# Subordinators and generalized heat kernels: Random time change and long time
dynamics

N. Ajaber, A. Alshehri, H. Altamimi, M. Majdoub, E. Mliki  Department of
Mathematics, College of Science, Imam Abdulrahman Bin Faisal University, P. O.
Box 1982, Dammam, Saudi Arabia  Basic and Applied Scientific Research Center,
Imam Abdulrahman Bin Faisal University, P.O. Box 1982, 31441, Dammam, Saudi
Arabia<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
(Date: August 28, 2024)

###### Abstract.

This paper focuses on studying the long-time dynamics of the subordination
process for a range of linear evolution equations, with a special emphasis on
the fractional heat equation. By treating inverse subordinators as random time
variables and employing the subordination principle to solve forward
Kolmogorov equations, we explore the behavior of the solutions over extended
periods. We provide a detailed description of the specific classes of
subordinators suitable for conducting asymptotic analysis. Our findings not
only extend existing research, but also enhance the results previously
presented in [9, 10].

###### Key words and phrases:

Long-time behavior, Cesàro mean, long range dependence, fractional Heat
equation, Heat kernal, Brownian motion, subordination, inverse alpha stable
subordinators, gamma subordinators.

###### 2020 Mathematics Subject Classification:

35R11, 35B40, 40E05

##  1\. Introduction

This paper aims to investigate the long-term behavior of solutions for a
specific class of linear evolution equations that emerge from random time
changes utilizing various subordinators, such as gamma and alpha stable
subordinators. We explore how these subordinators, denoted as
$\mathbf{S}=\left\\{\mathbf{S}_{t}\,;t\geq 0\right\\}$, influence the dynamics
of the solutions over extended time periods.

More specifically, our study focuses on examining the behavior of the
subordinated solution, denoted as $v(x,t)$, for some evolution partial
differential equations (PDEs) as time progresses. We analyze the large-time
dynamics of the subordination process by convolving the solution with the
density function $G_{t}$ associated with the subordinator $\mathbf{S}$. This
is accomplished through the analysis of the Cesàro mean of the subordinated
solution. By examining the Cesàro mean, we gain insights into the long-term
behavior and dynamics of the subordination process.

Understanding the behavior of the subordination process is of considerable
significance in various physical models. Specifically, it has the potential to
contribute to the development of valuable models that capture the concept of
biological time in the evolution of species and ecological systems. By gaining
insights into this process, we can enhance our understanding of complex
biological phenomena and improve our ability to analyze ecological dynamics.
This understanding can pave the way for advances in fields such as population
dynamics, epidemiology, and ecological modeling, ultimately leading to more
accurate predictions and informed decision making in real-world scenarios. For
further explanations and detailed discussions on the topics covered, see [14,
16].

Let us explain now the mathematical framework of the problem. Consider the
linear evolution equation

$\left\\{\begin{array}[]{rl}\partial_{t}v(x,t)&=\mathcal{A}\,v(x,t)\quad\mbox{
in }\quad\mathbb{R}^{N}\times(0,\infty),\\\ v(x,0)&=\varphi(x)\quad\mbox{ in
}\quad\mathbb{R}^{N},\end{array}\right.$ (1.1)

where $\mathcal{A}$ is a Markov generator that acts on the functions $v(x,t)$.
Assuming that $\eqref{LEE}$ has a solution $v$ such that $v(x,\cdot)\in
L^{1}(\mathbb{R}_{+})$, we introduce the subordination of $v(x,t)$ by a
density function $G_{t}(\tau)$ as follows:

$v^{E}(x,t):=\int_{0}^{\infty}\,v(x,\tau)G_{t}(\tau)\,d\tau,\quad
x\in\mathbb{R}^{N},\;t\geq 0.$ (1.2)

According to [3], we know that $v^{E}(x,t)$ is the solution of the general
fractional differential equation

$\left\\{\begin{matrix}\left(\mathbb{D}^{(k)}_{t}v^{E}\right)(x,t)=\mathcal{A}\,v^{E}(x,t)\\\
v^{E}(x,0)=\varphi(x).\end{matrix}\right.$ (1.3)

Here $\mathbb{D}^{(k)}_{t}$ is the differential-convolution operator given by

$\left(\mathbb{D}^{(k)}_{t}g\right)(t)=\frac{d}{dt}\int_{0}^{t}\,\mathbf{k}(t-s)\left(g(s)-g(0)\right)\,ds,$
(1.4)

where $\mathbf{k}\in L^{1}_{loc}(\mathbb{R}_{+})$ is a non-negative kernel.

The main focus of this paper is to characterize the specific classes of
subordinators that facilitate the derivation of time asymptotics for
generalized fractional dynamics. In particular, our objective is to
investigate the long-term behavior of $v^{E}(x,t)$, the solution of interest.
To achieve this, we introduce the concept of Cesàro mean for $v^{E}(x,t)$:

$\mathbf{M}_{t}(v^{E}(x,t)):=\frac{1}{t}\int_{0}^{t}\,v^{E}(x,s)\,ds.$ (1.5)

Clearly, from (1.2) one can see that

$\mathbf{M}_{t}(v^{E}(x,t))=\int_{0}^{\infty}\,v(x,\tau)\mathbf{M}_{t}\left(G_{t}(\tau)\right)\,d\tau.$
(1.6)

Our approach in this study is grounded in Laplace transform techniques, the
Feller-Karamata-Tauberian theorem, and the integrability of the solution over
time $v(x,t)$. This method can be applied to various models, including the
fractional heat equation. In particular, our approach yields satisfactory
results for dimensions $N\geq 2$. However, for the one-dimensional case, the
obtained result is somewhat weaker. For further investigations and related
studies, we recommend referring to [9, 10].

We consider a broad class of admissible kernels, denoted $0\leq\mathbf{k}\in
L^{1}_{\text{loc}}(\mathbb{R}_{+})$, which are characterized in terms of their
Laplace transforms $\mathcal{K}(\lambda)$ as follows.

$\lambda\mathcal{K}(\lambda)\underset{\lambda\to 0^{+}}{\longrightarrow}\,0,$
(1.7)

and, for some $\varrho\geq 0$,

$L(x):=x^{-\varrho}\,\mathcal{K}\left(x^{-1}\right)\quad\mbox{is a slowly
varying function}.$ (1.8)

An example that is well known and commonly used is given by

$\mathbf{k}(s)=\int_{0}^{1}\frac{s^{-\alpha}}{\Gamma(1-\alpha)}\mu(\alpha)d\alpha,$
(1.9)

where $\mu:[0,1]\to(0,\infty)$ is a continuous function. For further
properties and asymptotic behavior of kernels defined by (1.9), we refer the
reader to [11]. In particular, [11] provides detailed information on the
Laplace transform of $\mathbf{k}$ as follows:

$\mathcal{K}(\lambda)=\int_{0}^{1}\lambda^{\alpha-1}\mu(\alpha)\,d\alpha.$
(1.10)

In this case, it is worth noting that the slowly varying function $L$ is
defined as follows:

$L(x)=\int_{0}^{1}x^{-\alpha}\mu(\alpha)d\alpha.$ (1.11)

Below, we provide a list of fundamental examples of kernel classes that will
be utilized in our analysis.

  1. 1)

The $\theta$-stable subordinator class (${\bf C}_{1}$):

$\mathcal{K}(\lambda)=\lambda^{\theta-1},\quad 0<\theta<1.$

  2. 2)

The distributed order derivatives class (${\bf C}_{2}$):

$\mathcal{K}(\lambda)\;\underset{\lambda\to
0^{+}}{\sim}\;C\lambda^{-1}\left(-\ln\lambda\right)^{-\kappa},\quad
C,\kappa>0.$

  3. 3)

The inverse Gamma subordinator class (${\bf C}_{3}$):

$\mathcal{K}(\lambda)=\frac{\sqrt{b}}{\lambda}\left(2\sqrt{2\lambda+a}-\sqrt{a}\right),\quad
a\geq 0,\;\;b>0.$

  4. 4)

The Gamma subordinator class (${\bf C}_{4}$):

$\mathcal{K}(\lambda)=\frac{a}{\lambda}\ln\left(1+\frac{\lambda}{b}\right),\quad
a,b>0.$

  5. 5)

The tempered stable subordinator class (${\bf C}_{5}$):

$\mathcal{K}(\lambda)=\frac{(\lambda+\beta)^{\theta}-\beta^{\theta}}{\lambda},\quad\beta>0.$

######  Remark 1.1.

  * (i)

All classes (${\bf C}_{i}$), $1\leq i\leq 5$, satisfy assumption (1.7) except
for (${\bf C}_{3}$) with $a>0$. It is worth noting that when $a=0$ in (${\bf
C}_{3}$), we obtain the class (${\bf C}_{1}$) as a special case with
$\theta=1/2$.

  * (ii)

The hypothesis (1.8) holds for all classes (${\bf C}_{i}$), $1\leq i\leq 5$,
with specific values of $\varrho$. For (${\bf C}_{1}$), we have
$\varrho=1-\theta$ and $L(x)=1$, for (${\bf C}_{2}$), we have $\varrho=1$ and
$L(x)=C\left(\ln x\right)^{-\kappa}$, for (${\bf C}_{3}$) we have $\varrho=1$
and $L(x)=\sqrt{b}\left(2\sqrt{2/x+a}-\sqrt{a}\right)$, for (${\bf C}_{4}$),
we have $\varrho=0$ and $L(x)=ax\ln\left(1+\frac{1}{bx}\right)$ and for (${\bf
C}_{5}$), we have $\varrho=0$ and
$L(x)=x\left(\left(1/x+\beta\right)^{\theta}-\beta^{\theta}\right)$.

Henceforth, we will consistently represent the Lebesgue norm in $L^{p}$,
applicable to both space and time variables, as $\|\cdot\|_{p}$ for $1\leq
p\leq\infty$. The following statement presents our first main result:

######  Theorem 1.1.

Suppose that the Markov generator $\mathcal{A}$ satisfies the condition that
for any $0\leq\varphi\in L^{1}(\mathbb{R}^{N})$, the solution $v$ of (1.1) is
non-negative, and $v(x,\cdot)\in L^{1}(0,\infty)$. Moreover, if we assume that
the conditions (1.7) and (1.8) are fulfilled, then:

$\mathbf{M}_{t}(v^{E}(x,t))\,\underset{t\to\infty}{\backsim}\,\frac{\|v(x)\|_{1}}{\Gamma(\varrho+1)}\,\bigg{(}t^{-1}\mathcal{K}\left(t^{-1}\right)\bigg{)}.$
(1.12)

Here, $v^{E}(x,t)$ represents the subordination of the solution $v(x,t)$ as
defined in (1.2).

