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Analogical proportions in monounary algebras

Christian Antić [email protected]
Vienna, Austria
Abstract.

This paper studies analogical proportions in monounary algebras consisting only of a universe and a single unary function. We show that the analogical proportion relation is characterized in the infinite monounary algebra formed by the natural numbers together with the successor function via difference proportions.

1. Introduction

Analogical proportions are expressions of the form “aa is to bb what cc is to dd” written a:b::c:da:b::c:d at the core of analogical reasoning which itself is at the core of artificial general intelligence (e.g. ?, ?, ?, ?, ?, ?, ?, ?, ?, ?).

The purpose of this paper is to study ?’s (?) abstract algebraic framework of analogical proportions — recently introduced in the general setting of universal algebra — in monounary algebras containing only a single unary function.

The motivation for studying proportions in that specific context is that monounary algebras are simple enough to provide a convenient context to analyze interesting concepts like congruences in combination with proportions, and rich enough to yield interesting novel insights.

The rest of the paper is structured as follows.

In §3, we repeat that every monounary algebra satisfies — as an instance of the general framework — for all elements of the domain (cf. Theorem 4):

  • p-symmetry: a:b::c:dc:d::a:ba:b::c:d\quad\Leftrightarrow\quad c:d::a:b,

  • inner p-symmetry: a:b::c:db:a::d:ca:b::c:d\quad\Leftrightarrow\quad b:a::d:c,

  • inner p-reflexivity: a:a::c:ca:a::c:c,

  • p-reflexivity: a:b::a:ba:b::a:b,

  • p-determinism: a:a::a:dd=aa:a::a:d\quad\Leftrightarrow\quad d=a.

To the contrary, we will provide counterexamples showing that there are monounary algebras and elements such that:

  • central permutation: a:b::c:d⇎a:c::b:da:b::c:d\quad\not\Leftrightarrow\quad a:c::b:d,

  • strong inner p-reflexivity: a:a::c:d⇏d=ca:a::c:d\quad\not\Rightarrow\quad d=c,

  • strong p-reflexivity: a:b::a:d⇏d=aa:b::a:d\quad\not\Rightarrow\quad d=a,

  • p-commutativity: a:b:b:aa:b\not::b:a,

  • p-transitivity: a:b::c:dandc:d::e:f⇏a:b::e:fa:b::c:d\quad\text{and}\quad c:d::e:f\quad\not\Rightarrow\quad a:b::e:f,

  • inner p-transitivity: a:b::c:dandb:e::d:f⇏a:e::c:fa:b::c:d\quad\text{and}\quad b:e::d:f\quad\not\Rightarrow\quad a:e::c:f.

This is in line with what we have observed in the general context of arbitrary algebras in Theorem 28 in ? (?).

Moreover, in §4 we shall prove that analogical proportions and congruences are in general not compatible in the following sense:111This observation was the motivation for introducing proportional congruences in ? (?). In each of the following cases, there is a monounary algebra 𝔄=(A,S)\mathfrak{A}=(A,S), a congruence θ\theta on 𝔄\mathfrak{A}, and elements a,b,c,dAa,b,c,d\in A such that:

  • a:b::c:da:b::c:d whereas a/θ:b/θ:c/θ:d/θa/\theta:b/\theta\not::c/\theta:d/\theta (Theorem 7).

  • a/θ:b/θ::c/θ:d/θa/\theta:b/\theta::c/\theta:d/\theta whereas a:b:c:da:b\not::c:d (Theorem 8).

  • a:b:a/θ:b/θa:b\not::a/\theta:b/\theta (Theorem 9).

  • aθba\theta b and cθdc\theta d whereas a:b:c:da:b\not::c:d (Theorem 10).

In §5, on the other hand, we shall prove a positive result showing that in the monounary algebra (,S)(\mathbb{N},S) consisting of the natural numbers together with the unary successor function, we obtain the Difference Proportion Theorem 11 saying that within the abstract framework we have

a:b::c:d holds in (,S)ab=cd.\displaystyle\text{$a:b::c:d$\quad holds in $(\mathbb{N},S)$}\quad\Leftrightarrow\quad a-b=c-d.

This is more surprising than it might appear, as the abstract framework is not explicitly geared towards proportions in monounary algebras.

In a broader sense, this paper is a further step towards a mathematical theory of analogical proportions and analogical reasoning in general.

2. Analogical proportions in monounary algebras

In this section, we interpret the abstract algebraic framework of analogical proportions in ? (?) within monounary algebras of the form 𝔄=(A,S)\mathfrak{A}=(A,S), where AA is a universe and S:AAS:A\to A is a unary function.

We shall now recall the abstract algebraic framework of analogical proportions in ? (?) where we restrict ourselves to monounary algebras of the form 𝔄=(A,S)\mathfrak{A}=(A,S), where AA is a set and S:AAS:A\to A is a unary function on AA (we can imagine SS to be a generalized “successor” function). Terms in 𝔄\mathfrak{A} have the form SkzS^{k}z, for k0k\geq 0.

In what follows, let 𝔄=(A,SA)\mathfrak{A}=(A,S_{A}) and 𝔅=(B,SB)\mathfrak{B}=(B,S_{B}) be two monounary algebras — we will always omit the subscripts from notation and simply write SS instead of SAS_{A} and SBS_{B}.

We define the analogical proportion entailment relation in two steps:222We use here the updated notation of ? (?) where we write \uparrow instead of JusJus for sets of justifications; see ? (?) for motivation.

