3)$ hold. Then on the one hand we have
$$
x \wedge x^{\star}=x \wedge\left(x \wedge 0^{\star}\right)^{\star}=x \wedge 0^{\star \star}=x \wedge 0=0 ;
$$
and on the other, if $x \wedge y=0$ then
$$
y \wedge x^{\star}=y \wedge(y \wedge x)^{\star}=y \wedge 0^{\star}=y
$$
and so $y \leqslant x^{\star}$. Thus $x \mapsto x^{\star}$ is the pseudocomplementation.