Exponential Object in Proposition Logic, Adjunction Between two logic operator

In discrete mathematics courses, we need to deal with the truth table of propositions, but it is quite annoying.

The truth table is just like the graph of a function. So I use the method of Boolean Ring to solve it. That is much more clear.

You can search Boolean Ring on my Blog and see sth.

In term 1, I trans $p\vee q$$p\vee q$ to $p+q+pq$$p+q+pq$ and $p\wedge q$$p\wedge q$ to $pq$$pq$. $\mathrm{\neg }p$$\neg p$ to $p+1$$p+1$... So $p\to q=\mathrm{\neg }p\vee q=(p+1)+q+(p+1)q=p+1+pq$$p\rightarrow q=\neg p\vee q=(p+1)+q+(p+1)q=p+1+pq$.

But as you see, $p\to q=p+1+pq$$p\rightarrow q=p+1+pq$ is not good enough, when you only need to check $p\to q$$p\rightarrow q$.

Actually, the most natural way to deal with $p\to q$$p\rightarrow q$ is consider $q^p$$q^p$. It is not hard to see that $p+1+pq$$p+1+pq$ is identically equal to the $q^p$$q^p$.

But what is the deep reason? Why do I say it is natural?

It comes from exponential object in Category Theory.

Let ${\mathrm{Prop}}_{\mathcal{L}}$$\mathrm{Prop}_{\mathcal L}$ be the Category of first-order propositional over the formal language $\mathcal{L}$$\mathcal L$ . The object is the proposition, and the morphism is $\models$$\models$.

Remember the $\wedge$$\wedge$ is the product, thus we have

$$\begin{array}{}\text{(1)}& (B\to C)\wedge B\models C\end{array}$$

is the final object.

In other word, if $A\wedge B\models C$$A\wedge B\models C$, $A$$A$ should be unique through $B\to C$$B\rightarrow C$ .

Thus $B\to C$$B\to C$ is the exponential object $C^B$$C^B$. That is the deep reason why we have $q^p$$q^p$!

In general, like in intuitionistic logic, we can not define $p\to q$$p\to q$ as $\mathrm{\neg }p\vee q$$\neg p\vee q$. The exponential object is a much more general and natural definition.

From the universal property of the exponential object, we can induce that:

$$\begin{array}{}\text{(2)}& A\wedge B\models C⟺A\models (B\to C)\end{array}$$

Using the notation in Category Theory, we get

$$\begin{array}{}\text{(3)}& {\mathrm{Hom}}_{\mathcal{L}}(B\wedge (-),(-))\cong {\mathrm{Hom}}_{\mathcal{L}}(-,B\to (-))\end{array}$$

In other words, $B\wedge (-)$$B\wedge(-)$ is the left adjoint of $B\to (-)$$B\to(-)$.

In general, we will have $(-)\times B$$(-)\times B$ is the left adjoint of $(-{)}^B$$(-)^B$ in Cartesian Closed Category.

In other world ${\mathrm{Hom}}_{\mathcal{C}}(A\times B,C)\cong {\mathrm{Hom}}_{\mathcal{C}}(A,C^B)$$\mathrm{Hom}_{\mathcal C}(A\times B,C)\cong \mathrm{Hom}_{\mathcal C}( A,C^B)$.

Thus we might have an interesting explanation for conditional probability.

$$\begin{array}{}\text{(4)}& \mu {|}_B=\frac{\mu (B\cap (-))}{\mu (B)}\end{array}$$

Since $\mu {|}_B$$\mu|_B$ is talking about something like $B\to (-)$$B\to(-)$, but the formula has $B\cap (-)$$B\cap(-)$.

I do not know. I will come back to it after I have some idea about it.