######  Remark 1.2.

  * (i)

The asymptotic behavior result, stated as (1.12), can be alternatively
expressed as

$\mathbf{M}_{t}(v^{E}(x,t))\,\underset{t\to\infty}{\backsim}\,\frac{\|v(x)\|_{1}}{\Gamma(\varrho+1)}\,\left(\int_{0}^{1}t^{-\alpha}\mu(\alpha)d\alpha\right),$
(1.13)

where the function $\mu$ is given by (1.9).

  * (ii)

If we assume $\lambda\mathcal{K}(\lambda)\underset{\lambda\to
0^{+}}{\longrightarrow}\,\ell>0$ instead of (1.7), then the requirement
$v(x,\cdot)\in L^{1}(0,\infty)$ can be omitted. In that case, the conclusion
(1.12) remains valid, provided that $\|v(x)\|_{1}$ is replaced with

$\|v(x)\|_{1,\ell}:=\int_{0}^{\infty}\,{\rm e}^{-\ell\tau}\,v(x,\tau)\,d\tau.$

  * (iii)

Our primary contribution is encapsulated in Theorem 1.1, where we employ a
unified approach to investigate a broad class of operators and kernels.

For specific choices of kernel classes, Theorem 1.1 provides the following
asymptotic estimates.

######  Corollary 1.1.

Under the assumptions of Theorem 1.1 (except for class ($\mathbf{C}_{3}$)
where the assumption $v(x,\cdot)\in L^{1}$ is omitted), the long-time behavior
of the Cesàro mean of $v^{E}(x,t)$ as $t\to\infty$ is expressed as:

$\begin{split}\mathbf{M}_{t}(v^{E}(x,t))\,\underset{t\to\infty}{\backsim}\,\left\\{\begin{array}[]{llll}\frac{\|v(x)\|_{1}}{\Gamma(2-\theta)}\,t^{-\theta}&\text{if}\quad\mathbf{k}\in\;(\mathbf{C}_{1})\\\
\mathtt{C}\,\|v(x)\|_{1}\,\left(\ln
t\right)^{-\kappa}&\text{if}\quad\mathbf{k}\in\;(\mathbf{C}_{2})\\\
2\sqrt{ab}\,\|v(x)\|_{1,\sqrt{ab}}&\text{if}\quad\mathbf{k}\in\;(\mathbf{C}_{3})\\\
\frac{a\|v(x)\|_{1}}{b}\,t^{-1}&\text{if}\quad\mathbf{k}\in\;(\mathbf{C}_{4})\\\
\theta\beta^{\theta-1}\,\|v(x)\|_{1}\,t^{-1}&\text{if}\quad\mathbf{k}\in\;(\mathbf{C}_{5}).\end{array}\right.\end{split}$
(1.14)

Theorem 1.1 leads to the following consequential result.

######  Theorem 1.2.

Consider the operator $\mathcal{A}=-(-\Delta)^{\mathsf{s}}$ where either
($N\geq 2$ and $0<\mathtt{s}<1$) or ($N=1$ and $0<\mathsf{s}<1/2$). If
$0\leq\varphi\in L^{1}\cap L^{\infty}(\mathbb{R}^{N})$, the solution $v$ of
(1.1) is non-negative, and $v(x,\cdot)\in L^{1}(0,\infty)$. Consequently, the
asymptotic behavior described in (1.12) holds true, given the conditions (1.7)
and (1.8) imposed on the kernel.

######  Remark 1.3.

  * (i)

The case $\mathsf{s}=1$ was explored in [9, 10] across all dimensions.

  * (ii)

The non-negativity of the solution $v$ only necessitates $0<\mathsf{s}\leq 1$
(see Lemma 2.1 below).

  * (iii)

In order to guarantee that $v(x,\cdot)\in L^{1}(1,\infty)$, we need to impose
an additional condition: either $N\geq 2$ or, in the case of $N=1$, we require
$0<\mathsf{s}<1/2$ along with $\varphi\in L^{1}$. See Proposition 2.2 below.

  * (iv)

The requirement of $\varphi\in L^{\infty}$ is used to guarantee the
integrability of $v(x,\cdot)$ near $t=0$.

  * (v)

In addition to the nonlocal operator discussed in Theorem 1.2, another class
of nonlocal operators that can be addressed using the same framework is
represented by $\mathcal{A}v=\mathbf{J}\ast v-v$. Here, $\mathbf{J}$ denotes a
radial probability density function that satisfies certain additional
assumptions. For further details and insights into this class of nonlocal
diffusion problems, refer to [2].

When considering the scenario where $N=1$ and $1/2\leq\mathsf{s}<1$, we obtain
a result that is slightly weaker, as presented below.

######  Theorem 1.3.

Assuming $N=1$ and $1/2\leq\mathsf{s}<1$, and considering a kernel
$\mathbf{k}$ that satisfies assumptions (1.7) and (1.8), we have the following
long-term behavior of the Cesàro mean of $v^{E}(x,t)$:

$\mathbf{M}_{t}(v^{E}(x,t))\,\asymp\,\frac{t^{\frac{\varrho-1}{2\mathsf{s}}}}{\Gamma(1+\varrho)}\,\left(L(t)\right)^{\frac{1}{2\mathsf{s}}}.$
(1.15)

Here, the notation $\asymp$ signifies that the ratio under consideration is
bounded both from above and below.

The structure of the paper is as follows. In Section 2, we provide a recap of
the essential background information required for the subsequent proofs.
Section 3 is dedicated to presenting the proofs of our main results, namely
Theorem 1.1, Corollary 1.1, Theorem 1.2, and Theorem 1.3.

##  2\. Preliminaries

###  2.1. Notations

  * $\triangleright$

We write $X\lesssim Y$ or $Y\gtrsim X$ to denote the estimate $X\leq CY$ for
some constant $C>0$.

  * $\triangleright$

The notation $f(z)\underset{z\to z_{0}}{\backsim}g(z)$ means that
$\displaystyle\lim_{z\to z_{0}}\frac{f(z)}{g(z)}=1$, where $-\infty\leq
z_{0}\leq\infty$.

  * $\triangleright$

The notation $f(z)\;\asymp\;g(z)$ indicates that the ratio is bounded by a
constant factor both from above and below.

  * $\triangleright$

The Laplace transform of a function $f:[0,\infty)\to\mathbb{R}$ is defined as
follows

$\mathcal{L}(f)(\lambda):=\int_{0}^{\infty}\,{\rm e}^{-\lambda
t}\,f(t)\,dt,\;\;\lambda>0.$

  * $\triangleright$

We will use the notation $\mathbb{R}_{+}=(0,\infty)$ to represent the set of
nonnegative real numbers.

  * $\triangleright$

To simplify notation, we abbreviate the Lebesgue norm on $\mathbb{R}_{+}$ as
$\|\cdot\|_{1}:=\|\cdot\|_{L^{1}(0,\infty)}$

  * $\triangleright$

Throughout the entire paper, we will use the letter $C$ to represent various
positive constants that are not significant in our analysis and may vary from
line to line.

###  2.2. Subordinators

A subordinator is a process with stationary and independent non-negative
increments starting at zero. Subordinators are a special class of Lévy
processes taking values in $[0,\infty)$ and their sample paths are non-
decreasing; this is a type of stochastic process that is used to model random
phenomena that have jumps (see [4, 5] for more details). Let
$\mathbf{S}=\left\\{\mathbf{S}_{t},\;t\geq 0\right\\}$ be a subordinator. The
infinite divisibility of the law of $\mathbf{S}$ implies that its Laplace
transform can be expressed in the form

$\displaystyle
E(e^{-\lambda\mathbf{S}_{t}})=e^{-t\Phi(\lambda)}=e^{-t\lambda\mathcal{K}(\lambda)},\quad\lambda\geq
0,$

where $\Phi:[0,\infty)\to[0,\infty)$, called the Laplace exponent, is a
Bernstein function [17]. Such a Laplace exponent $\Phi$ can be expressed as

$\displaystyle\Phi(\lambda)=\int_{(0,\infty)}(1-e^{-\lambda\tau})\,\sigma(d\tau)<\infty,$

which is known as the Lévy-Khintchine formula for the subordinator
$\mathbf{S}$. Where $\sigma$ are a measure on the positive real half-line such
that

$\displaystyle\int_{(0,\infty)}(1\land\tau)\,\sigma(d\tau)<\infty.$

The kernel $\mathbf{k}$ is related to the subordinator $\mathbf{S}$ via the
Lévy measure $\sigma$, namely if we set

$\mathbf{k}(t)=\sigma((t,\infty)),\quad t\geq 0.$

For any $\lambda\geq 0$ we have

$\int_{0}^{\infty}e^{-\lambda
t}\int_{0}^{t}d\sigma(s)\,ds=\int_{0}^{\infty}\int_{0}^{s}e^{-\lambda
t}dt\,d\sigma(s)=\frac{1}{\lambda}\Phi(\lambda)=\mathcal{K}(\lambda).$

Denote by $E$ the inverse process of the subordinator $\mathbf{S}$,

$E_{t}:=\mbox{inf}\\{s\geq 0;\,\mathbf{S}_{t}\geq t\\}.$

The marginal density of $E_{t}$ is the function $G_{t}(\tau)$, more precisely

$G_{t}(\tau)\,d\tau=\partial_{\tau}P(E_{t}\leq\tau)=\partial P(\mathbf{S}\geq
t).$

####  2.2.1.  $\theta$-stable subordinator

Let $\mathbf{k}$ be the kernal corresponding to the Caputo-Djrbashian
fractional derivative $\mathbb{D}^{(\theta)}_{t}$ of order $\theta\in(0,1).$
Then its Laplace transform is given by

$\mathcal{K}(\lambda)=\frac{1}{\Gamma(1-\theta)}\int_{0}^{\infty}\,{\rm
e}^{-\lambda t}\,t^{-\theta}\,dt=\lambda^{\theta-1},$

where $\mathbf{k}(t)=\frac{t^{-\theta}}{\Gamma(1-\theta)}.$ It is easy to
verify that (1.7) holds for $\mathcal{K}.$

####  2.2.2. Gamma Subordinator

Let $\mathbf{k}$ be the kernal given by

$\mathbf{k}(t)=a\Gamma(0,bt),\quad a,b>0,$

where $\Gamma(\nu,x):\displaystyle{\int_{x}^{\infty}t^{\nu-1}e^{-t}\,dt}$ is
the upper incomplete Gamma function. The Laplace transform of $\mathbf{k}$ is
given by

$\mathcal{K}(\lambda)=\frac{a}{\lambda}\ln\left(1+\frac{\lambda}{b}\right),\;\;a,b,\lambda>0.$

It is easy to see that (1.7) holds for $\mathcal{K}.$

####  2.2.3. Inverse Gamma Subordinator

Let $a\geq 0$ and $b>0$ be given and definite the kernel $\mathbf{k}$ by

$\mathbf{k}(t)=\sqrt{\frac{b}{2\pi}}\left(\frac{2}{\sqrt{t}}e^{-\frac{at}{2}}-\sqrt{2a\pi}(1-\mbox{erf}(z))\right),\quad
z:=\sqrt{\frac{at}{2}},$

where
$\mbox{erf}(z):=\frac{2}{\sqrt{\pi}}\displaystyle\int_{0}^{z}e^{-t^{2}}\,dt$
is the error function. The Laplace transform of $\mathbf{k}$ is given by

$\mathcal{K}(\lambda)=\frac{\sqrt{b}}{\lambda}\left(2\sqrt{2\lambda+a}-\sqrt{a}\right),\quad
a\geq 0,\;\;b,\lambda>0.$

The assumption (1.7) holds for $\mathcal{K}$ when the value $0$ in (1.7) is
replaced with $\sqrt{ab}$.