  1. (1)

    Define the set of justifications of an arrow aba\to b in 𝔄\mathfrak{A} by

    𝔄(ab):={SkzSz|ab=SkoSo, for some oA},\displaystyle\uparrow_{\mathfrak{A}}(a\to b):=\left\{S^{k}z\to S^{\ell}z\;\middle|\;a\to b=S^{k}o\to S^{\ell}o,\text{ for some $o\in A$}\right\},

    extended to an arrow proportion ab:⋅cda\to b:\joinrel\cdot\,c\to d333Read as “aa transforms into bb as cc transforms into dd”. in (𝔄,𝔅)(\mathfrak{A,B}) by

    (𝔄,𝔅)(ab:⋅cd):=𝔄(ab)𝔅(cd).\displaystyle\uparrow_{(\mathfrak{A,B})}(a\to b:\joinrel\cdot\,c\to d):=\ \uparrow_{\mathfrak{A}}(a\to b)\ \cap\uparrow_{\mathfrak{B}}(c\to d).

    A justification is trivial in (𝔄,𝔅)(\mathfrak{A,B}) iff it justifies every arrow proportion in (𝔄,𝔅)(\mathfrak{A,B}), and we say that JJ is a trivial set of justifications in (𝔄,𝔅)(\mathfrak{A,B}) iff every justification in JJ is trivial.

    Now we say that ab:⋅cda\to b:\joinrel\cdot\,c\to d holds in (𝔄,𝔅)(\mathfrak{A,B}) — in symbols,

    (𝔄,𝔅)ab:⋅cd\displaystyle(\mathfrak{A,B})\models a\to b:\joinrel\cdot\,c\to d

    iff

    1. (a)

      either 𝔄(ab)𝔅(cd)\uparrow_{\mathfrak{A}}(a\to b)\ \cup\uparrow_{\mathfrak{B}}(c\to d) consists only of trivial justifications, in which case there is neither a non-trivial relation from aa to bb in 𝔄\mathfrak{A} nor from cc to dd in 𝔅\mathfrak{B}; or

    2. (b)

      (𝔄,𝔅)(ab:⋅cd)\uparrow_{(\mathfrak{A,B})}(a\to b:\joinrel\cdot\,c\to d) is maximal with respect to subset inclusion among the sets (𝔄,𝔅)(ab:⋅cd)\uparrow_{(\mathfrak{A,B})}(a\to b:\joinrel\cdot\,c\to d^{\prime}), dBd^{\prime}\in B, containing at least one non-trivial justification, that is, for any element dBd^{\prime}\in B,444We ignore trivial justifications and write “\emptyset\subsetneq\ldots” instead of “{trivial justifications}\{\text{trivial justifications}\}\subsetneq\ldots” et cetera.

      (𝔄,𝔅)(ab:⋅cd)\displaystyle\emptyset\subsetneq\ \uparrow_{(\mathfrak{A,B})}(a\to b:\joinrel\cdot\,c\to d) (𝔄,𝔅)(ab:⋅cd)\displaystyle\subseteq\ \uparrow_{(\mathfrak{A,B})}(a\to b:\joinrel\cdot\,c\to d^{\prime})

      implies

      (𝔄,𝔅)(ab:⋅cd)(𝔄,𝔅)(ab:⋅cd).\displaystyle\emptyset\subsetneq\ \uparrow_{(\mathfrak{A,B})}(a\to b:\joinrel\cdot\,c\to d^{\prime})\subseteq\ \uparrow_{(\mathfrak{A,B})}(a\to b:\joinrel\cdot\,c\to d).

      We abbreviate the above definition by simply saying that (𝔄,𝔅)(ab:⋅cd)\uparrow_{(\mathfrak{A,B})}(a\to b:\joinrel\cdot\,c\to d) is dd-maximal.

  2. (2)

    Finally, the analogical proportion entailment relation is most succinctly defined by

    a:b::c:d:\displaystyle a:b::c:d\quad:\Leftrightarrow\quad ab:⋅cdandba:⋅dc\displaystyle a\to b:\joinrel\cdot\,c\to d\quad\text{and}\quad b\to a:\joinrel\cdot\,d\to c
    cd:⋅abanddc:⋅ba.\displaystyle c\to d:\joinrel\cdot\,a\to b\quad\text{and}\quad d\to c:\joinrel\cdot\,b\to a.

    This means that in order to prove (𝔄,𝔅)a:b::c:d(\mathfrak{A,B})\models a:b::c:d, we need to check the first two relations in the first line with respect to \models in (𝔄,𝔅)(\mathfrak{A,B}), and the last two relations in the same line in (𝔅,𝔄)(\mathfrak{B,A}).

We will always write 𝔄\mathfrak{A} instead of (𝔄,𝔄)(\mathfrak{A,A}) and we often omit the explicit reference to the underlying algebras.

Let :={n,n+1,n+2,}\mathbb{N}:=\{n,n+1,n+2,\ldots\}. Given some justification SkzSz𝔄(ab)S^{k}z\to S^{\ell}z\in\ \uparrow_{\mathfrak{A}}(a\to b), k,0k,\ell\geq 0, we can depict the two cases kk\leq\ell and k\ell\leq k as follows:

ooa=Skoa=S^{k}ok:k\leq\ell:b=Sob=S^{\ell}oSkS^{k}SkS^{\ell-k}oob=Sob=S^{\ell}ofor some oAo\in A.k:\ell\leq k:a=Skoa=S^{k}oSS^{\ell}SkS^{k-\ell}

This is an abstract version of the situation in (,S)(\mathbb{N},S) where, for example, SzS2z(,S)(12)Sz\to S^{2}z\in\ \uparrow_{(\mathbb{N},S)}(1\to 2) and S2zSz(,S)(21)S^{2}z\to Sz\in\ \uparrow_{(\mathbb{N},S)}(2\to 1) both have the following pictorial representation:

01=S01=S02=S202=S^{2}0SSSS
Example 1.

