Verification: Heyting Algebra : Show that $(a \to b)$ is the largest element of $H$ s.t. $a \wedge (a \to b) \le b$

Question

Heyting Algebra : Show that $ (a \to b)$ is the largest element of $H$ s.t. $a \wedge (a \to b) \le b$

Let $H$ be a Heyting algebra. there is an identity which states

$x \wedge (x \to y) = x \wedge y$ and lets say $a \to b = c$ for $a,b \in H$

so we have $a \wedge c = a \wedge (a \to b) = a \wedge b$ and so by definition we know $a \wedge b \le b$ thus $a \wedge c \le b$. Need to show that this $c = a \to b$ is the largest such element.

Suppose $d \in H$ is s.t. $a \wedge d \le b$ what makes $d \le c = a \to b$

I'm not sure if there is another identity I need to use or if it is right in front of me.

EDIT: there is another identity that states $b \wedge (a \to b) = b$

so let $c = (a \to b)$ and assume there exists $d \in H$ s.t. $d > c$ and $a\wedge d \le b$ then we get that $a \wedge c < a \wedge d$ and $b \wedge c < b \wedge d$

all put together:

$a \wedge b = a \wedge (a \to b) = a \wedge c \le b \wedge c = b \wedge (a \to b) = b < b \wedge d $ and thus $\rightarrow \leftarrow$

And so $c = a \to b$ is the largest such element

Answer 1

I think you are correct, but I find your exposition a bit confusing, so let me start it over...

You want to prove that $a \to b = \max \{ u \in H : a \wedge u \leq b \}$.
I don't know what are your axioms, so I'll start by saying the ones I'm using (among the ones of the equational base given in Burris & Sankappanavar).

1. $x \to x = 1$;
2. $x \wedge ( x \to y ) = x \wedge y$;
3. $(x \to y) \wedge y = y$;
4. $x \to (y \wedge z) = (x \to y) \wedge (x \to z)$;
5. (Consequence of 4) $\to$ is isotone in the second coordinate.

By (2), we get that $a \to b \in \{ u \in H : a \wedge u \leq b \}$, and you got this right.

Now take $c \in H$ such that $a \wedge c \leq b$. Then \begin{align} a \wedge b \leq c &\Rightarrow a \to (a \wedge c) \leq a \to b \tag{5}\\ &\Rightarrow (a \to a) \wedge (a \to c) \leq a \to b \tag{4}\\ &\Rightarrow a \to c \leq a \to b \tag{1} \end{align} By $(3)$, $(a \to c) \wedge c = c$, whence $c \leq a \to c \leq a \to b$.