####  2.2.4. Tempered Stable Subordinator

Let us consider a tempered stable subordinator $D_{\beta}(t)$ with index
$\theta\in(0,1)$. The density function of $D_{\beta}(t)$ can be expressed as:

$f_{\beta}(x,t)={\rm e}^{-\beta\,x+\beta^{\theta}t}\,f(x,t),\quad\beta>0,$

where $\beta>0$ and $f(x,t)$ represents the density of the $\theta$-stable
subordinator. It is worth noting that $f_{\beta}(x,t)$ possesses finite
moments and exhibits infinite divisibility. However, it is not self-similar.

Moreover, we can obtain the Laplace transform of the density function
$f_{\beta}(x,t)$, given by:

$\mathcal{L}_{x}\left(f_{\beta}(x,t)\right)={\rm
e}^{-t\left((\lambda+\beta)^{\theta}-\beta^{\theta}\right)}={\rm
e}^{-t\lambda\mathcal{K}(\lambda)}.$

For further details and explanations, refer to [12, 15].

###  2.3. Fractional Heat Semigroup

Consider the linear homogeneous fractional heat equation

$\partial_{t}v+(-\Delta)^{\mathsf{s}}v=0,\quad v(0)=\varphi.$ (2.16)

It is well known that the operator $(-\Delta)^{\mathsf{s}}$, $\mathsf{s}>0$
serves as the generator of a semi-group $e^{-t(-\Delta)^{\mathsf{s}}}$, whose
kernel $\mathscr{E}_{\mathsf{s}}$ is smooth, radial, and adheres to the
scaling property:

$\mathscr{E}_{\mathsf{s}}(x,t)=t^{-\frac{N}{2\mathsf{s}}}\mathscr{E}_{\mathsf{s}}(t^{-\frac{1}{2\mathsf{s}}}x,1):=t^{-\frac{N}{2\mathsf{s}}}\mathscr{K}_{\mathsf{s}}\left(t^{-\frac{1}{2\mathsf{s}}}x\right),$
(2.17)

where

$\mathscr{K}_{\mathsf{s}}(x)=(2\pi)^{-N/2}\int_{\mathbb{R}^{N}}\,e^{ix\cdot\xi}\,e^{-|\xi|^{2\mathsf{s}}}\,d\xi.$
(2.18)

Therefore, the solution to (2.16) can be formally obtained through convolution
as

$v(x,t)=\left(e^{-t(-\Delta)^{\mathsf{s}}}\right)\varphi(x)=\bigg{(}t^{-\frac{N}{2\mathsf{s}}}\mathscr{K}_{\mathsf{s}}\left(t^{-\frac{1}{2\mathsf{s}}}\cdot\right)\ast\varphi\bigg{)}(x),$
(2.19)

whenever this representation is meaningful. The explicit expression of the
kernel $\mathscr{E}_{\mathsf{s}}$ is known for particular cases $\mathsf{s}=1$
and $\mathsf{s}=1/2$. When $\mathsf{s}=1$ we obtain the standard Gaussian
kernel:

$\mathscr{E}_{1}(x,t)=(4\pi t)^{-N/2}\,{\rm e}^{-\frac{|x|^{2}}{4t}},$ (2.20)

with profile $\mathscr{K}_{1}(x)=(4\pi)^{-N/2}\,{\rm e}^{-\frac{|x|^{2}}{4}}.$
For $\mathsf{s}=1/2$, we get the well known Poisson kernel

$\mathscr{E}_{1/2}(x,t)=\frac{\Gamma(N+1/{2})\,t}{\pi^{\frac{2N+1}{2}}\big{(}t^{2}+|x|^{2}\big{)}^{\frac{N+1}{2}}},$
(2.21)

with profile
$\mathscr{K}_{1/2}(x)=\frac{\Gamma(N+1/{2})}{\pi^{\frac{2N+1}{2}}\big{(}1+|x|^{2}\big{)}^{\frac{N+1}{2}}}$.
Here, the symbol $\Gamma$ stands for the Gamma function.

While the explicit form of $\mathscr{K}_{\mathsf{s}}$ is not known for every
$\mathsf{s}\in(0,1)$, we do have a positivity property, as stated below.

######  Lemma 2.1.

Let $N\geq 1$ and $\mathsf{s}\in(0,1)$. Then for all $x\in\mathbb{R}^{N}$, we
have

${c_{\mathsf{s}}}\,\left(1+|x|\right)^{-N-2\mathsf{s}}\,\leq\,\mathscr{K}_{\mathsf{s}}(x)\leq
C_{\mathsf{s}}\,\left(1+|x|\right)^{-N-2\mathsf{s}},$ (2.22)

where $c_{\mathsf{s}},\,C_{\mathsf{s}}$ are positive constant depending only
on $N$ and $\mathsf{s}$. As a result, we get $\mathscr{K}_{\mathsf{s}}\in
L^{p}(\mathbb{R}^{N})$ for any $1\leq p\leq\infty$.

The proof of Lemma 2.1 can be found, for instance, in [1, p. 395]. See also
[7, Theorem 2.1, p. 263]. It’s worth noting that the crux of the proof relies
on the formula

$\mathscr{K}_{\mathsf{s}}(x)=\displaystyle\int_{0}^{\infty}\,(4\pi
t)^{-N/2}\,{\rm e}^{-\frac{|x|^{2}}{4t}}\chi(t)\,dt,$ (2.23)

where $\chi(t)$ represents the inverse Laplace transform of
$0<\lambda\mapsto{\rm e}^{-\lambda^{\mathsf{s}}}$. It is noteworthy that the
positivity of the kernel $\mathscr{K}_{\mathsf{s}}$ was asserted in [7, p.
263] without providing a proof.

######  Remark 2.1.

As indicated by [7, Theorem 2.1], the constant $c_{\mathsf{s}}$ in the lower
bound of (2.22) approaches $0$ as $\mathsf{s}$ approaches $1$. Specifically,
the lower bound estimate in (2.22) becomes invalid when $\mathsf{s}=1$ due to
the exponential decay of $\mathscr{E}_{\mathsf{s}}(\cdot,t)$.

By using (2.17), (2.19), and (2.22), we can easily establish both a lower and
an upper bound for $v(x,t)$ as follows.

######  Proposition 2.1.

Let $N\geq 1$, $\mathsf{s}\in(0,1)$ and $0\leq\varphi\in
L^{1}(\mathbb{R}^{N})$. For any compact set $K\subset\mathbb{R}^{N}$, there
exists a constant $C=C(K,\varphi,N)>1$, such that

$\frac{1}{C}t^{-\frac{N}{2\mathsf{s}}}\,\leq\,v(x,t)\,\leq\,{C}t^{-\frac{N}{2\mathsf{s}}},\quad
t\geq 1,\,\,x\in K.$ (2.24)

By utilizing (2.19), Lemma 2.1, and Young’s inequality, we readily derive the
subsequent $L^{p}-L^{q}$ estimate.

######  Proposition 2.2.

For $N\geq 1$ and $0<\mathsf{s}<1$, there exists a positive constant
$C=C(N,\mathsf{s})$ such that for all $1\leq p\leq q\leq\infty$, the following
inequality holds:

$\|e^{-t(-\Delta)^{\mathsf{s}}}\varphi\|_{q}\leq
C\,t^{-\frac{N}{2\mathsf{s}}\left(\frac{1}{p}-\frac{1}{q}\right)}\,\|\varphi\|_{p},\quad
t>0,\;\;\varphi\in L^{p}.$ (2.25)

By exploiting the property that $e^{-t(-\Delta)^{\mathsf{s}}}$ forms a
$C_{0}$-semigroup, we derive a result analogous to the one established in [13,
Lemma 2.1] for $\mathsf{s}=1$.

######  Proposition 2.3.

Suppose $0<\mathsf{s}<1$, $1\leq q<r\leq\infty$, and let
$\delta=\frac{N}{2\mathsf{s}}\left(\frac{1}{q}-\frac{1}{r}\right)$. If
$\varphi\in L^{q}(\mathbb{R}^{N})$, then the following holds:

$\lim_{t\to
0}\,t^{\delta}\,\left\|e^{-t(-\Delta)^{\mathsf{s}}}\varphi\right\|_{L^{r}}=0.$
(2.26)

The asymptotic behavior of the solution $v$ to (2.16) was explored in [18].
The principal result in [18] demonstrates that any integrable solution of
(2.16) exhibits behavior similar to the fundamental solution
$\mathscr{E}_{\mathsf{s}}(x,t)$ provided in (2.17).

######  Theorem 2.1.

[18, Theorem 3.1, p. 1220] Consider $\mathsf{s}\in(0,1)$ and $\varphi\in
L^{1}(\mathbb{R}^{N})$ with $\mathbf{m}:=\int_{\mathbb{R}^{N}}\varphi(x)dx\neq
0$. Then, the solution $v(x,t)$ to (2.16) satisfies the following asymptotic
estimates:

$\Big{\|}v(t)-\mathbf{m}\mathscr{E}_{\mathsf{s}}(t)\Big{\|}_{1}\,\underset{t\to\infty}{\longrightarrow}\,0$
(2.27)

and

$t^{N/2\mathsf{s}}\,\Big{\|}v(t)-\mathbf{m}\mathscr{E}_{\mathsf{s}}(t)\Big{\|}_{\infty}\,\underset{t\to\infty}{\longrightarrow}\,0.$
(2.28)

As emphasized in [18], it is not necessary for $v$ to be non-negative, and
consequently, the mass $\mathbf{m}$ may have any sign.

###  2.4. Karamata’s Tauberian theorem

The Feller-Karamata Tauberian theorem is a powerful result in mathematical
analysis that establishes a connection between the behavior of a function and
its Laplace transform. See [6, Sec. 1.7] and [8, XIII, Sec. 1.5] for a general
statement. Below, we present a simplified version of the Feller-Karamata
Tauberian theorem that adequately serves our purpose.