Consider the monounary algebra

aabbccddSSSSSSSS

We expect a:b::c:da:b::c:d to fail as it has no non-trivial justification. In fact,

(ab)(cd)={zSz|1}\displaystyle\uparrow(a\to b)\ \cup\uparrow(c\to d)=\left\{z\to S^{\ell}z\;\middle|\;\ell\geq 1\right\}\neq\emptyset

and

(ab:⋅cd)=\displaystyle\uparrow(a\to b:\joinrel\cdot\,c\to d)=\emptyset

show

a:b:c:d.\displaystyle a:b\not::c:d.
Example 2.

Consider the monounary algebra

11223344SSSSSS

The relation between 11 and 22 is a “loop”, whereas the relation between 33 and 44 is not as there is no edge from 44 back to 33; instead, there is a loop at 44. We therefore expect the following relations:

(1) 1:2:3:4,\displaystyle 1:2\not::3:4,
(2) 1:2::4:4.\displaystyle 1:2::4:4.

To prove the first item, we shall show

(3) 21:⋅ 43.\displaystyle 2\to 1\not:\joinrel\cdot\,4\to 3.

For this, compute

(21)\displaystyle\uparrow(2\to 1) ={SkzSz| 21=SkoSo, for some o,k,}\displaystyle=\left\{S^{k}z\to S^{\ell}z\;\middle|\;2\to 1=S^{k}o\to S^{\ell}o,\text{ for some $o$},k,\ell\in\mathbb{N}\right\}
={SkzSz|(k is even and  is odd) or (k is odd and  is even)}\displaystyle=\left\{S^{k}z\to S^{\ell}z\;\middle|\;\text{($k$ is even and $\ell$ is odd) or ($k$ is odd and $\ell$ is even)}\right\}

and

(43)={Skzz|k1}{SkzSz|k,}=(44).\displaystyle\uparrow(4\to 3)=\left\{S^{k}z\to z\;\middle|\;k\geq 1\right\}\subsetneq\left\{S^{k}z\to S^{\ell}z\;\middle|\;k,\ell\in\mathbb{N}\right\}=\ \uparrow(4\to 4).

We thus have

(21:⋅ 44)=(21)(44)=𝔄(21)\displaystyle\uparrow(2\to 1:\joinrel\cdot\,4\to 4)=\ \uparrow(2\to 1)\ \cap\uparrow(4\to 4)=\ \uparrow_{\mathfrak{A}}(2\to 1)

whereas

(21:⋅ 43)=(21)(43)={Skzz|k is odd}(21:⋅ 44).\displaystyle\uparrow(2\to 1:\joinrel\cdot\,4\to 3)=\ \uparrow(2\to 1)\ \cap\uparrow(4\to 3)=\left\{S^{k}z\to z\;\middle|\;\text{$k$ is odd}\right\}\subsetneq\ \uparrow(2\to 1:\joinrel\cdot\,4\to 4).

This shows (3) and therefore (1).

To prove (2), we need to show the following relations:

(4) 12:⋅ 44,21:⋅ 44,44:⋅ 12,44:⋅ 21.\displaystyle 1\to 2:\joinrel\cdot\,4\to 4,\quad 2\to 1:\joinrel\cdot\,4\to 4,\quad 4\to 4:\joinrel\cdot\,1\to 2,\quad 4\to 4:\joinrel\cdot\,2\to 1.

The first two relations are immediate consequences of the following observation:

(ab:⋅ 44)=(ab),for all a,b{1,2}.\displaystyle\emptyset\subsetneq\ \uparrow(a\to b:\joinrel\cdot\,4\to 4)=\ \uparrow(a\to b),\quad\text{for all $a,b\in\{1,2\}$}.

To prove the last two relations, we first compute

(11)={SkzSz|(k and  are even) or (k and  are odd)}\displaystyle\uparrow(1\to 1)=\left\{S^{k}z\to S^{\ell}z\;\middle|\;\text{($k$ and $\ell$ are even) or ($k$ and $\ell$ are odd)}\right\}
(12)={SkzSz|(k is even and  is odd) or (k is odd and  is even)}\displaystyle\uparrow(1\to 2)=\left\{S^{k}z\to S^{\ell}z\;\middle|\;\text{($k$ is even and $\ell$ is odd) or ($k$ is odd and $\ell$ is even)}\right\}
(13)=,\displaystyle\uparrow(1\to 3)=\emptyset,
(14)=,\displaystyle\uparrow(1\to 4)=\emptyset,
(21)=(12),\displaystyle\uparrow(2\to 1)=\ \uparrow(1\to 2),
(22)=(11),\displaystyle\uparrow(2\to 2)=\ \uparrow(1\to 1),
(23)=,\displaystyle\uparrow(2\to 3)=\emptyset,
(24)=.\displaystyle\uparrow(2\to 4)=\emptyset.

Now

(44:⋅ab)=(ab),for all a,b{1,2},\displaystyle\uparrow(4\to 4:\joinrel\cdot\,a\to b)=\ \uparrow(a\to b),\quad\text{for all $a,b\in\{1,2\}$},

shows that

(44:⋅ 12)=(12)and(44:⋅ 21)=(21)\displaystyle\uparrow(4\to 4:\joinrel\cdot\,1\to 2)=\ \uparrow(1\to 2)\quad\text{and}\quad\uparrow(4\to 4:\joinrel\cdot\,2\to 1)=\ \uparrow(2\to 1)

are both non-empty and maximal with respect to the last argument. This shows (4) and thus (2).

