Is the World Itself Fuzzy? Physical Arguments and Unexpected Computational Consequences of Zadeh’s Vision

Abstract

Fuzzy methodology has been invented to describe imprecise (“fuzzy”) human statements about the world, statements that use imprecise words from natural language like “small” or “large”. Usual applications of fuzzy techniques assume that the world itself is “crisp”, that there are exact equations describing the world, and fuzziness of our statements is caused by the incompleteness of our knowledge. But what if the world itself is fuzzy? What if there is no perfect system of equations describing the physical world – in the sense that no matter what system of equations we try, there will always be cases when this system leads to wrong predictions? This is not just a speculation: this idea is actually supported by many physicists. At first glance, this is a pessimistic idea: no matter how much we try, we will never be able to find the the Ultimate Theory of Everything. But it turns out that this idea also has its optimistic aspects: namely, in this chapter, we show (somewhat unexpectedly), that if such a no-perfect-theory principle is true, then the use of physical data can drastically enhance computations.

Appendices

A A Formal Definition of Definable Sets

Definition A1. Let $\mathcal L$ be a theory, and let $P(x)$ be a formula from the language of the theory $\mathcal L$, with one free variable $x$ for which the set $\{x\,|\,P(x)\}$ is defined in the theory $\mathcal L$. We will then call the set $\{x\,|\,P(x)\}$ $\mathcal L$-definable.

Crudely speaking, a set is $\mathcal L$-definable if we can explicitly define it in $\mathcal L$. The set of all real numbers, the set of all solutions of a well-defined equation, every set that we can describe in mathematical terms: all these sets are $\mathcal L$-definable.

This does not mean, however, that every set is $\mathcal L$-definable: indeed, every $\mathcal L$-definable set is uniquely determined by formula $P(x)$, i.e., by a text in the language of set theory. There are only denumerably many words and therefore, there are only denumerably many $\mathcal L$-definable sets. Since, e.g., in a standard model of set theory ZF, there are more than denumerably many sets of integers, some of them are thus not $\mathcal L$-definable.

Our objective is to be able to make mathematical statements about $\mathcal L$-definable sets. Therefore, in addition to the theory $\mathcal L$, we must have a stronger theory $\mathcal M$ in which the class of all $\mathcal L$-definable sets is a set – and it is a countable set.

Denotation.  For every formula $F$ from the theory $\mathcal L$, we denote its Gödel number by $\lfloor F\rfloor$.

Comment. A Gödel number of a formula is an integer that uniquely determines this formula. For example, we can define a Gödel number by describing what this formula will look like in a computer. Specifically, we write this formula in LaTeX, interpret every LaTeX symbol as its ASCII code (as computers do), add 1 at the beginning of the resulting sequence of 0 s and 1s, and interpret the resulting binary sequence as an integer in binary code.

Definition A2. We say that a theory $\mathcal M$ is stronger than $\mathcal L$ if it contains all formulas, all axioms, and all deduction rules from $\mathcal L$, and also contains a special predicate $\mathrm{def}(n,x)$ such that for every formula $P(x)$ from $\mathcal L$ with one free variable, the formula $\forall y\,(\mathrm{def}(\lfloor P(x)\rfloor , y)\leftrightarrow P(y))$ is provable in $\mathcal M$.

The existence of a stronger theory can be easily proven: indeed, for $\mathcal L$=ZF, there exists a stronger theory $\mathcal M$. As an example of such a stronger theory, we can simply take the theory $\mathcal L$ plus all countably many equivalence formulas as described in Definition A2 (corresponding to all possible formulas $P(x)$ with one free variable). This theory clearly contains $\mathcal L$ and all the desired equivalence formulas, so all we need to prove is that the resulting theory $\mathcal M$ is consistent (provided that $\mathcal L$ is consistent, of course). Due to compactness principle, it is sufficient to prove that for an arbitrary finite set of formulas $P_1(x),\ldots ,P_m(x)$, the theory $\mathcal L$ is consistent with the above reflection-principle-type formulas corresponding to these properties $P_1(x),\ldots ,P_m(x)$.

This auxiliary consistency follows from the fact that for such a finite set, we can take

$$ \mathrm{def}(n,y)\leftrightarrow (n=\lfloor P_1(x)\rfloor \, \& \,P_1(y))\vee \ldots \vee (n=\lfloor P_m(x)\rfloor \, \& \,P_m(y)).$$

This formula is definable in $\mathcal L$ and satisfies all $m$ equivalence properties. The statement is proven.

Important comments. In the main text, we will assume that a theory $\mathcal M$ that is stronger than $\mathcal L$ has been fixed; proofs will mean proofs in this selected theory $\mathcal M$.

An important feature of a stronger theory $\mathcal M$ is that the notion of an $\mathcal L$-definable set can be expressed within the theory $\mathcal M$: a set $S$ is $\mathcal L$-definable if and only if $\exists n\in \mathrm{I\!N}\,\forall y(\mathrm{def}(n,y)\leftrightarrow y\in S).$

In the chapter, when we talk about definability, we will mean this property expressed in the theory $\mathcal M$. So, all the statements involving definability become statements from the theory $\mathcal M$ itself, not statements from metalanguage.