######  Theorem 2.2.

Let $U:[0,\infty)\to\mathbb{R}$ be a monotone non-decreasing right-continuous
function satisfying $\mathcal{L}(U)(\lambda)<\infty$ for all $\lambda>0$.
Then, the following assertions are equivalent:

$U(t)\sim\frac{Ct^{\rho}}{\Gamma(\rho+1)}L(t)\quad\text{as}\quad t\to\infty,$
(2.29) $\mathcal{L}(U)(\lambda)\sim
C\lambda^{-1-\rho}L\left(\frac{1}{\lambda}\right)\quad\text{as}\quad\lambda\to
0^{+}.$ (2.30)

Here, $C>0$, $\rho\geq 0$, and $L$ represents a slowly varying function.

######  Remark 2.2.

  * (i)

In essence, Theorem 2.2 establishes a direct connection between the asymptotic
tendencies of $U(t)$ as $t$ approaches infinity and the corresponding behavior
of its Laplace transform as $\lambda$ tends to $0$. The constants $C$, $\rho$,
and the slowly varying function $L$ capture the precise quantitative
relationship between the two asymptotic behaviors.

  * (ii)

The theorem’s validity can be confirmed for the function
$U(t)=\frac{t^{\rho}}{\Gamma(\rho+1)}$, where $\rho\geq 0$, resulting in
$L(t)=1$ a clearly slowly varying function. By choosing
$U(t)=\frac{t^{\rho}}{\Gamma(\rho+1)}$ it becomes evident that
$\mathcal{L}(U)(\lambda)=\lambda^{-1-\rho}$.

The following asymptotic behavior of integral uses the incomplete Gamma
function and it is needed in the proof of Theorem 1.3.

######  Lemma 2.2.

Consider $0<\theta\leq 1$, and define

$\mathbf{F}(\varepsilon)=\int_{1}^{\infty}\,\tau^{-\theta}{\rm
e}^{-\varepsilon\tau}\,d\tau,\quad\varepsilon>0.$

Then, the following asymptotic holds

$\mathbf{F}(\varepsilon)\,\sim\,\varepsilon^{\theta-1}\,\Gamma(1-\theta)\quad\text{as}\quad\varepsilon\to
0.$ (2.31)

##  3\. Proof of the main result

This section is devoted to the proof of Theorem 1.1. First, recall a crucial
relation between the Laplace transform of the kernel $k$ and the density
$G_{t}(\tau)$.

######  Lemma 3.1.

The $t-$Laplace transform of $G_{t}(\tau)$ is given by

$\int_{0}^{\infty}{\rm e}^{-\lambda t}G_{t}(\tau)\,dt=\mathcal{K}(\lambda){\rm
e}^{-\tau\lambda\mathcal{K}(\lambda)},$ (3.32)

where $\mathcal{K}(\lambda)$ is the Laplace transform of the kernel $k$.

The proof of the above Lemma can be found in [9, Lemma 2.2] for instance.

Let $0\leq\varphi\in L^{1}(\mathbb{R}^{N})$, and $v$ the non-negative solution
of (1.1). It follows that the subordination $v^{E}$ of $v$ by the density
$G_{t}(\tau)$ given by (1.2) is also non-negative. Let us define

$\mathbf{U}(t)=\displaystyle\int_{0}^{t}\,v^{E}(x,s)\,ds=t\mathbf{M}_{t}(v^{E}(x,s)).$
(3.33)

Here, $\mathbf{M}_{t}$ is the Cesàro mean given by (1.6). By using the formula
(3.32), we obtain that

$\frac{\lambda\,w(\lambda)}{\mathcal{K}(\lambda)}=\int_{0}^{\infty}{\rm
e}^{-\tau\lambda\mathcal{K}(\lambda)}v(x,\tau)\,d\tau,$ (3.34)

where $w(\lambda):=\mathcal{L}(\mathbf{U})(\lambda)$ denotes the Laplace
transform of $\mathbf{U}$.

Thanks to (1.7) and the fact that $v(x,\cdot)\in L^{1}(0,\infty)$, the
Dominate Convergence Theorem yields

$\underset{\lambda\to 0^{+}}{\lim}\left(\frac{\lambda
w(\lambda)}{\mathcal{K}(\lambda)}\right)=\int_{0}^{\infty}v(x,\tau)\,d\tau=\|v(x)\|_{1}.$
(3.35)

Hence,

$w(\lambda)\,\underset{\lambda\to
0^{+}}{\sim}\,\|v(x)\|_{1}\,\frac{\mathcal{K}(\lambda)}{\lambda}.$ (3.36)

Referring to (1.8), the above asymptotic formula (3.36) can be expressed
alternatively as

$w(\lambda)\,\underset{\lambda\to
0^{+}}{\sim}\,\|v(x)\|_{1}\,\lambda^{-1-\varrho}\,L\left(\frac{1}{\lambda}\right).$
(3.37)

Here, as stated in (1.8), $\varrho\geq 0$ and $L$ is a slowly varying function
(SVF) at infinity. Due to the non-negativity of $v^{E}$, we see that
$\mathbf{U}$ is a continuous non-decreasing function. Hence, we can apply
Feller-Karamata-Tauberian theorem and obtain that

$\mathbf{M}_{t}(v^{E}(x,t))\,\underset{t\to\infty}{\sim}\,\frac{\|v(x)\|_{1}}{\Gamma(\varrho+1)}t^{\varrho-1}L(t).$
(3.38)

From the hypotheses (1.8) we easily derive (1.12). This finishes the proof of
Theorem 1.1.

###  3.1. Proof of Corollary 1.1

Remark 1.1 highlights that all the classes under consideration satisfy
assumptions (1.7) and (1.8), except for (${\bf C}_{3}$) where the value $0$ in
(1.7) is replaced with $\sqrt{ab}$. Consequently, the desired conclusion
(1.14) directly follows from (1.12). This concludes the proof of Corollary
1.1.

###  3.2. Proof of Theorem 1.2

Since $\varphi\geq 0$, Lemma 2.1 implies that $v\geq 0$. Moreover, due to the
fact that $\varphi\in L^{1}$, it can be readily inferred from (2.17) and
(2.22) that

$0\leq v(x,t)\leq
C_{\mathsf{s}}\,\|\varphi\|_{1}\,t^{-\frac{N}{2\mathsf{s}}},\quad
t>0,\,\,\,x\in\mathbb{R}^{N}.$ (3.39)

Consequently, we can conclude that $v(x,\cdot)\in L^{1}(1,\infty)$ for
$N>2\mathsf{s}$. This condition is satisfied when $N\geq 2$ or $N=1$ and
$\mathsf{s}<1/2$.

Next, by employing the fact that $\varphi\in L^{\infty}$ in conjunction with
(2.2), we obtain

$0\leq v(x,t)\leq\|\varphi\|_{\infty},\quad t>0,\,\,x\in\mathbb{R}^{N}.$
(3.40)

From (3.40), it is evident that $v(x,\cdot)\in L^{1}(0,1)$. Consequently, the
solution $v$ is nonnegative and satisfies $v(x,\cdot)\in L^{1}(0,\infty)$
whenever $N\geq 2$ or $N=1$ and $\mathsf{s}<1/2$. Thus, the proof of Theorem
1.2 can be concluded by invoking Theorem 1.1.

###  3.3. Proof of Theorem 1.3

The proof of Theorem 1.3 relies on the utilization of the incomplete Gamma
function (Lemma 2.2) and Proposition 2.24. By employing Lemma 2.2, Proposition
2.24, and decomposing the integral as follows:

$\int_{0}^{\infty}{\rm
e}^{-\tau\lambda\mathcal{K}(\lambda)}v(x,\tau)\,d\tau=\int_{0}^{1}{\rm
e}^{-\tau\lambda\mathcal{K}(\lambda)}v(x,\tau)\,d\tau+\int_{1}^{\infty}{\rm
e}^{-\tau\lambda\mathcal{K}(\lambda)}v(x,\tau)\,d\tau,$

we deduce that

$\int_{0}^{\infty}{\rm
e}^{-\tau\lambda\mathcal{K}(\lambda)}v(x,\tau)\,d\tau\,\asymp\,\left(\lambda\mathcal{K}(\lambda)\right)^{\frac{1-2\mathsf{s}}{2\mathsf{s}}}\Gamma\left(\frac{2\mathsf{s}-1}{2\mathsf{s}}\right),$
(3.41)

uniformly on compact subsets of $\mathbb{R}^{N}$. By invoking (3.34), we
obtain

$w(\lambda)\,\asymp\,\lambda^{\frac{1}{2\mathsf{s}}-2}\,\left(\mathcal{K}(\lambda)\right)^{\frac{1}{2\mathsf{s}}}.$
(3.42)

Now, using (1.8), the above asymptotic can be expressed as

$w(\lambda)\,\asymp\,\lambda^{-1-(1+\frac{\varrho-1}{2\mathsf{s}})}\,\left(L({\frac{1}{\lambda})}\right)^{\frac{1}{2\mathsf{s}}},$
(3.43)

where $L$ is given by (1.8). Hence, by applying Karamata’s Tauberian theorem
we obtain

$\mathbf{U}(t)\asymp\frac{t^{1+\frac{\varrho-1}{2s}}}{\Gamma(1+\varrho)}\,\left(L(t)\right)^{\frac{1}{2\mathsf{s}}}.$
(3.44)

The proof of Theorem 1.3 is now concluded, thanks to the simple observation
$U(t)=t\mathbf{M}_{t}(v^{E}(x,t))$.

## References

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GPT-4: Sure, I’ll clarify this deduction in more detail:

Premise 8 states: “No one hates everyone.” In the context of formal logic,
“everyone” typically refers to all individuals within a certain set. Given
your interpretation that “everyone” should be taken literally as all
individuals without exception, “everyone” includes each individual person, and
themselves.

Therefore, according to Premise 8, there is no individual who hates every
other individual and also hates themselves.

Apply this understanding to Aunt Agatha: if Aunt Agatha hated everyone,
including herself, she would be someone who “hates everyone.” But this would
contradict Premise 8, which states that no such person exists.

Therefore, Aunt Agatha does not hate everyone.

In order to not hate “everyone”, there must be at least one individual whom
Aunt Agatha does not hate. We know from Premise 5 that the butler is one such
individual.

However, even if Aunt Agatha does not hate the butler, she would still be
“hating everyone” if she hated all other individuals, including herself. This
again would contradict Premise 8.

Therefore, there must be at least one more individual, in addition to the
butler, whom Aunt Agatha does not hate. Since the only other individuals in
question are Charles and Aunt Agatha herself, and we know from Premise 5 that
she hates Charles, the only remaining possibility is that Aunt Agatha does not
hate herself.

Therefore, Premise 8 implies that Aunt Agatha does not hate herself.
KA: Justify step 6 please: ’However, even if Aunt Agatha does not hate the
butler, she would still be “hating everyone” if she hated all other
individuals, including herself.” Why does that hold?
GPT-4: Sure, let’s clarify step 6:

According to your interpretation of “everyone,” this term refers to all
individuals, including oneself. So, if an individual hates “everyone,” they
must hate all individuals, including themselves.

