Categorical Propositional Logic, Algebraic Geometry and Fields extension.

The blog is a review of propositional logic.

Let us consider the category of proposition, $\mathcal{P}$$\mathcal P$.

The objects in this category are propositions, and morphism is $⟹$$\implies$.

The initial object is $F$$F$ and the final object is $T$$T$.

Recall that the initial object chases the colimit of $\mathrm{\varnothing }$$\empty$, and the final object chases the limit of $\mathrm{\varnothing }$$\empty$.

The product of $p,q$$p,q$ is $p\wedge q$$p\wedge q$, the coproduct is $p\vee q$$p\vee q$.

$\mathcal{P}$$\mathcal P$ is an internal category. Since $p⟹q$$p\implies q$ is a proposition as well.

$(\mathcal{P},\wedge ,T)$$(\mathcal P,\wedge,T)$ or $(\mathcal{P},\vee ,F)$$(\mathcal P,\vee,F)$ form a tensor category.

For $(\mathcal{P},\wedge ,T)$$(\mathcal P,\wedge,T)$, we have tensor-com adjoint.

$$\begin{array}{}\text{(1)}& {\mathrm{Hom}}_{\mathcal{P}}(p\otimes q,r)\cong {\mathrm{Hom}}_{\mathcal{P}}(p,{\mathrm{Hom}}_{\mathcal{P}}(q,r))\end{array}$$

That is

$$\begin{array}{}\text{(2)}& (p\wedge q)⟹r⟺p⟹(q⟹r)\end{array}$$

By duality, we have

$$\begin{array}{}\text{(3)}& (p\vee q)⟸r⟺p⟸(q⟸r)\end{array}$$

Let us consider the Boolean Ring now.

The polynomial ring

$$\begin{array}{}\text{(4)}& {\mathbb{F}}_2[X_1,...,X_n]\end{array}$$

could give us all the propositions generated from $Q$$Q$.

Since

$$\begin{array}{}\text{(5)}& p\vee q=p+q+pq\end{array}$$

$$\begin{array}{}\text{(6)}& p\wedge q=pq\end{array}$$

$$\begin{array}{}\text{(7)}& \mathrm{\neg }p=p+1\end{array}$$

$$\begin{array}{}\text{(8)}& p⟹q=\mathrm{\neg }p\vee q=(p+1)+q+(p+1)q=p+q+pq+p+1=q+pq+1\end{array}$$

Hence we could view those forms of propositions $P$$P$ as polynomials in ${\mathbb{F}}_2[X_1,...,X_n]$$\mathbb F_2[X_1,...,X_n]$.

The evaluation map $X_i\mapsto q_i$$X_i\mapsto q_i$ gives you some propositions. For example, let $P(X,Y)=XY,$$P(X,Y)=XY,$ evaluate it at $(p,q)$$(p,q)$ gives you $p\wedge q$$p\wedge q$.

Here the propositions are either true or false.

The solutions of $P(X_1,...,X_n)=1\subseteq A_{{\mathbb{F}}_2}^n$$P(X_1,...,X_n)=1\subseteq A_{\mathbb F_2}^n$ is an algebraic variety $V(P)$$V(P)$, tells you where the $P$$P$ is true.

We could define the dual variety $V^{\ast }(P)$$V^*(P)$ be the solutions of $P(X_1,...,X_n)=0$$P(X_1,...,X_n)=0$

Remark. $V(P)$$V(P)$ is the equalizer of $P$$P$ and $1$$1$, $V^{\ast }(P)$$V^*(P)$ is the equalizer of $P$$P$ and $0$$0$.

An interesting observation here is ${\mathbb{F}}_2$$\mathbb F_2$ is not algebraic closed, (it is the initial object of the category of the fields characteristic 2)

Hence we have some irreducible polynomials.

For example.

$$\begin{array}{}\text{(9)}& P=X^2+X=1\in {\mathbb{F}}_2[X]\end{array}$$

$$\begin{array}{}\text{(10)}& V(P)=\mathrm{\varnothing }\end{array}$$

Hence in propositional logic, we get a contradiction here.

But if we consider the fields extension, It admits a solution in the field

$$\begin{array}{}\text{(11)}& {\mathbb{F}}_2[X]/(X^2+X+1)\end{array}$$

What if we consider propositions lie on ${\mathbb{F}}_2[X]/(X^2+X+1)$$\mathbb F_2[X]/(X^2+X+1)$?

The propositions are:

$$\begin{array}{}\text{(12)}& 0,1,X,1+X\end{array}$$

A really wired things here is

$$\begin{array}{}\text{(13)}& X\wedge X=X^2=X+1=\mathrm{\neg }X\end{array}$$

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