I have been trying to find if there are other matrix type representations logical circuits, in the example below, $$\begin{bmatrix} 1 & 0\\ 1 & 1\\ \end{bmatrix} \equiv \, \, \Rightarrow$$ where:

$$ \begin{bmatrix} X, \ \neg X \end{bmatrix}\begin{bmatrix} 1 & 0\\ 1 & 1\\ \end{bmatrix} \begin{bmatrix} Y \\ \neg Y \\ \end{bmatrix} = \begin{bmatrix} X \oplus \neg X, \ 0 \oplus \neg X \end{bmatrix} \begin{bmatrix} Y \\ \neg Y \\ \end{bmatrix} = \begin{bmatrix} 1, \ \neg X \end{bmatrix} \begin{bmatrix} Y \\ \neg Y \\ \end{bmatrix}$$ $$ = Y \oplus (\neg X \neg Y) = X \Rightarrow Y$$

Note that the juxtaposition of two variables without an operator corresponds to conjunction i.e. $XY = X \wedge Y$ and of course $1 \wedge X = X$ and $0 \wedge X = 0$ and are written accordingly. Actually all of the propositional logical operators can be done with a unique binary $2 \times 2$ matrix - it is fascinating.


Yes, "all of the propositional logical operators can be done with a unique binary 2×2 matrix".

In fact, every possible map $g: \{0,1\}\times\{0,1\}\to\{0,1\}$ can be done with a unique binary 2x2 matrix.

Here is a proof.

$$\begin{align} &\quad \begin{bmatrix} X, \ \neg X \end{bmatrix}\begin{bmatrix} A & B\\ C & D\\ \end{bmatrix} \begin{bmatrix} Y \\ \neg Y \\ \end{bmatrix} \\&= \begin{bmatrix} XA \oplus \neg XC, \ XB \oplus \neg XD \end{bmatrix} \begin{bmatrix} Y \\ \neg Y \\ \end{bmatrix} \\ &= XAY \oplus \neg XCY \oplus XB\neg Y \oplus \neg XD\neg Y\\ \end{align}$$

  • Suppose the above expression equals $g(X,Y)$.
    • Let $X=0, Y=0$, we get $0\oplus0\oplus0\oplus D=D=g(0,0)$.
    • Let $X=0, Y=1$, we get $0\oplus C\oplus0\oplus 0=C=g(0,1)$.
    • Let $X=1, Y=0$, we get $0\oplus0\oplus B\oplus 0=B=g(1,0)$.
    • Let $X=1, Y=1$, we get $A\oplus0\oplus0\oplus 0=A=g(1,1)$.
  • Conversely, if we have $A= g(1,1)$, $B= g(1,0)$, $C=g(0,1)$ and $D=g(0,0)$, the above expression can be verified to be equal to $g(X,Y)$ for all $X, Y$.

So we have proved that

$$ \begin{bmatrix} X, \ \neg X \end{bmatrix}\begin{bmatrix} A & B\\ C & D\\ \end{bmatrix} \begin{bmatrix} Y \\ \neg Y \\ \end{bmatrix} = g(X,Y) \\ \Leftrightarrow A= g(1,1),\ B= g(1,0),\ C=g(0,1),\ D=g(0,0)$$

If we use in-fix notation for $g$, we have $$ \begin{bmatrix} X, \ \neg X \end{bmatrix} \begin{bmatrix} 1g1 & 1g0\\ 0g1 & 0g0\\ \end{bmatrix} \begin{bmatrix} Y \\ \neg Y \\ \end{bmatrix} = XgY, $$ or $$ \begin{bmatrix} 1g1 & 1g0\\ 0g1 & 0g0\\ \end{bmatrix} \equiv g. $$ In particular, we can substitute $g$ by anyone of $\land, \lor, \oplus, \odot, \Rightarrow, \Leftarrow, \equiv$.

For more related concepts and computation, please check boolean rings.

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  • $\begingroup$ Wow, I really appreciate this, I will add it to a paper I am working and and cite this answer. $\endgroup$ – Relative0 Jan 16 '19 at 23:30
  • $\begingroup$ For the paper I am working on I originally posted this question to find if something similar had been discovered so as to give credit where it is due and cite it in my paper. $\endgroup$ – Relative0 Jan 16 '19 at 23:32
  • 1
    $\begingroup$ You are welcome and thanks. If you are asking for the origin, I cannot recall whether I have read my answer before or not. It looks like folklore to me. $\endgroup$ – John L. Jan 16 '19 at 23:40

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