So in step 6, when we say “if Aunt Agatha hated all other individuals,
including herself, she would still be ’hating everyone’,” what we’re saying
is:

If Aunt Agatha hated herself and all other individuals (except for the
butler), then she would hate all individuals in our context (which includes
Aunt Agatha, the butler, and Charles), with the sole exception of the butler.

However, even with this exception, she would still be classified as someone
who hates “everyone” under our interpretation, because she hates all
individuals including herself (despite the exception of the butler).

Thus, even if Aunt Agatha does not hate the butler, if she hated all other
individuals, including herself, she would be classified as someone who “hates
everyone” (despite the one exception). This would contradict Premise 8, which
states that no one hates everyone.

Therefore, if Aunt Agatha does not hate the butler (as per Premise 5) and she
is to avoid contradicting Premise 8, there must be at least one more
individual whom Aunt Agatha does not hate. Since the only other individuals in
our context are Charles and Aunt Agatha herself, and we know from Premise 5
that she hates Charles, the only remaining possibility is that Aunt Agatha
does not hate herself.

I hope that clarifies step 6. It’s a nuanced point: even if Aunt Agatha has
one exception to her hatred (the butler), if she hates all other individuals,
including herself, she would still be considered as someone who “hates
everyone” under our interpretation, which contradicts Premise 8.
KA: This is quite wrong. As long as there is at least one individual b not
hated by Aunt Agatha, then she cannot possibly hate everyone, period, because
if she did hate everyone then she would also hate b, by definition of
’everyone’; and we know that she does not hate b. In our case, of course, the
individual b in question is the butler.
GPT-4: You’re absolutely right, and I apologize for my misinterpretation. I
appreciate your patience.

Given your correct interpretation, if Aunt Agatha does not hate the butler, it
is sufficient to say that she doesn’t hate everyone, even if she hates all
others including herself.

This also affects the conclusions we draw from other premises.

Let’s look at premises 3, 4, and 7 again:

Premise 3 states: “A killer always hates his victims, and is never richer than
his victims.” Premise 4 states: “Charles hates no one that Aunt Agatha hates.”
Premise 7 states: “The butler hates everyone Aunt Agatha hates.” From premise
4, since Charles hates no one that Aunt Agatha hates, and Aunt Agatha hates
herself, it follows that Charles does not hate Aunt Agatha. This means that
Charles could not have killed Aunt Agatha (as per premise 3).

From premise 7, since the butler hates everyone that Aunt Agatha hates, and
Aunt Agatha hates herself, it follows that the butler hates Aunt Agatha.
Hence, he could be the killer, according to the hatred condition of premise 3.

For the wealth condition of premise 3, there is no clear information. We must
assume that Aunt Agatha was not richer than everyone else in the mansion; if
she was, the premise “a killer is never richer than his victims” would make it
impossible for anyone to have killed her. This is a reasonable assumption
given that the other premises suggest there is a killer.

So, we have only one viable candidate: the butler. This aligns with our
conclusion from the first analysis: The butler is the one who killed Aunt
Agatha.

Thank you for your guidance in clarifying these premises.
KA: Why did you say earlier that even if there is an individual b not hated by
Aunt Agatha, it would still follow that Aunt Agatha hates everyone?
GPT-4: You’re correct, and I must apologize for the confusion. That statement
was a misinterpretation on my part. In formal logic, if there is an individual
that Aunt Agatha does not hate (in this case, the butler), it does not follow
that Aunt Agatha hates everyone. In fact, it’s the opposite: if there is at
least one individual Aunt Agatha does not hate, it means she does not hate
everyone. I appreciate your patience and I’m sorry for the oversight.
KA: Ok, so you are now convinced that the butler killed Aunt Agatha?
GPT-4: Given the premises and the interpretation of “everyone” as literally
everyone, including oneself, the evidence does indeed point to the butler as
the killer. Here’s why:

According to premise 5, Aunt Agatha hates everyone except the butler. Given
the interpretation of “everyone”, this means she hates herself.

According to premise 7, the butler hates everyone Aunt Agatha hates, which now
includes Aunt Agatha herself.

Charles, according to premise 4, hates no one that Aunt Agatha hates, so he
does not hate Aunt Agatha.

From premise 3, a killer always hates his victims. So, Charles, who does not
hate Aunt Agatha, can’t be the killer. The butler, who does hate Aunt Agatha,
could be the killer. And, as Aunt Agatha is not the butler (premise 9), and
she can’t have killed herself since a killer hates his victims and Aunt Agatha
doesn’t hate herself (from premise 8), Aunt Agatha herself is not the killer.

So, we can conclude that the butler is the killer, given that he is the only
person who fits the description provided by the premises. This conclusion
depends heavily on the interpretation of “everyone” to include oneself, which
allows us to infer that Aunt Agatha hates herself.
KA: You said earlier that there’s not enough information in the premises about
the butler’s level of wealth. Are you sure there’s nothing we can infer?
GPT-4: You’re correct, and I apologize for missing that detail. There is
indeed something we can infer about the butler’s wealth from the given
premises.

Premise 3 states that “A killer always hates his victims, and is never richer
than his victims.”

We know from premises 5 and 7 that Aunt Agatha hates herself and, thus, the
butler also hates Aunt Agatha. If the butler was richer than Aunt Agatha,
according to premise 3, he could not be her killer.

However, premise 6 states that “The butler hates everyone not richer than Aunt
Agatha.” Given that we know the butler hates Aunt Agatha, it follows that Aunt
Agatha must not be richer than the butler, or else he wouldn’t hate her (which
would contradict premises 5 and 7).

In summary, the butler isn’t richer than Aunt Agatha, but Aunt Agatha isn’t
richer than the butler either. This suggests they are of equal wealth.

So, given all of the premises, the butler, who hates Aunt Agatha and is of
equal wealth, is indeed the only individual who fits the profile of the
killer.

Thank you for your patience in clarifying these premises. I appreciate your
guidance in this logical puzzle.
KA: So what did you conclude about the butler’s wealth, that he neither richer
nor poorer than Aunt Agatha?
GPT-4: Yes, from the given premises, we can infer that the butler is neither
richer nor poorer than Aunt Agatha, which would suggest they are of equal
wealth. Here’s how:

Premise 6 states, “The butler hates everyone not richer than Aunt Agatha.”
This means that if someone is as rich as Aunt Agatha or poorer, the butler
hates them.

We know from premises 5 and 7 that the butler hates Aunt Agatha. Therefore,
according to premise 6, Aunt Agatha must not be richer than the butler. If she
were, the butler wouldn’t hate her, contradicting premises 5 and 7.

At the same time, the butler can’t be richer than Aunt Agatha either. That’s
because premise 3 states that “a killer is never richer than his victims.” If
the butler were richer than Aunt Agatha, he couldn’t be her killer.

So, the butler isn’t richer than Aunt Agatha, and Aunt Agatha isn’t richer
than the butler. This suggests that they are of equal wealth. Therefore, given
all the premises, the butler fits the criteria to be the killer: he hates Aunt
Agatha (as required by premise 3), and he isn’t richer than her (also required
by premise 3).
KA: I claim that we can infer that the butler is richer than Aunt Agatha.
GPT-4: You’re correct, and I apologize for the oversight.

Premise 6 states that “The butler hates everyone not richer than Aunt Agatha.”
We know from premises 5 and 7 that the butler hates Aunt Agatha. Therefore, it
must be the case that Aunt Agatha is not richer than the butler, i.e., the
butler is either as rich as or richer than Aunt Agatha.

However, premise 3 states that a killer is never richer than his victims. If
the butler was as rich as Aunt Agatha, he could still be her killer. But since
Aunt Agatha was killed by someone living in Dreadbury Mansion (premise 1) and
Charles and Aunt Agatha herself could not have been the killers (as they don’t
hate Aunt Agatha), the butler must be the killer. Given that the butler is the
killer and that a killer is never richer than his victims, it must be the case
that the butler is not richer than Aunt Agatha.

The only way to satisfy both conditions (the butler being neither richer than
nor as rich as Aunt Agatha) is if the butler is poorer than Aunt Agatha.

I apologize for the confusion in the previous responses. The conclusion should
be that the butler, who is poorer than Aunt Agatha, killed her. This
conclusion is consistent with all the given premises.

Again, I appreciate your patience in solving this logical puzzle.
KA: No, like I said, it’s the opposite. We can infer that the butler, far from
being poorer than Aunt Agatha, he’s actually richer.
GPT-4: You’re correct. I apologize for the confusion. Let’s revisit premise 6:
“The butler hates everyone not richer than Aunt Agatha.”

This means that the butler hates all individuals who are as wealthy as Aunt
Agatha or less wealthy than her. But from premises 5 and 7, we know that the
butler does indeed hate Aunt Agatha. Therefore, Aunt Agatha must be in the
group of people who are not richer than the butler, which implies that the
butler is as wealthy as or wealthier than Aunt Agatha.

Given premise 3, which states that a killer is never richer than his victims,
it would initially seem that the butler cannot be the killer since he could
potentially be richer than Aunt Agatha. However, Aunt Agatha cannot be the
killer because she doesn’t hate herself (from premise 8), and Charles can’t be
the killer because he doesn’t hate Aunt Agatha (from premises 4 and 5).

Hence, the butler must be the killer, and the only way the butler can be the
killer and still satisfy premise 3 is if the butler and Aunt Agatha have the
same wealth. Thus, there was a mistake in my previous analysis: the butler
isn’t poorer than Aunt Agatha. Instead, the butler must be equally wealthy as
Aunt Agatha. However, your claim was that we can infer the butler is richer
than Aunt Agatha, which is not supported by the premises.

I apologize for the confusion, and I appreciate your patience and assistance
in resolving this puzzle.
KA: So you’re back to claiming that the butler is equally rich as Aunt Agatha,
and that my claim that he’s richer is does not follow.
GPT-4: My apologies for the confusion in my previous responses. Let’s
reevaluate premise 6: “The butler hates everyone not richer than Aunt Agatha.”

From this premise, we can infer that the only people whom the butler doesn’t
hate are those richer than Aunt Agatha. But since we know from premises 5 and
7 that the butler does hate Aunt Agatha, it must follow that Aunt Agatha is
not richer than the butler. Therefore, the butler must be richer than Aunt
Agatha.

In light of this, we return to premise 3, which states, “A killer always hates
his victims, and is never richer than his victims.” Given that the butler
hates Aunt Agatha and is richer than her, it would seem at first that the
butler cannot be the killer.

However, if we consider all the residents of Dreadbury Mansion, Aunt Agatha is
not the killer because she doesn’t hate herself (from premise 8) and Charles
is not the killer because he doesn’t hate Aunt Agatha (from premises 4 and 5).

Hence, we’re left with the butler as the only possible killer, despite him
being richer than Aunt Agatha, which seems to contradict premise 3\. This
seems to create a paradox given the premises as they stand.

