I encountered the following problem in Burris' A Course in Universal Algebra:

If $\langle H,\vee,\wedge,\rightarrow,0,1\rangle$ is a Heyting algebra and $a,b\in H$ show that $a\rightarrow b$ is the largest element $c$ of $H$ such that $a\wedge c\leq b$.

The definition given for a Heyting algebra is a distributive bounded lattice with a binary operation $\rightarrow$ where the following hold:

  1. $x\rightarrow x=1$,
  2. $(x\rightarrow y)\wedge y=y$,
  3. $x\wedge (x\to y)=x\wedge y$,
  4. $x\to (y\wedge z)=(x\to y)\wedge(x\to z)$,
  5. $(x\vee y)\to z=(x\to z)\wedge(y\to z)$.

It is clear that $(a\to b)\wedge a=a\wedge b\leq b$, but if $x\wedge a\leq b$ I am struggling to show that $x\leq a\to b$. I want to manipulate $x\wedge (a\to b)$ to $x$, but I cannot seem to find any way to do this.

For example I tried $$x\wedge (a\to b)=x\wedge(a\to((x\wedge a)\vee b))=x\wedge (a\to((x\vee b)\wedge(a\vee b)))\\=x\wedge((a\to(x\vee b))\wedge(a\to(a\vee b))),$$ and I get stuck. I figure $a\to(a\vee b)$ ought to be $1$ (which I cannot show), which would give me $x\wedge(a\to (x\vee b))$, which I don't know what to do with. Any help would be greatly appreciated.


From (4) you get immediately that $$y \leq z \Rightarrow x\to y \leq x\to z,$$ whence \begin{equation} x\wedge a \leq b \Rightarrow a\to (x\wedge a) \leq a\to b. \tag{*} \end{equation} Now, \begin{align} x &\leq a \to x \tag{by (2)}\\ &= (a\to x) \wedge (a\to a) \tag{by (1)}\\ &= a \to (x \wedge a) \tag{by (4)}\\ &\leq a \to b. \tag{by (*)} \end{align} I think that the first observation (that $\to$ is order preserving in the second coordinate) is the main trick here.


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