这是一份durham杜伦大学MATH2581-WE01作业代写的成功案例
\begin{aligned}
\varphi_{L}(a) \vee \varphi_{L}(b)=\left(a^{\star \uparrow} \vee b^{\star \dagger}\right) \cap D(L) &=\left(a^{\star} \wedge b^{\star}\right)^{\dagger} \cap D(L) \
&=(a \vee b)^{\star \uparrow} \cap D(L) \
&=\varphi_{L}(a \vee b)
\end{aligned}
$$
\begin{aligned}
\varphi_{L}(a) \cap \varphi_{L}(b)=a^{\star \uparrow} \cap b^{\star \uparrow} \cap D(L) &=\left(a^{\star} \vee b^{\star}\right)^{\dagger} \cap D(L) \
&=(a \wedge b)^{\star \uparrow} \cap D(L) \
&=\varphi_{L}(a \wedge b) .
\end{aligned}
$$
Since also $\varphi_{L}(0)={1}$ and $\varphi_{L}(1)=D(L)$ it follows that $\varphi_{L}$ is a $(0,1)$-lattice morphism.
Now, by the distributivity, we have $x=x^{\star \star} \wedge\left(x \vee x^{\star}\right)$. It follows that
$$
(\forall x \in L) \quad x^{\dagger}=x^{\star \star \uparrow} \vee\left(x^{\dagger} \cap D(L)\right) \text {. }
$$
In fact, if $t \in x^{\dagger}$ then $t=t \vee x=\left(t \vee x^{\star \star}\right) \wedge\left(t \vee x \vee x^{\star}\right)$ where $x \vee x^{\star} \in D(L)$; and conversely if $t \in x^{\star \star \uparrow} \vee\left(x^{\dagger} \cap D(L)\right)$ then $t \geqslant y \wedge z$ where $y \geqslant x^{\star \star} \geqslant x$ and $z \geqslant x$ and thus $t \geqslant x$.
MATH2581-WE01 COURSE NOTES ：
Proof $\Rightarrow$ : If $L$ is a Heyting algebra follow from the above whereas (3) is immediate from the fact that $x:(x \wedge y)=1$.
$\Leftarrow$ : Conversely, if the identites hold then $x \wedge(y: x)=x \wedge y \leqslant y$; and if $x \wedge z \leqslant y$ then
$$
z \wedge(y: x)=z \wedge[(z \wedge y):(z \wedge x)]=z \wedge[(y \wedge z):(x \wedge y \wedge z)]=z
$$
and so $z \leqslant y: x$. Thus $(L ;:)$ is a Heyting algebra.