A property of Heyting implication

Question

Let $H$ be a Heyting algebra with $\Rightarrow$ being its Heyting implication and $0$ its bottom element. Define $x^\ast:= x\Rightarrow 0$. Since: \[ x\cdot (x^\ast+y)=(x\cdot x^\ast)+(x\cdot y)=x\cdot y\leqslant y \] it must be the case that for any $x,y\in H$: $x^\ast+y\leqslant x\Rightarrow y$.

Is the converse inclusion generally true? And if not could you please give me an example of a Heyting algebra in which it fails?

Answer 1

Take from this section from wikipedia article on Heyting algebras the second and the third items:

• Every totally ordered set that is a bounded lattice is also a Heyting algebra, where $p\Rightarrow q$ is equal to $q$ when $p>q$, and $1$ otherwise.
• The simplest Heyting algebra that is not already a Boolean algebra is the totally ordered set $\{0, \frac{1}{2}, 1\}$ with $\Rightarrow$ defined as above (...)

From this you can see that for such a smallest Heyting algebra $\frac{1}{2}^* \lor \frac{1}{2} = \frac{1}{2}$, but $\frac{1}{2} \Rightarrow \frac{1}{2} = 1$.