Define the monounary algebra ({0,1},S)(\{0,1\},S) by S0:=1S0:=1 and S1:=0S1:=0. We can imagine 0 and 11 to be boolean truth values and SS to be negation. On the other hand, in (,S)(\mathbb{N},S) let Sa:=a+1Sa:=a+1 be the unary successor function on the natural numbers ={0,1,2,}\mathbb{N}=\{0,1,2,\ldots\} starting at 0. We can depict the algebras as:

01122\vdots011SSSSSSSS

The next result shows how we can capture evenness and oddness via analogical proportions between the two algebras.

Theorem 3.
((,S),({0,1},S))a:b::c:d(c=d\displaystyle((\mathbb{N},S),(\{0,1\},S))\models a:b::c:d\quad\Leftrightarrow\quad(c=d andab0mod2)\displaystyle\quad\text{and}\quad a-b\equiv 0\mod 2)
or(cdandab1mod2).\displaystyle\text{or}\quad(c\neq d\quad\text{and}\quad a-b\equiv 1\mod 2).
Proof.

We have

(,S)(ab)={{SkzSk+baz| 0ka}ab{SkzSk+baz|abka}a>b.\displaystyle\uparrow_{(\mathbb{N},S)}(a\to b)=\begin{cases}\left\{S^{k}z\to S^{k+b-a}z\;\middle|\;0\leq k\leq a\right\}&a\leq b\\ \left\{S^{k}z\to S^{k+b-a}z\;\middle|\;a-b\leq k\leq a\right\}&a>b.\end{cases}

Moreover, we have

({0,1},S)(cd)={{SkzSz|(k, even) or (k, odd)}if c=d,{SkzSz|(k even and  odd) or (k odd and  even)}if cd.\displaystyle\uparrow_{(\{0,1\},S)}(c\to d)=\begin{cases}\left\{S^{k}z\to S^{\ell}z\;\middle|\;(\text{$k,\ell$ even})\text{ or }(\text{$k,\ell$ odd})\right\}&\text{if $c=d$},\\ \left\{S^{k}z\to S^{\ell}z\;\middle|\;(\text{$k$ even and $\ell$ odd})\text{ or }(\text{$k$ odd and $\ell$ even})\right\}&\text{if $c\neq d$}.\\ \end{cases}

Hence,

\displaystyle\uparrow (ab:⋅cd)((,S),({0,1},S))=(,S)(ab)({0,1},S)(cd){}_{((\mathbb{N},S),(\{0,1\},S))}(a\to b:\joinrel\cdot\,c\to d)=\ \uparrow_{(\mathbb{N},S)}(a\to b)\ \cap\uparrow_{(\{0,1\},S)}(c\to d)
={{SkzSk+baz|0kak and k+ba are both even or odd}if ab and c=d,{SkzSk+baz|0kak even and k+ba odd or vice versa}if ab and cd,{SkzSk+baz|abkak and k+ba are both even or odd}if a>b and c=d,{SkzSk+baz|abkak even and k+ba odd or vice versa}if a>b and cd.\displaystyle=\begin{cases}\left\{S^{k}z\to S^{k+b-a}z\;\middle|\;\begin{array}[]{c}0\leq k\leq a\\ \text{$k$ and $k+b-a$ are both even or odd}\end{array}\right\}&\text{if $a\leq b$ and $c=d$},\\ \left\{S^{k}z\to S^{k+b-a}z\;\middle|\;\begin{array}[]{c}0\leq k\leq a\\ \text{$k$ even and $k+b-a$ odd or vice versa}\end{array}\right\}&\text{if $a\leq b$ and $c\neq d$},\\ \left\{S^{k}z\to S^{k+b-a}z\;\middle|\;\begin{array}[]{c}a-b\leq k\leq a\\ \text{$k$ and $k+b-a$ are both even or odd}\end{array}\right\}&\text{if $a>b$ and $c=d$},\\ \left\{S^{k}z\to S^{k+b-a}z\;\middle|\;\begin{array}[]{c}a-b\leq k\leq a\\ \text{$k$ even and $k+b-a$ odd or vice versa}\end{array}\right\}&\text{if $a>b$ and $c\neq d$}.\end{cases}

From elementary algebra we know that

even+even=evenandeven+odd=oddandodd+odd=even.\displaystyle\text{even}+\text{even}=\text{even}\quad\text{and}\quad\text{even}+\text{odd}=\text{odd}\quad\text{and}\quad\text{odd}+\text{odd}=\text{even}.

Therefore,

k is even[k+ba is evenba is even],\displaystyle k\text{ is even}\quad\Rightarrow\quad[k+b-a\text{ is even}\quad\Leftrightarrow\quad b-a\text{ is even}],
k is odd[k+ba is oddba is even],\displaystyle k\text{ is odd}\quad\Rightarrow\quad[k+b-a\text{ is odd}\quad\Leftrightarrow\quad b-a\text{ is even}],
k is even[k+ba is oddba is odd],\displaystyle k\text{ is even}\quad\Rightarrow\quad[k+b-a\text{ is odd}\quad\Leftrightarrow\quad b-a\text{ is odd}],
k is odd[k+ba is evenba is odd].\displaystyle k\text{ is odd}\quad\Rightarrow\quad[k+b-a\text{ is even}\quad\Leftrightarrow\quad b-a\text{ is odd}].

