Proving a property of relatively pseudocomplemented lattices

Question

Goldblatt's "Topoi" suggests proving $(a \Rightarrow b) \sqcap (b \Rightarrow c) \leq (a \sqcup b) \Rightarrow c$, where $a \Rightarrow b$ is the pseudocomplement of $a$ relative to $b$ (exercise 8.3.16). I see why this holds in a distributive lattice, but how to prove this for a general lattice?

Answer 1

Let $L$ be a lattice in which the expression $$x \to y = \max \{ z \in L : x \wedge z \leq y \}$$ defines an operation, that is, for every $a,b \in L$, the set $\{ c \in L : a \wedge c \leq b\}$ has a maximum element.

Suppose that $L$ has a copy of either $N_5$ or $M_3$ as sub-lattices.
You can see versions of these lattices here.
With the same notation, let us try to compute $x \to z$, in $N_5$. We have $$x \to z = \max\{ u \in N_5 : x \wedge u \leq z \}.$$ Now, $$\{ u \in N_5 : x \wedge u \leq z \} = \{ 0, y, z \},$$ and this set has no maximum element.

Also with the same notation, in $M_3$, we see that $x \to z$ is the maximum of the set $\{ 0, y, z \}$ which also has no maximum (with the order of $M_3$). This means that these lattices are not pseudo-complemented.

Now, in the same section linked above, a characterization of distributive lattices is given has those lattices $L$ such that neither $M_3$ nor $N_5$ are isomorphic to sub-lattices of $L$.

So if $L$ is not distributive, it has either $M_3$ or $N_5$ as a sub-lattice (or an isomorphic copy).
Then, as we saw above, there is no pseudo-complement operation definable in the interval given by $[u,v]$, where $u$ is the minimum of that sub-lattice and $v$ the maximum.
But then $L$ is not relatively pseudo-complemented.

Hence, if $L$ is relatively pseudo-complemented, then (by definition) every interval in $L$ is pseudo-complemented, and so neither $N_5$ nor $M_3$ are sub-lattices of that interval, and therefore, not sub-lattices of $L$, and so $L$ is distributive.

Now, as you claim that you can prove the desired equality for distributive lattices, this is enough.