B Proofs

Proof of Proposition 1. Let us fix an integer $C$. To prove the desired property for this $C$, let us prove that the set $T$ of all the sequences which do not satisfy this property, i.e., for which $K(\alpha _1\ldots \alpha _n\,|\,\omega )\ge K(\alpha _1\ldots \alpha _n)-C$ for all $n$, is a physical theory in the sense of Definition 1. For this, we need to prove that this set $T$ is non-empty, closed, nowhere dense, and definable. Then, from Definition 2, it will follow that the sequence $\omega$ does not belong to this set and thus, that the conclusion of Proposition 1 is true.

The set $T$ is clearly non-empty: it contains, e.g., a sequence $\omega =00\ldots 0\ldots$ which does not affect computations. The set $T$ is also clearly definable: we have just defined it.

Let us prove that the set $T$ is closed. For that, let us assume that $\omega ^{(m)}\rightarrow \omega$ and $\omega ^{(m)}\in T$ for all $m$. We then need to prove that $\omega \in T$. Indeed, let us fix $n$, and let us prove that $K(\alpha _1\ldots \alpha _n\,|\,\omega )\ge K(\alpha _1\ldots \alpha _n)-C$. We will prove this by contradiction. Let us assume that $K(\alpha _1\ldots \alpha _n\,|\,\omega )< K(\alpha _1\ldots \alpha _n)-C$. This means that there exists a program $p$ of length $\mathrm{len}(p)<K(\alpha _1\ldots \alpha _n)-C$ which uses $\omega$ to compute $\alpha _1\ldots \alpha _n$. This program uses only finitely many bits of $\omega$; let $B$ be the largest index of these bits. Due to $\omega ^{(m)}\rightarrow \omega$, there exists $M$ for which, for all $m\ge M$, the first $B$ bits of $\omega ^{(m)}$ coincide with the first $B$ bits of the sequence $\omega$. Thus, the same program $p$ will work exactly the same way – and generate the same sequence $\alpha _1\ldots \alpha _n$ – if we use $\omega ^{(m)}$ instead of $\omega$. But since $\mathrm{len}(p)<K(\alpha _1\ldots \alpha _n)-C$, this would means that the shortest length $K(\alpha _1\ldots \alpha _n\,|\,\omega ^{(m)})$ of all the programs which use $\omega ^{(m)}$ to compute $\alpha _1\ldots \alpha _n$ also satisfies the inequality $K(\alpha _1\ldots \alpha _n\,|\,\omega ^{(m)})<K(\alpha _1\ldots \alpha _n)-C$. This inequality contradicts to our assumption that $\omega ^{(m)}\in T$ and thus, that $K(\alpha _1\ldots \alpha _n\,|\,\omega ^{(m)})\ge K(\alpha _1\ldots \alpha _n)-C$. The contradiction proves that the set $T$ is indeed closed.

Let us now prove that the set $T$ is nowhere dense, i.e., that for every finite sequence $\omega _1\ldots \omega _m$, there exists a continuation $\omega$ which does not belong to the set $T$. Indeed, as such a continuation, we can simply take a sequence $\omega =\omega _1\ldots \omega _m\alpha _1\alpha _2\ldots$ obtained by appending $\alpha$ at the end. For this new sequence, computing $\alpha _1\ldots \alpha _n$ is straightforward: we just copy the values $\alpha _i$ from the corresponding places of the new sequence $\omega$. Here, the relative Kolmogorov complexity $K(\alpha _1\ldots \alpha _n\,|\,\omega )$ is very small and is, thus, much smaller than the complexity $K(\alpha _1\ldots \alpha _n)$ which – since ZF is not decidable – grows with $n$.

The proposition is proven.

Proof of Proposition 2.

$1^\circ$. As the desired ph-algorithm, we will, given an instance $i$, simply produce the result $\omega _i$ of the $i$-th experiment. Let us prove, by contradiction, that this algorithm satisfies the desired property.

$2^\circ$. We want to prove that for every $\varepsilon >0$ and for every $n$, there exists an integer $N\ge n$ for which

$$ \#\{i\le N:i\in S_\mathcal{P}\, \& \,\omega _i=s_{\mathcal{P},i}\}>(1-\varepsilon )\cdot \#\{i\le N:i\in S_\mathcal{P}\}.$$

The assumption that this property is not satisfied means that for some $\varepsilon >0$ and for some integer $n$, we have

$$ \#\{i\le N:i\in S_\mathcal{P}\, \& \,\omega _i=s_{\mathcal{P},i}\}\le (1-\varepsilon )\cdot \#\{i\le N:i\in S_\mathcal{P}\} \text{ for } \text{ all } N\ge n.{(1)}$$

Let $T$ denote the set of all the sequences $x$ that satisfy the property (1), i.e., let

$$T\mathop {=}\limits ^\mathrm{def}$$

$$ \{x:\#\{i\le N:i\in S_\mathcal{P}\, \& \,x_i=s_{\mathcal{P},i}\}\le (1-\varepsilon )\cdot \#\{i\le N:i\in S_\mathcal{P}\} \text{ for } \text{ all } N\ge n\}.$$

We will prove that this set $T$ is a physical theory in the sense of Definition 1.