Hence,

((,S),({0,1},S))\displaystyle\uparrow_{((\mathbb{N},S),(\{0,1\},S))} (ab:⋅cd)\displaystyle(a\to b:\joinrel\cdot\,c\to d)
={(,S)(ab)if (c=d and ba is even) or (cd and ba is odd)otherwise.\displaystyle=\begin{cases}\uparrow_{(\mathbb{N},S)}(a\to b)&\text{if ($c=d$ and $b-a$ is even) or ($c\neq d$ and $b-a$ is odd)}\\ \emptyset&\text{otherwise.}\end{cases}

This proves

((,S),({0,1},S))ab:⋅cd(c=d and\displaystyle((\mathbb{N},S),(\{0,1\},S))\models a\to b:\joinrel\cdot\,c\to d\quad\Leftrightarrow\quad\text{($c=d$ and } ba is even)\displaystyle b-a\text{ is even)}
or (cd and ba is odd).\displaystyle\text{or ($c\neq d$ and $b-a$ is odd)}.

Since bab-a is even iff aba-b is and the same holds for oddness, analogous arguments prove the remaining arrow proportions and thus the theorem. ∎

3. Proportional axioms

In the tradition of the ancient Greeks, ? (?) (cf. ?, pp. 796-797) introduced (in the linguistic context) a set of proportional axioms as a guideline for formal models of analogical proportions — his list has since been extended by a number of authors (e.g. ?, ?, ?, ?, ?):555? (?) uses different names for the axioms — we have decided to remain consistent with the nomenclature in ? (?, §4.2).

a:b::c:dc:d::a:b(p-symmetry),\displaystyle a:b::c:d\quad\Leftrightarrow\quad c:d::a:b\quad\text{(p-symmetry)},
a:b::a:b(p-reflexivity),\displaystyle a:b::a:b\quad\text{(p-reflexivity)},
a:b::c:db:a::d:c(inner p-symmetry),\displaystyle a:b::c:d\quad\Leftrightarrow\quad b:a::d:c\quad\text{(inner p-symmetry)},
a:b::c:da:c::b:d(central permutation),\displaystyle a:b::c:d\quad\Leftrightarrow\quad a:c::b:d\quad\text{(central permutation)},
a:a::a:dd=a(p-determinism),\displaystyle a:a::a:d\quad\Leftrightarrow\quad d=a\quad\text{(p-determinism)},
a:a::c:dd=c(strong inner p-reflexivity),\displaystyle a:a::c:d\quad\Rightarrow\quad d=c\quad\text{(strong inner p-reflexivity)},
a:a::c:c(inner p-reflexivity),\displaystyle a:a::c:c\quad\text{(inner p-reflexivity)},
a:b::a:dd=b(strong p-reflexivity),\displaystyle a:b::a:d\quad\Rightarrow\quad d=b\quad\text{(strong p-reflexivity)},

  a:b::c:da:b::c:d         c:d::e:fc:d::e:f    (p-transitivity),              a:b::e:fa:b::e:f

  a:b::c:da:b::c:d         b:e::d:fb:e::d:f    (inner p-transitivity).              a:e::c:fa:e::c:f

We have the following analysis of the proportional axioms in monounary algebras:

Theorem 4.

The analogical proportion relation, restricted to monounary algebras, satisfies

  • p-symmetry,

  • inner p-symmetry,

  • p-reflexivity,

  • p-determinism,

and, in general, does not satisfy

  • central permutation,

  • strong inner p-reflexivity,

  • strong p-reflexivity,

  • p-commutativity,

  • p-transitivity,

  • inner p-transitivity.

Proof.

We have the following proofs:

  • The proof of p-symmetry, inner p-symmetry, p-reflexivity, and p-determinism is the same as the corresponding proofs of Theorem 28 in ? (?).

  • The disproof of central permutation is the same as in Theorem 28 in ? (?): it fails, for example, in the monounary algebra given by (we omit the loops So:=oSo:=o, for o{b,c,d}o\in\{b,c,d\}, in the figure)

    aabbccddSS
  • The disproof of strong inner p-reflexivity is the same as in Theorem 28 in ? (?): it fails, for example, in the monounary algebra

    aaccddSSSS
  • Strong p-reflexivity fails, for example, in the monounary algebra

    aabbddSSSSSS
  • p-Transitivity fails, for example, in the monounary algebra

    aabb\astccddeeffff^{\prime}\astSSSSSSSSSSSSSSSSSS

    We first prove the relations

    (5) a:b::c:dandc:d::e:f.\displaystyle a:b::c:d\quad\text{and}\quad c:d::e:f.

    Since

    (ab)={Skzz|k1}=(cd)=(ef),\displaystyle\uparrow(a\to b)=\left\{S^{k}z\to z\;\middle|\;k\geq 1\right\}=\ \uparrow(c\to d)=\ \uparrow(e\to f),

    we have

    ab:⋅cdandcd:⋅ab,\displaystyle a\to b:\joinrel\cdot\,c\to d\quad\text{and}\quad c\to d:\joinrel\cdot\,a\to b,
    cd:⋅efandef:⋅cd.\displaystyle c\to d:\joinrel\cdot\,e\to f\quad\text{and}\quad e\to f:\joinrel\cdot\,c\to d.

    Moreover, since

    (ba)={zSkz|k1}=(dc)=(fe),\displaystyle\uparrow(b\to a)=\left\{z\to S^{k}z\;\middle|\;k\geq 1\right\}=\ \uparrow(d\to c)=\ \uparrow(f\to e),

    we have

    ba:⋅dcanddc:⋅ba,\displaystyle b\to a:\joinrel\cdot\,d\to c\quad\text{and}\quad d\to c:\joinrel\cdot\,b\to a,
    dc:⋅feandfe:⋅dc,\displaystyle d\to c:\joinrel\cdot\,f\to e\quad\text{and}\quad f\to e:\joinrel\cdot\,d\to c,

    which thus proves (5).