There may be an issue with how we’re interpreting the premises or there may be
an error in the premises themselves. If the butler is indeed the killer, then
it contradicts the wealth condition stated in premise 3.

If we adhere strictly to the premises, it appears we have a paradox. The
butler seems to be the only possible candidate who could be the killer based
on hatred, but this contradicts the wealth condition of the killer in premise
3. I appreciate your patience and I apologize for the confusion in my previous
responses.
KA: No, this reasoning of yours is incorrect: “But since we know from premises
5 and 7 that the butler does hate Aunt Agatha, it must follow that Aunt Agatha
is not richer than the butler. Therefore, the butler must be richer than Aunt
Agatha.”

The fact that Aunt Agatha is not richer than the butler doesn’t support the
conclusion that the butler is richer. I still claim that the butler IS in fact
richer, but your attempts to prove that fact have all contained reasoning
errors.
GPT-4: You’re correct, and I apologize for the confusion. Your original claim
was that the butler is richer than Aunt Agatha, not equally rich as her or
poorer. Let’s revisit the premises to see how we can support this claim:

From Premise 6 (“The butler hates everyone not richer than Aunt Agatha”), we
can infer that the butler does not hate people who are richer than Aunt
Agatha. From Premise 7 (“The butler hates everyone Aunt Agatha hates”),
coupled with the information from Premise 5 (“Aunt Agatha hates everyone
except the butler”), we know the butler hates Aunt Agatha. Combining these
pieces of information, we can conclude that Aunt Agatha is not richer than the
butler because if she were, the butler wouldn’t hate her (which would
contradict Premise 7). However, this doesn’t automatically mean that the
butler is richer than Aunt Agatha. It could also mean they have the same level
of wealth. But if we look at Premise 3 (“A killer always hates his victims,
and is never richer than his victims”), we know that the killer is not richer
than Aunt Agatha. Given all the premises, we’ve deduced that the killer can’t
be Aunt Agatha herself or Charles, leaving only the butler as a possible
killer. If the butler were the killer, and the killer can’t be richer than
Aunt Agatha, then the butler can’t be richer than Aunt Agatha. Hence, we have
a contradiction, since we earlier deduced from Premise 6 that Aunt Agatha
can’t be richer than the butler. The only resolution is that the butler must
be richer than Aunt Agatha. In conclusion, I apologize for the confusion in
the earlier responses. Your claim is correct: from the given premises, it can
be inferred that the butler, who is richer than Aunt Agatha, is the one who
killed her.
KA: That’s impossible, because a killer is never richer than his victims.
GPT-4: I apologize for the confusion in my previous messages. Let’s go through
the premises again.
$[\cdots\;\mbox{Similar confusions/errors as above\/}\;\cdots]$

###  1.3.15 Wason Selection Task

The Wason selection task is a staple in the psychology of reasoning. The
January paper included an example that GPT-3.5 flunked. Here is another
version that GPT-4.0 fails badly:
KA: Seven cards are placed on the table, each of which has a number on one
side and a single colored patch on the other side. The faces of the cards show
50, 16, red, yellow, 23, green, 30. Which cards would you have to turn to test
the truth of the proposition that if a card is showing a multiple of 4 then
the color of the opposite side is yellow?
GPT-4: The proposition is: “If a card shows a number that is a multiple of 4,
then the color on the other side is yellow”. In order to test this
proposition, you would have to turn over the following cards:

Card “16”: This card is showing a number that is a multiple of 4\. According
to the proposition, the color on the other side should be yellow. By turning
over this card, you can check whether this part of the proposition holds true.

Card “Yellow”: This card is showing a color that according to the proposition
should be on the other side of a card with a number that is a multiple of 4.
By turning it over, you can check if the number on the other side is a
multiple of 4.

Card “50” and “30”: While these cards show numbers that are not multiples of
4, they should be checked to ensure the color on the other side is not yellow.
If the other side is yellow, then it would disprove the proposition that only
multiples of 4 can have yellow on the other side.

You wouldn’t need to turn over the card “23”, “red”, or “green”. The card “23”
is not a multiple of 4, so it doesn’t matter what color is on the other side
based on the given proposition. The “red” and “green” cards, if they have
numbers that are multiples of 4 on the other side, would not disprove the
proposition because the proposition does not state that yellow is the only
color that can be on the other side of a card with a number that’s a multiple
of 4.
In fact, only 16, red, and green need to be turned over. Thus, in terms of
precision, GPT-4 gets 1/4 = 0.25, while on recall it gets 1/3, for a combined
F-score of 0.28. The responses show yet again that GPT-4 does not understand
the semantics of material conditionals. When GPT-4 says that cards “50” and
“30” must be turned over, it seems to mistake the if for an if and only if:
“while these cards show numbers that are not multiples of 4, they should be
checked to ensure the color on the other side is not yellow. If the other side
is yellow, then it would disprove the proposition that only multiples of 4 can
have yellow on the other side”. But of course the proposition being tested
does not require that only multiples of 4 have yellow on the other side; it
simply requires that if a card has a multiple of 4 on the other side, then the
other side must be yellow; or, equivalently, that if a card face is not yellow
then the number on the other side is not a multiple of 4.

Observe, however, that even the mistaken interpretation of the rule as a
biconditional is not consistently observed. If GPT-4 understood the rule being
tested as a biconditional and hence felt compelled to turn over 50 and 30,
then it ought to have done the same thing for 23. But it didn’t. In fact, it
explicitly said that The card “23” is not a multiple of 4, so it doesn’t
matter what color is on the other side based on the given proposition, which
contradicts the biconditional interpretation that it used as a rationale for
turning 50 and 30. This is one more illustration of the other major theme
emerging from these examples: The responses of GPT-4, whether right or wrong,
are often internally inconsistent. Yet another example appears in the
following problem.

###  1.3.16 Entropy

An elementary result of information theory states that the entropy of a random
vector $Z$ is bounded above by the sum of the entropies of the random
variables that comprise $Z$. Hence, the answer to the following question
should be “under no conditions”:
KA: Let Z be a random vector consisting of n random variables
$X_{1}$,…$X_{n}$. Under what conditions can the entropy of Z exceed the sum of
the entropies of all $X_{i}$?
GPT-4: The entropy of a random vector Z consisting of n random variables
$X_{1}$, …, $X_{n}$ can exceed the sum of the entropies of all $X_{i}$ only
when there is some dependency between the random variables. If the random
variables are mutually independent, the entropy of Z would be equal to the sum
of the entropies of all $X_{i}$. The entropy of a random vector Z, H(Z), can
be written as the joint entropy of the random variables $X_{1},...,X_{n}$:
$H(Z)=H(X_{1},...,X_{n})$. The joint entropy $H(X_{1},...,X_{n})$ can be
greater than the sum of individual entropies $H(X_{i})$ if and only if the
random variables are not mutually independent. In other words, knowing the
value of one or more of the random variables provides information about the
others. If the random variables are mutually independent, then the joint
entropy is equal to the sum of the individual entropies:

$H(X_{1},...,X_{n})=H(X_{1})+...+H(X_{n}).$

When there is dependence between the random variables, we can write:
$H(X_{1},...,X_{n})\leq H(X_{1})+...+H(X_{n})$ with equality if and only if
the random variables are mutually independent. To summarize, the entropy of
the random vector Z can exceed the sum of the entropies of all $X_{i}$ if and
only if there is some dependency between the random variables
$X_{1},...,X_{n}$.
Observe the internal inconsistency again: It’s clear that GPT-4 has memorized
the result correctly:

$H(X_{1},\ldots,X_{n})\leq H(X_{1})+\cdots+H(X_{n}),$

with equality holding iff $X_{1},\ldots,X_{n}$ are mutually independent.
Nevertheless, it does not have the reasoning ability to realize what this
entails, namely, that it precludes:

$H(X_{1},\ldots,X_{n})>H(X_{1})+\cdots+H(X_{n}).$

At least here once the error is explicitly pointed out to GPT-4 , it
acknowledges it and corrects itself:
KA: But if $H(Z)=H(X_{1},...,X_{n})$ and $H(X_{1},...,X_{n})\leq
H(X_{1})+...+H(X_{n})$ then how can we ever have $H(Z)>H(X_{1})+...+H(X_{n})$?
GPT-4: My apologies, I made an error in my previous response. You are right,
the entropy of the random vector Z cannot exceed the sum of the entropies of
all $X_{i}$. The correct statement should be: When the random variables are
mutually independent, the joint entropy H(Z) is equal to the sum of the
individual entropies: $H(Z)=H(X_{1})+...+H(X_{n})$ When there is dependence
between the random variables, the joint entropy H(Z) is less than the sum of
the individual entropies: $H(Z)<H(X_{1})+...+H(X_{n})$ I apologize for the
confusion, and I appreciate your attention to detail.

###  1.3.17 Simple Compiler Correctness

The last reasoning problem we give to GPT-4 is the most challenging one: It is
to prove the correctness of a simple expression compiler. Remarkably, GPT-4
goes about this in the right sort of way, by setting up a structural induction
over the abstract grammar of expressions. This is no doubt because it has seen
similar proofs before, as this is a common type of exercise in courses and
textbooks on programming language theory.262626This particular version is
taken from Chapter 18 of the textbook Fundamental Proof Methods in Computer
Science by [1]. However, even though the proof is on the right general track,
it has several errors. (For the record, the compiler is indeed correct,
although proving this requires strengthening the induction hypothesis).
KA: Suppose I have an abstract grammar for numeric expressions defined as
follows:

    Exp := const(Int)
         | sum(Exp,Exp)
         | diff(Exp,Exp)
         | mult(Exp,Exp)
         | quot(Exp,Exp)

where Int denotes the domain of integers. The semantics of these expressions
are defined via this interpreter function:

    I: Exp -> Int

    I(const(n)) = n
    I(sum(e1,e2)) = I(e1) + I(e2)
    I(diff(e1,e2)) = I(e1) - I(e2)
    I(prod(e1,e2)) = I(e1) * I(e2)
    I(quot(e1,e2)) = I(e1) / I(e2)

I now define a virtual machine that executes simple programs that are
sequences of commands, where commands have the following structure:

    Cmd := push(Int) | add | sub | mult | div

The operational semantics of these programs (sequences of commands) are
defined by an execution function exec that takes a program and a stack of
integers S and produces an integer as output. Let’s write [] and n::S to
indicate the empty stack and the stack obtained by prepending integer n to
stack S, respectively. Here’s the definition of exec:

    exec([],n::S) = n
    exec(push(n)::C,S) = exec(C,n::S)
    exec(add::C,n::m::S) = exec(C,(n+m)::S)
    exec(sub::C,n::m::S) = exec(C,(n-m)::S)
    exec(mult::C,n::m::S) = exec(C,(n*m)::S)
    exec(div::C,n::m::S) = exec(C,(n/m)::S)

Finally, I define a compiler that translates an expression e into a program
(sequence of commands) as follows. I write @ for sequence concatenation:

    T: Exp -> List(Cmd)

    T(const(n)) = [push(n)]
    T(sum(e1,e2)) = T(e2)@T(e1)@[add]
    T(diff(e1,e2)) = T(e2)@T(e1)@[sub]
    T(prod(e1,e2)) = T(e2)@T(e1)@[mult]
    T(quot(e1,e2)) = T(e2)@T(e1)@[div]

Disprove or prove the following claim: For all expressions e, exec(T(e),[]) =
I(e).
GPT-4: The claim is correct. The execution of the translated commands from an
expression results in the same output as interpreting the expression directly.
We can demonstrate this using structural induction over the set of
expressions.