Then, due to Definition 2 and the fact that the sequence $\omega$ satisfies the no-perfect-theory principle, we will be able to conclude that $\omega \not \in T$, and thus, that the property (1) is not satisfied for the given sequence $\omega$. This will conclude the proof by contradiction.

$3^\circ$. By definition of a physical theory $T$, it is a set which is non-empty, closed, nowhere dense, and definable. Let us prove these four properties one by one.

$3.1^\circ$. Non-emptiness comes from the fact that the sequence $x_i$ for which $x_i=\lnot s_{\mathcal{P},i}$ for $i\in S_\mathcal{P}$ and $x_i=0$ otherwise clearly belongs to this set: for this sequence, for every $N$, we have $\#\{i\le N:i\in S_\mathcal{P}\, \& \,x_i=s_{\mathcal{P},i}\}=0$ and thus, the desired property is satisfied.

$3.2^\circ$. Let us prove that the set $T$ is closed, i.e., that if we have a family of sequences $x^{(m)}\in T$ for which $x^{(m)}\rightarrow x$, then $x\in T$.

Indeed, let us take any $N\ne n$, and let us prove that

$$ \#\{i\le N:i\in S_\mathcal{P}\, \& \,x_i=s_{\mathcal{P},i}\}\le (1-\varepsilon )\cdot \#\{i\le N:i\in S_\mathcal{P}\}$$

for this $N$. Due to $x^{(m)}\rightarrow x$, there exists $M$ for which, for all $m\ge M$, the first $N$ bits of $x^{(m)}$ coincide with the first $N$ bits of the sequence $x$: $x^{(m)}_i=\omega _i$ for all $i\le N$. Thus,

$$ \#\{i\le N:i\in S_\mathcal{P}\, \& \,x_i=s_{\mathcal{P},i}\}=\#\{i\le N:i\in S_\mathcal{P}\, \& \,x^{(m)}_i=s_{\mathcal{P},i}\}.$$

Since $x^{(m)}\in T$, we have

$$ \#\{i\le N:i\in S_\mathcal{P}\, \& \,x^{(m)}_i=s_{\mathcal{P},i}\}\le (1-\varepsilon )\cdot \#\{i\le N:i\in S_\mathcal{P}\},$$

thus

$$ \#\{i\le N:i\in S_\mathcal{P}\, \& \,x_i=s_{\mathcal{P},i}\}\le (1-\varepsilon )\cdot \#\{i\le N:i\in S_\mathcal{P}\}.$$

So, the set $T$ is indeed closed.

$3.3^\circ$. Let us now prove that the set $T$ is nowhere dense, i.e., that for every finite sequence $x_1\ldots x_m$, there exists a continuation $x$ which does not belong to the set $T$.

Indeed, as such a continuation, we can simply take a sequence

$$x=x_1\ldots x_m x_{m+1}x_{m+2}\ldots $$

where for $i>m$, we take $x_i=s_{\mathcal{P},i}$ if $i\in S_\mathcal{P}$ and $x_i=0$ otherwise. For this new sequence, for every $N$, at most $m$ first instances may lead to results different from $s_{\mathcal{P},i}$, so we have

$$ \#\{i\le N: i\in S_\mathcal{P}\, \& \,x_i=s_{\mathcal{P},i}\}\ge \#\{i\le N:i\in S_\mathcal{P}\}-m.$$

When $N\rightarrow \infty$, then $\#\{i\le N:i\in S_\mathcal{P}\}\rightarrow \infty$, so for sufficiently large $N$, we have

$$\#\{i\le N:i\in S_\mathcal{P}\}-m>(1-\varepsilon )\cdot \#\{i\le N:i\in S_\mathcal{P}\},$$

thus,

$$ \#\{i\le N:i\in S_\mathcal{P}\, \& \,x_i=s_{\mathcal{P},i}\}>(1-\varepsilon )\cdot \#\{i\le N:i\in S_\mathcal{P}\},$$

and we cannot have

$$ \#\{i\le N:i\in S_\mathcal{P}\, \& \,x_i=s_{\mathcal{P},i}\}\le (1-\varepsilon )\cdot \#\{i\le N:i\in S_\mathcal{P}\}.$$

Therefore, this continuation does not belong to the set $T$.

$3.4^\circ$. Finally, since the formula (1) explicitly defines the set $T$, this set $T$ is clearly definable.

So, $T$ is a physical theory, hence $\omega \not \in T$, and the proposition is proven.