    On the other hand,

    (ab:⋅ef)\displaystyle\emptyset\neq\ \uparrow(a\to b:\joinrel\cdot\,e\to f) ={Skzz|k1}\displaystyle=\left\{S^{k}z\to z\;\middle|\;k\geq 1\right\}
    {Skzz|k1}{SkzSz|k2}\displaystyle\subsetneq\left\{S^{k}z\to z\;\middle|\;k\geq 1\right\}\cup\left\{S^{k}z\to Sz\;\middle|\;k\geq 2\right\}
    =(ab:⋅ef)\displaystyle=\ \uparrow(a\to b:\joinrel\cdot\,e\to f^{\prime})

    shows

    ab:⋅ef\displaystyle a\to b\not:\joinrel\cdot\,e\to f

    and thus

    a:b:e:f.\displaystyle a:b\not::e:f.
  • The disproof of inner p-transitivity is the same as in Theorem 28 in ? (?): it fails, for example, in the monounary algebra given by (we omit the loops So:=oSo:=o, for o{b,e,c,d,f}o\in\{b,e,c,d,f\}, in the figure)

    aabbeeccddffSS

Remark 5.

Since p-transitivity fails in general, the relation :::: is in general not an equivalence relation.

Problem 6.

Characterize those monounary algebras which satisfy p-transitivity.

4. Congruences

This section is a collection of results relating congruences to analogical proportions in monounary algebras. Recall that an equivalence relation θ\theta on 𝔄=(A,S)\mathfrak{A}=(A,S) is a congruence iff

aθbSaθSb,for all a,bA.\displaystyle a\theta b\quad\Rightarrow\quad Sa\theta Sb,\quad\text{for all $a,b\in A$.}

The factor algebra obtained from 𝔄\mathfrak{A} with respect to θ\theta is given by

𝔄/θ:=(A/θ,S/θ),\mathfrak{A}/\theta:=(A/\theta,S/\theta),

where

A/θ:={a/θaA}A/\theta:=\{a/\theta\mid a\in A\}

contains the congruence classes

a/θ:={bAaθb}a/\theta:=\{b\in A\mid a\theta b\}

with respect to θ\theta, and S/θ:A/θA/θS/\theta:A/\theta\to A/\theta is defined by

S/θ(a/θ):=Sa/θ.\displaystyle S/\theta(a/\theta):=Sa/\theta.
Theorem 7.

There is a monounary algebra 𝔄=(A,S)\mathfrak{A}=(A,S), a congruence θ\theta on 𝔄\mathfrak{A}, and elements a,b,c,dAa,b,c,d\in A such that

(6) 𝔄a:b::c:dwhereas𝔄/θ⊧̸a/θ:b/θ::c/θ:d/θ.\displaystyle\mathfrak{A}\models a:b::c:d\quad\text{whereas}\quad\mathfrak{A}/\theta\not\models a/\theta:b/\theta::c/\theta:d/\theta.
Proof.

Define the monounary algebra 𝔄\mathfrak{A} by

aaaa^{\prime}bbbb^{\prime}ccddSSSSSSSSSSSS

The identities

𝔄(ab)𝔄(cd)=and𝔄(ba)𝔄(dc)=\displaystyle\uparrow_{\mathfrak{A}}(a\to b)\ \cup\uparrow_{\mathfrak{A}}(c\to d)=\emptyset\quad\text{and}\quad\uparrow_{\mathfrak{A}}(b\to a)\ \cup\uparrow_{\mathfrak{A}}(d\to c)=\emptyset

imply 𝔄a:b::c:d\mathfrak{A}\models a:b::c:d.

Now define the congruence θ:={{a,a},{b,b},{c},{d}}\theta:=\{\{a,a^{\prime}\},\{b,b^{\prime}\},\{c\},\{d\}\} yielding the factor algebra 𝔄/θ\mathfrak{A}/\theta given by

{a,a}=a/θ\{a,a^{\prime}\}=a/\theta{b,b}=b/θ\{b,b^{\prime}\}=b/\theta{c}=c/θ\{c\}=c/\theta.{d}=d/θ\{d\}=d/\thetaSSSSSSSS

Now

𝔄/θ(a/θb/θ)𝔄/θ(c/θd/θ)={zSz,}\displaystyle\uparrow_{\mathfrak{A}/\theta}(a/\theta\to b/\theta)\ \cup\uparrow_{\mathfrak{A}/\theta}(c/\theta\to d/\theta)=\{z\to Sz,\ldots\}\neq\emptyset

and

𝔄/θ(a/θb/θ:⋅c/θd/θ)=\displaystyle\uparrow_{\mathfrak{A}/\theta}(a/\theta\to b/\theta:\joinrel\cdot\,c/\theta\to d/\theta)=\emptyset

imply

𝔄/θ⊧̸a/θb/θ:⋅c/θd/θ.\displaystyle\mathfrak{A}/\theta\not\models a/\theta\to b/\theta:\joinrel\cdot\,c/\theta\to d/\theta.

Theorem 8.

There is a monounary algebra 𝔄=(A,S)\mathfrak{A}=(A,S), a congruence θ\theta on 𝔄\mathfrak{A}, and elements a,b,c,dAa,b,c,d\in A such that

𝔄/θa/θ:b/θ::c/θ:d/θwhereas𝔄⊧̸a:b::c:d.\displaystyle\mathfrak{A}/\theta\models a/\theta:b/\theta::c/\theta:d/\theta\quad\text{whereas}\quad\mathfrak{A}\not\models a:b::c:d.
Proof.