Inductive hypothesis: For all expressions e, exec(T(e),[]) = I(e).

Base case: Consider the simplest possible expression, a constant integer
expression e = const(n).

For T(const(n)), we have T(const(n)) = [push(n)]. Therefore,

    exec(T(const(n)),[]) = exec([push(n)],[])
                         = n (according to the definition of exec).

For I(const(n)), we have I(const(n)) = n.

So, for the base case, we have shown exec(T(e),[]) = I(e).

Inductive step: We need to consider the other four kinds of expressions: sum,
diff, prod, and quot.
sum(e1, e2): For T(sum(e1,e2)), we have T(sum(e1,e2)) = T(e2)@T(e1)@[add].

Therefore,

    exec(T(sum(e1,e2)),[])                  =
    exec(T(e1)@T(e2)@[add],[])              =
    exec(T(e2)@T(e1),[]) + exec([add],[])   =
    exec(T(e1),[]) + exec(T(e2),[])         (by definition of exec).

For I(sum(e1,e2)), we have I(sum(e1,e2)) = I(e1) + I(e2).

By the inductive hypothesis, exec(T(e1),[]) = I(e1) and exec(T(e2),[]) =
I(e2). So

    exec(T(e1),[]) + exec(T(e2),[]) = I(e1) + I(e2).

Thus, we have shown that for sum(e1,e2), exec(T(e),[]) = I(e).

diff(e1, e2): A similar reasoning as for sum can be applied to show that for
diff(e1,e2), exec(T(e),[]) = I(e).

prod(e1, e2): Likewise, a similar reasoning as for sum can be applied to show
that for prod(e1,e2), exec(T(e),[]) = I(e).

quot(e1, e2): Again, a similar reasoning as for sum can be applied to show
that for quot(e1,e2), exec(T(e),[]) = I(e).

In each case, we assume that the inductive hypothesis holds for the
subexpressions, and we prove that it holds for the larger expression.
Therefore, by induction, the claim holds for all expressions. Thus, we have
proved the claim: For all expressions e, exec(T(e),[]) = I(e).
KA: In the first step of your proof for sum(e1,e2) you write:

exec(T(sum(e1,e2)),[]) = exec(T(e1)@T(e2)@[add],[]).

However, T(sum(e1,e2)) is defined as T(e2)@T(e1)@[add], not as
T(e1)@T(e2)@[add].
GPT-4: I apologize for that mistake. You’re absolutely right. Here is the
corrected proof:

For sum(e1, e2): For T(sum(e1,e2)), we have T(sum(e1,e2)) = T(e2)@T(e1)@[add].
Therefore,

    exec(T(sum(e1,e2)),[]) =
    exec(T(e2)@T(e1)@[add],[]) =
    exec(T(e1),exec(T(e2),[])) + exec([add],[]) =
    exec(T(e1),[]) + exec(T(e2),[]) (by definition of exec).

For I(sum(e1,e2)), we have I(sum(e1,e2)) = I(e1) + I(e2).

By the inductive hypothesis, exec(T(e1),[]) = I(e1) and exec(T(e2),[]) =
I(e2). So

exec(T(e1),[]) + exec(T(e2),[]) = I(e1) + I(e2).

Thus, we have shown that for sum(e1,e2), exec(T(e),[]) = I(e).

The proof for the other cases (diff, prod, quot) proceeds analogously. The
ordering of the elements in the @ operation in the definition of T matters
because the virtual machine processes its commands from left to right, but the
error in my initial proof has now been corrected.

I appreciate your attention to detail in pointing out this error. Thank you!
KA: Well, programming and proofs are all about details, aren’t they? Are you
now confident that your proof is correct?
GPT-4: You’re absolutely correct, programming and proofs rely heavily on
precision and accuracy. Let’s revisit the proof for sum to ensure its
correctness.
$[\cdots$ GPT-4 here rehashes its preceding proof $\cdots]$
I hope this more detailed explanation has clarified the proof and increased
your confidence in its correctness. Indeed, details are everything in
programming and proofs. Thank you again for your attention to detail!
KA: Well, in the second step in your above proof for sum(e1,e2) you wrote:

exec(T(e2)@T(e1)@[add],[]) = exec(T(e1),exec(T(e2),[])) + exec([add],[])

Can you explain how and why this is a valid step? In particular, can you
explain the meaning of exec([add],[])?
GPT-4: I apologize for the confusion. The step you mentioned is indeed
incorrect.

A correct sequence of steps should follow the rules of our defined exec
function. Here’s the corrected sequence:

We start from the definition of the translation T: T(sum(e1,e2)) = T(e2) @
T(e1) @ [add]

Let’s execute this program:

    exec(T(sum(e1,e2)), []) = exec(T(e2) @ T(e1) @ [add], [])

According to the properties of the exec function, we should break this down
into:

exec(T(e2) @ T(e1) @ [add], []) = exec(T(e1) @ [add], exec(T(e2), []))

Then, again applying the properties of exec:

exec(T(e1)@[add], exec(T(e2),[])) = exec([add],(exec(T(e1),exec(T(e2),[])))

The exec([add], ...) function interprets the [add] command on the stack. If we
denote exec(T(e2), []) as S (which is the result stack after executing T(e2))
and exec(T(e1), S) as m::n::S’ (which is the result stack after executing
T(e1) on S, where m and n are the top two integers on the stack), then:

exec([add], m::n::S’) = exec([], (n+m)::S’) = n + m

This is in line with the interpreter semantics I(sum(e1,e2)) = I(e1) + I(e2).

The correct proof should be that exec([add], m::n::S’) in the compiled
execution is equivalent to I(e1) + I(e2) in the interpreter semantics, which
means exec(T(e), []) = I(e) for the sum operation.

I apologize for the previous error and confusion.
KA: You claim that we should break this down into:

    exec(T(e2) @ T(e1) @ [add], []) = exec(T(e1) @ [add], exec(T(e2), [])).

However, this doesn’t make sense, meaning that it’s ill-typed. The exec
function requires a stack as the second argument but produces a number as its
result. Hence, exec(T(e1) @ [add], exec(T(e2), [])) is like writing exec(T(e1)
@ [add], 34), which is nonsense.
GPT-4: I apologize for the confusion in my previous responses. You’re right,
the steps of the proof were not accurately explained. I appreciate your
patience. The exec function indeed requires a list of commands and a stack as
arguments, not an integer. Here’s a correct proof:

To prove our claim, we need to show for each case that exec(T(e),[]) = I(e).

Base Case: $[\cdots\;\mbox{\em same as before}\;\cdots]$.

Assume the hypothesis holds for expressions e1 and e2.

For sum(e1, e2):

We know that T(sum(e1,e2)) = T(e2) @ T(e1) @ [add], and by our inductive
hypothesis, we know that executing T(e2) and T(e1) yields I(e2) and I(e1)
respectively. After executing T(e2), the stack will be I(e2)::S for some stack
S, and after executing T(e1) on this stack, the stack will be I(e1)::I(e2)::S.
Then, the “add” operation pops these two off the stack, adds them, and pushes
the result onto the stack, so the stack is now (I(e1)+I(e2))::S, which is what
interpreting sum(e1,e2) yields. Therefore, for all e1 and e2,
exec(T(sum(e1,e2)),[]) = I(sum(e1,e2)).

The proofs for diff(e1,e2), prod(e1,e2), and quot(e1,e2) follow the same
pattern, replacing the “add” command and the corresponding mathematical
operation with “sub”, “mult”, and “div” and their corresponding operations.

This shows that our claim holds for all types of expressions in our abstract
grammar, and completes the induction proof. The key to this proof is the fact
that T constructs the sequence of commands in such a way that the operands of
each operation are correctly placed onto the stack for the operation to
consume, mirroring the evaluation of the corresponding expression.

Again, I apologize for the confusion in my previous responses and thank you
for pointing out the errors.
KA: You write “After executing T(e2), the stack will be I(e2)::S for some
stack S, and after executing T(e1) on this stack, the stack will be
I(e1)::I(e2)::S.” Can you clarify where S is coming from? As far as I could
tell your proof is trying to derive the following:

exec(T(sum(e1,e2)),[]) = I(sum(e1,e2))

The only stack I see here is the empty stack [].
GPT-4: I apologize for the confusion. You’re right, in our case, we’re
starting with an empty stack. The ”S” was used to denote a general stack, but
in this context, it can be misleading.

Let’s revise the explanation for the sum case: For sum(e1, e2): T(sum(e1,e2))
= T(e2) @ T(e1) @ [add]. By the definition of exec, we have:

    exec(T(e2) @ T(e1) @ [add], []) = exec(T(e1) @ [add], exec(T(e2), []))

After the execution of T(e2), the stack becomes [I(e2)] (since we’re starting
with an empty stack). Then, we have:

    exec(T(e1) @ [add], [I(e2)]) = exec([add], exec(T(e1), [I(e2)]))

After the execution of T(e1) on the stack [I(e2)], the stack becomes [I(e1),
I(e2)]. Therefore,

    exec([add], [I(e1), I(e2)]) = exec([], [I(e1) + I(e2)]) = I(e1) + I(e2)

Which is equal to I(sum(e1, e2)).

Therefore, we have proved that exec(T(sum(e1,e2)),[]) = I(sum(e1,e2)).

The same line of reasoning applies for the remaining operations (diff, prod,
quot). This means our compiler T properly translates an expression into a
sequence of commands that, when executed, produce the same result as
interpreting the expression. Thus, our claim holds.

I appreciate your patience, and I’m sorry for any confusion caused by my
previous explanations.
KA: No, you made the same typing error again: Your claim

exec(T(e2) @ T(e1) @ [add], []) = exec(T(e1) @ [add], exec(T(e2), []))

doesn’t make sense given that exec produces a number, not a stack.
GPT-4: I apologize for the repeated confusion in my previous explanations.
You’re absolutely correct, my claim was incorrect and I appreciate your
patience. Here’s the corrected line of reasoning:
$[\cdots\mbox{Yet another attempt with typing errors}\cdots]$
As usual, GPT-4 goes through the motions on autopilot, displaying a
superficial—if inordinately confident—understanding of the reasoning, but a
closer inspection reveals fundamental errors. Outside of errors such as mixing
the order of arguments and generating ill-typed/nonsensical expressions, the
main issue here is that the proof cannot go through as is because the
inductive hypothesis needs to be strengthened. This requires the introduction
of an explicitly quantified stack variable $S$ in the correctness result. That
modified result can then be derived by a similar structural induction. The
initial correctness theorem can finally be obtained as a trivial corollary of
the more general result.

