The Classical Logic and the Continuous Logic

Abstract

The truth value of a proposition in classical logic is either 0 or 1, where 0 stands for falsity and 1 stands for truth. In the real world, however, there exist many propositions with variable answers that are neither false nor true. In this paper, we present the continuous logic with the truth value of a proposition falling into the continuous range [0, 1], where 0 stands for complete falsity and 1 for complete truth. To compare the continuous logic with the classical logic, fistly, we define three primitive logic operators not, and, or, and several compound logic operators not-and, not-or, exclusive-or, not-exclusive-or, and implication from $[0, 1]^{n}$ to [0, 1], where n is an integer and $n\ge 1$. Secondly, we discuss various laws and inference rules in both the classical logic and the continuous logic. We show that the continuous logic is consistent with the classic logic, and that the classical logic is simply a special case of the continuous logic.

Appendices

Appendix A

Proof of Not-exclusive-or Equation

$$\begin{aligned} \begin{array}{ll} (x\oplus y)' &{} = (x'y + xy')' \\ &{} = (x'y)'(xy')' \\ &{} = (x+y')(x'+y) \\ &{} = xx' + xy + y'x' + y'y \end{array} \end{aligned}$$

• 1. Case of $x\ge y$

  In this case, we have

  $$\begin{aligned} \begin{array}{ll} xy &{} = y \\ x'y' &{} = x' \end{array} \end{aligned}$$

  It follows that

  $$\begin{aligned} \begin{array}{ll} (x\oplus y)' &{} = xx' + y + x' + y'y \\ &{} = (x+1)x' + y(1+y') \\ &{} = x' + y \\ &{} = x'y' + xy \end{array} \end{aligned}$$

• 2. Case of $x<y$

  In this case, we have

  $$\begin{aligned} \begin{array}{ll} xy &{} = x \\ x'y' &{} = y' \end{array} \end{aligned}$$

  It follows that

  $$\begin{aligned} \begin{array}{ll} (x\oplus y)' &{} = xx' + x + y' + y'y \\ &{} = x(x'+1) + y'(1+y) \\ &{} = x + y' \\ &{} = xy + x'y' \end{array} \end{aligned}$$

Therefore, in all cases, we have

$$\begin{aligned} (x\oplus y)' = x'y' + xy \end{aligned}$$

It is equivalent to say that $\forall x, y \in C$,

$$\begin{aligned} \begin{array}{ll} &{} \max \{\min (x,1-x), \min (x, y), \min (1-x,1-y), \min (y,1-y)\} \\ = &{} \max \{\min (x,y), \min (1-x, 1-y)\} \end{array} \end{aligned}$$

Appendix B

Proof of the Distributivity Law

• 1. $x(y+z)=xy+xz$

  • (a) Case of $x\ge y$

    if $y \ge z$, we have

    $x(y+z) = xy = y$

    $xy+xz=y+xz=y+z=y$

    otherwise, we have

    $x(y+z) = xz$

    $xy+xz=y+xz=xz$.

  • (b) Case of $x<y$

    if $y \ge z$, we have

    $x(y+z) = xy = x$

    $xy+xz=x+xz=x$

    Otherwise, we have

    $x(y+z) = xz = x$

    $xy+xz=x+x=x$

  In all cases, we have $x(y+z)=xy+xz$ .

• 2. $x+yz=(x+y)(x+z)$ This equation can be proved in a way similar to the proof of $x(y+z)=xy+xz$.

Appendix C

Proof of the De Morgan Law

• 1. $(x+y)'=x'y'$

  • (a) Case of $x\ge y$

    In this case, we have $x '\le y'$. and hence

    (x+y)’=x’=x’y’

  • (b) Case of $x<y$

    In this case, we have $x'>y'$, and hence

    (x+y)’=y’=x’y’

  In all cases, we have $(x+y)'=x'y'$ .

• 2. $(xy)'=x'+y'$ This equation can be proved in a way similar to the proof of $(x+y)'=x'y'$.