Define the monounary algebra 𝔄\mathfrak{A} by

aaaa^{\prime}bbbb^{\prime}ccddSSSSSSSSSSSS

Since

𝔄(ab)𝔄(cd)and𝔄(ab:⋅cd)=,\displaystyle\uparrow_{\mathfrak{A}}(a\to b)\ \cup\uparrow_{\mathfrak{A}}(c\to d)\neq\emptyset\quad\text{and}\quad\uparrow_{\mathfrak{A}}(a\to b:\joinrel\cdot\,c\to d)=\emptyset,

we have

𝔄⊧̸ab:⋅cdand therefore𝔄⊧̸a:b::c:d.\displaystyle\mathfrak{A}\not\models a\to b:\joinrel\cdot\,c\to d\quad\text{and therefore}\quad\mathfrak{A}\not\models a:b::c:d.

Now define the congruence θ:={{a,a},{b,b},{c},{d}}\theta:=\{\{a,a^{\prime}\},\{b,b^{\prime}\},\{c\},\{d\}\} yielding the factor algebra 𝔄/θ\mathfrak{A}/\theta given by

{a,a}=a/θ\{a,a^{\prime}\}=a/\theta{b,b}=b/θ\{b,b^{\prime}\}=b/\theta{c}=c/θ\{c\}=c/\theta.{d}=d/θ\{d\}=d/\thetaSSSSSSSS

We clearly have

𝔄/θa/θ:b/θ::c/θ:d/θ.\displaystyle\mathfrak{A}/\theta\models a/\theta:b/\theta::c/\theta:d/\theta.

Theorem 9.

There is a monounary algebra 𝔄=(A,S)\mathfrak{A}=(A,S), a congruence θ\theta on 𝔄\mathfrak{A}, and elements a,bAa,b\in A such that

(𝔄,𝔄/θ)⊧̸a:b::a/θ:b/θ.\displaystyle(\mathfrak{A,A/\theta})\not\models a:b::a/\theta:b/\theta.
Proof.

Define the monounary algebra 𝔄\mathfrak{A} by

aa^{\prime}aabbbb^{\prime}SSSSSSSS

Now define the congruence θ:={{a,a},{b,b}}\theta:=\{\{a,a^{\prime}\},\{b,b^{\prime}\}\} yielding the factor algebra 𝔄/θ\mathfrak{A}/\theta given by

{a,a}=a/θ\{a,a^{\prime}\}=a/\theta{b,b}=b/θ\{b,b^{\prime}\}=b/\theta.SSSS

Since

𝔄(ab)𝔄/θ(a/θb/θ)={zSz,}\displaystyle\uparrow_{\mathfrak{A}}(a\to b)\ \cup\uparrow_{\mathfrak{A}/\theta}(a/\theta\to b/\theta)=\{z\to Sz,\ldots\}\neq\emptyset

whereas

𝔄(ab:⋅a/θb/θ)=,\displaystyle\uparrow_{\mathfrak{A}}(a\to b:\joinrel\cdot\,a/\theta\to b/\theta)=\emptyset,

we have

(𝔄,𝔄/θ)⊧̸ab:⋅a/θb/θand therefore(𝔄,𝔄/θ)⊧̸a:b::a/θ:b/θ.\displaystyle(\mathfrak{A,A/\theta})\not\models a\to b:\joinrel\cdot\,a/\theta\to b/\theta\quad\text{and therefore}\quad(\mathfrak{A,A/\theta})\not\models a:b::a/\theta:b/\theta.

Theorem 10.

There is a monounary algebra 𝔄=(A,S)\mathfrak{A}=(A,S), a congruence θ\theta on 𝔄\mathfrak{A}, and elements a,b,c,dAa,b,c,d\in A such that

aθbandcθdwhereas𝔄⊧̸a:b::c:d.\displaystyle a\theta b\quad\text{and}\quad c\theta d\quad\text{whereas}\quad\mathfrak{A}\not\models a:b::c:d.
Proof.

Define the monounary algebra 𝔄\mathfrak{A} by

aabbccddSSSSSSSS

and define the congruence θ:={{a,b},{c,d}}\theta:=\{\{a,b\},\{c,d\}\}. We then have

aθbandcθdwhereas𝔄⊧̸a:b::c:d.\displaystyle a\theta b\quad\text{and}\quad c\theta d\quad\text{whereas}\quad\mathfrak{A}\not\models a:b::c:d.

5. Difference proportion theorem

Arithmetic or difference proportions are characterized as

a:b::c:dab=cd\displaystyle a:b::c:d\quad\Leftrightarrow\quad a-b=c-d

and have been considered already by the ancient Greeks.666https://en.wikipedia.org/wiki/Proportion_(mathematics) In this section, we are going to see that difference proportions naturally occur in the prototypical infinite monounary algebra given by the natural numbers ={0,1,2}\mathbb{N}=\{0,1,2\ldots\} together with the unary successor function S:S:\mathbb{N}\to\mathbb{N} defined by Sa:=a+1Sa:=a+1. This is more surprising than it might appear, as the abstract framework is not geared towards proportions in monounary algebras — the forthcoming theorem is thus further conceptual evidence of the robustness of the underlying framework:

Theorem 11 (Difference Proportion Theorem).

For any natural numbers a,b,c,da,b,c,d\in\mathbb{N}, we have

(,S)a:b::c:dab=cd(difference proportion).\displaystyle(\mathbb{N},S)\models a:b::c:d\quad\Leftrightarrow\quad a-b=c-d\quad\text{{\em(difference proportion)}}.
Proof.