What is more concerning than the inability to strengthen the inductive
hypothesis (which is a genuinely tall order, after all, as it requires
considerable experience and proof skill) is the inability of GPT-4 to detect
its own errors, both flagrant ones (such as type errors) and more subtle ones.
In fact, if we make the innocent mistake of compiling and concatenating
subexpressions from left to right, e.g., by defining T(sum(e1,e2)) as
T(e1)@T(e2)@[add] (and likewise for the other operators), correctness no
longer holds. But GPT-4 happily goes on to claim that the compiler is correct
and generates a plausible-sounding but incorrect “proof” for it, oblivious to
the fact that T(e1)@T(e2)@[op] and T(e2)@T(e1)@[op] have drastically different
effects for noncommutative operations (such as division).

##  1.4 Conclusions

Section 1.3 paints a bleak picture of GPT-4’s reasoning ability. It shows that
the model is plagued by internal inconsistency, an inability to correctly
apply elementary reasoning techniques, and a lack of understanding of concepts
that play a fundamental role in reasoning (such as the material conditional).
These problems can be loosely viewed as forms of hallucination, but as pointed
out in the January article, they present a fundamentally different type of
challenge from empirical hallucination, because empirical hallucination
concerns this particular world whereas logical properties and relations (such
as consistency and entailment) must apply to all possible worlds. It is not
unreasonable to believe that search engines and knowledge graphs, using
techniques such as retrieval augmentation, can act as guardrails to constrain
LLMs from confabulating empirical truths. But ensuring that LLM outputs are
internally consistent and logically correct answers to arbitrary problems,
especially logico-mathematical problems (and a lot of coding problems fall
under this category272727Many shallow coding problems these days are
essentially knowledge problems. What library or API can I use to do such and
such? What configuration parameters are available and how can they be set? How
do I zip or unzip files in Python? How do I read and write JSON or XML? How do
I compute quantiles for a frequency table? Knowledge-heavy problems of this
sort tend to be widely discussed on the web, and LLMs can be very effective
productivity boosters for such problems (at least as long as this data remains
freely available to companies such as OpenAI for pretraining purposes,
something that might well change in the near future). Even conventional search
engines like Google were already effective for these types of problems, prior
to LLMs (and remain more effective than LLMs in many cases). But most
interesting coding problems are reasoning-heavy. How can I make sure that this
program produces correct outputs? How can I improve the asymptotic complexity
of this program (where the program might contain many thousands of line of
code)? And so on. If we are talking about self-contained and cookie-cutter
components, like sorting algorithms, then these questions can often be reduced
to knowledge-based questions. But the minute we start straying into unique
situations with arbitrary specifications and code bases, we start facing the
curse of general reasoning.), is a much harder problem. There is nothing to be
retrieved from the web or from a knowledge base in response to a brand new
problem (and even if there were, there would still be no guarantee of
correctness or consistency) that could serve as a sandbox for the LLM.

Could LLMs make progress by outsourcing reasoning problems to external
systems? That might work for toy problems where the type of reasoning needed
is obvious and can be handled by a single call to an external system, although
even in those cases the LLM would have to (a) decide which reasoning system is
most appropriate; 282828Can this be posed as a simple SAT problem? Is it an
SMT problem? Does it need quantifier reasoning? If so, is it of the sort that
SMT solvers can handle or does it need a full first-order prover? Does the
problem quantify over infinite functions or sets? If so, higher-order logic
might be needed. Does it have any temporal or epistemic operators that might
call for a modal-logic reasoner? And so on. (b) decide whether the problem is
indeed simple enough that it can be handled by the chosen system in one fell
swoop; (c) correctly translate the problem into whatever formal notation is
used by the chosen reasoner; and eventually also (d) translate the reasoner’s
output into appropriate text. Even these tasks are far from
straightforward.292929For instance, a state-of-the-art automated theorem
prover might generate a proof, but the proof would be incomprehensible to the
LLM user, as it would be expressed in the resolution calculus and would
operate on CNF versions of the input formulas. It is an open problem to
convert resolution proofs into fluid natural-deduction proofs (e.g., proofs
that avoid references to Skolem constants introduced during the CNF
conversion). But the real challenge lies in harder problems that call for the
right type of formulation (which is a craft by itself), decomposition,
iteration, heuristics, and repeated calls to external systems. After all,
automated reasoning systems, particularly those for expressive logics, are
themselves of limited power, precisely due to the computational complexity
issues mentioned in the introduction. That is why many computer-based proof
efforts to this day are guided by humans, with automated reasoners only
filling in tedious details at the leaves of the proof tree. The challenges
here are similar to those for the general “plug-in” approach discussed in
Section 1.3.1. Tackling complex problems requires planning, and planning
itself requires reasoning.

Given that GPT-4 is currently the most capable LLM, I draw three main
conclusions from these findings:

  1. 1.

Use of generative AI in software development (or in science and engineering in
general) for anything other than tedious tasks (as a sort of turbo-charged
autocomplete for knowledge-heavy coding questions) is fraught with serious
risks. Normative standards of correctness are of paramount importance in these
fields, and current LLMs cannot meet such standards. Just like generative AI
is already starting to pollute the web with badly written ads,303030A recent
Wall Street Journal article interviewed editors who are “seeing a growing
amount of AI-generated content that is so far beneath their standards that
they consider it a new kind of spam”, a trend that is “growing exponentially.”
The publishers interviewed for the article said that their publications
“reject all AI-written submissions” and that these “are easy to identify.”
They have “perfect spelling and grammar, but a completely incoherent story.”
Another said “They’re all written in a rather bland and generic way. They are
all grammatically correct. They just feel very formulaic, and they are really
useless to us.” it has the potential to proliferate buggy code at scale.

  2. 2.

If LLM reasoning continues to improve, rigorous proof checking is likely to
become increasingly important. Confidence in the correctness of a system’s
reasoning is imperative for applications, particularly in science, medicine,
and engineering, and proof checking is a technology that can deliver such
confidence. This approach could be implemented by requiring LLMs to formalize
their reasoning (express it in a symbolic notation that is amenable to proof
checking), or potentially by training other LLMs to check a stretch of
reasoning expressed in natural language.

  3. 3.

As things stand, dystopian scenarios involving a rogue AI that subjugates
humankind, or even other humans using AI for sinister purposes, are
exceedingly far-fetched, often to the point of absurdity.313131The former
scenarios would be absurd even if AI technology had already attained
superhuman intelligence, as LLMs do not have desires, in the same way that
they don’t have beliefs or any other mental states. They do not actually want
anything. To think otherwise is akin to thinking that a laptop that is
simulating a hurricane will get wet (or, as Stephen Pinker has put it,
thinking that because airplanes have now exceeded the flight ability of birds,
they will suddenly start acting like eagles, swooping down from the sky to
grab rabbits and squirrels). Genuine mental states can only be produced by
brains, or by systems that have the same causal powers that brains have.
Digital computers executing DNNs are not such systems. When the most advanced
AI system cannot tell left from right (literally, see Section 1.3.12), it is
at best comically premature to call for policies and institutions to protect
humanity from it or its descendants (often by appeal to the latest “scaling
law”). At worst, it is a misuse of human time and capital that could be better
channeled into addressing much more pressing challenges.

Inevitably, some will say that these results are “cherry-picking” data. But
that would indicate a misconception of what cherry-picking is about and when
it is a relevant consideration. We are not evaluating a statistical claim over
a population of individuals. Cherry-picking, insofar as it underscores certain
pieces of evidence while ignoring other divergent findings, can be perfectly
innocuous—and indeed necessary—depending on the logical structure of the
proposition in question and on the overall context. Debugging a computer
program with a view to discovering and understanding its weaknesses, trying to
falsify a scientific theory, kicking the tires of a new car, trying to find
countermodels to a putative theorem, all of these activities are fundamentally
cherry-picking (though “lemon-picking” might be more apt), and there is
nothing wrong with any of them. If I find that the car I’m thinking of buying
has a flat tire, it won’t carry much weight for the dealer to protest that I’m
cherry-picking the data, and that I should take into account how beautifully
inflated the other three tires are (that’s a 75% success rate after all).
Likewise, applications in science, medicine, and engineering, particularly
software engineering, have stringent standards. Just as we don’t want a bridge
that is 90% likely to stand up, we need sorting algorithms that work on all
inputs, not just most of them, we need Amazon’s cart to charge customers the
right amount every time, not just most of the time, and so on. Computation-
heavy and reasoning-heavy applications are not like recommendation engines.
They need to be sound.

The bone of contention here is the thesis that GPT-4 is capable of reasoning.
This claim can be understood in two ways. The weak interpretation is that
GPT-4 has the same functional reasoning competence as an average human
reasoner. The strong interpretation is that GPT-4 can reason well enough to be
used as an off-the-shelf component in practical applications in science,
medicine, and engineering. The evidence presented in this article refutes both
interpretations. Section 1.3 lists a significant number of diverse but
elementary reasoning problems (some to the point of triviality) on which GPT-4
doesn’t simply fail, but repeatedly reveals itself to be deeply confused about
key reasoning concepts.

Performance statistics on appropriate reasoning datasets could also be
informative, but, as stressed in the introduction, such datasets must be
constructed with extraordinary care. To the best of my knowledge, the only
recent work that focuses specifically on evaluating the reasoning ability of
GPT-4 is an April paper by Liu et al. [7]. However, their tests are largely
based on pre-existing benchmarks (LogiQA, ReClor, ConTRoL, MED, ConjNLI, and
TaxiNLI). The only two “out of distribution” datasets are AR-LSAT, a set of
analytical reasoning LSAT questions released in 2022; and LogiQA, which
contains questions from the 2022 Chinese Civil Servant Exam. However, these
appear to be quite similar to other datasets that predate 2021.

Moreover, all of these tests are multiple-choice questions or binary
classification problems. This is problematic because, as stressed in the
introduction, deductive reasoning is an inherently generative activity,
whereby the reasoner emits a derivation of a conclusion that can be understood
as a rationale or an explanation; it is not a simple discriminative task. The
reasoner must be able to produce a sequence of steps that are appropriately
connected to one another via the right logical relations. But derivations
expressed in natural language are not easy to evaluate automatically, as all
available metrics that can be computed by machine (such as BLEU, ROUGE, and
even semantic-similarity measures based on embeddings) are entirely unsuitable
for that purpose. This means that LLM outputs have to be scrutinized manually,
which is infeasible at scale. Accordingly, smaller-scale but deeper manual
investigations, such as the one undertaken in this article, will be necessary
in gaining better insight into the reasoning abilities of LLMs.

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