Notice that (,S)(ab)\uparrow_{(\mathbb{N},S)}(a\to b) always contains the non-trivial justification zSbazz\to S^{b-a}z in case aba\leq b, or SabzzS^{a-b}z\to z in case b<ab<a, which means that (,S)(ab)\uparrow_{(\mathbb{N},S)}(a\to b) is non-empty, for all a,ba,b\in\mathbb{N}. This implies that (,S)(ab)(,S)(cd)\uparrow_{(\mathbb{N},S)}(a\to b)\ \cup\uparrow_{(\mathbb{N},S)}(c\to d) is non-empty as well, which means that (,S)ab:⋅cd(\mathbb{N},S)\models a\to b:\joinrel\cdot\,c\to d cannot be justified by trivial justifications alone.

Since SS is injective in \mathbb{N}, every justification SkzSzS^{k}z\to S^{\ell}z of ab:⋅cda\to b:\joinrel\cdot\,c\to d in (,S)(\mathbb{N},S) is a characteristic justification by Uniqueness Lemma 23 in ? (?). By definition, SkzSzS^{k}z\to S^{\ell}z is a justification of ab:⋅cda\to b:\joinrel\cdot\,c\to d in (,S)(\mathbb{N},S) iff, for some o1,o2o_{1},o_{2}\in\mathbb{N},

a=Sko1andb=So1andc=Sko2andd=So2,\displaystyle a=S^{k}o_{1}\quad\text{and}\quad b=S^{\ell}o_{1}\quad\text{and}\quad c=S^{k}o_{2}\quad\text{and}\quad d=S^{\ell}o_{2},

which is equivalent to

a=k+o1andb=+o1andc=k+o2andd=+o2.\displaystyle a=k+o_{1}\quad\text{and}\quad b=\ell+o_{1}\quad\text{and}\quad c=k+o_{2}\quad\text{and}\quad d=\ell+o_{2}.

This holds iff ab=cda-b=c-d as desired. ∎

As a direct consequence of the Difference Proportion Theorem 11, we have the following analysis of the axioms in ? (?, §4.3) within the natural numbers with successor:

Theorem 12.

All the proportional axioms hold in (,S)(\mathbb{N},S) except for p-commutativity.777The monotonicity axiom is irrelevant here as we are interested in monounary algebras only.

Proof.

In addition to the positive part of Theorem 4, we have the following remaining proofs:

  • p-Commutativity fails since abbaa-b\neq b-a whenever aba\neq b.

  • Central permutation follows from ab=cdac=bda-b=c-d\Leftrightarrow a-c=b-d.

  • Strong inner p-reflexivity follows from aa=cdd=ca-a=c-d\Rightarrow d=c

  • Strong p-reflexivity follows from ab=add=ba-b=a-d\Rightarrow d=b.

  • p-Transitivity follows from

    ab=cdandcd=efab=ef.\displaystyle a-b=c-d\quad\text{and}\quad c-d=e-f\quad\Rightarrow\quad a-b=e-f.
  • Inner p-transitivity follows from

      ab=cda-b=c-d         be=dfb-e=d-f       ab+be=cd+dfa-b+b-e=c-d+d-f                  ae=cfa-e=c-f.

  • Central p-transitivity is a direct consequence of transitivity. Explicitly, we have

    ab=bcandbc=cdab=cd.\displaystyle a-b=b-c\quad\text{and}\quad b-c=c-d\quad\Rightarrow\quad a-b=c-d.

We call an “extern” unary function f:AAf:A\to A a proportional polymorphsism (or p-polymorphism) (?) on the monounary algebra 𝔄=(A,S)\mathfrak{A}=(A,S) iff, for all a,b,c,dAa,b,c,d\in A,

      𝔄a:b::c:d\mathfrak{A}\models a:b::c:d    .   𝔄fa:fb::fc:fd\mathfrak{A}\models fa:fb::fc:fd

Fact 13.

The successor function SS is a p-polymorphism on (,S)(\mathbb{N},S).

Proof.

We have the following derivation:

        (,S)a:b::c:d(\mathbb{N},S)\models a:b::c:d           ab=cda-b=c-d         SaSb=ScSdSa-Sb=Sc-Sd     (,S)Sa:Sb::Sc:Sd(\mathbb{N},S)\models Sa:Sb::Sc:Sd.

Acknowledgments

We would like to thank the reviewers for their thoughtful and constructive comments, and for their helpful suggestions to improve the presentation of the article.

Conflict of interest

The authors declare that they have no conflict of interest.

Data availability statement

The manuscript has no data associated.

References

  • Antić Antić, C. (2022). Analogical proportions.  Annals of Mathematics and Artificial Intelligence, 90(6), 595–644.
  • Antić Antić, C. (2023a). Analogical proportions II.  Journal of Artificial Intelligence Research, submitted, https://www.researchgate.net/publication/374265899_Analogical_proportions_II.
  • Antić Antić, C. (2023b). Proportional algebras.  Journal of Artificial Intelligence Research, under review, https://arxiv.org/pdf/2210.01751.pdf.
  • Barbot et al. Barbot, N., Miclet, L., and Prade, H. (2019). Analogy between concepts.  Artificial Intelligence, 275, 487–539.
  • Boden Boden, M. A. (1998). Creativity and artificial intelligence.  Artificial Intelligence, 103(1-2), 347–356.
  • Gentner Gentner, D. (1983). Structure-mapping: a theoretical framework for analogy.  Cognitive Science, 7(2), 155–170.
  • Gust et al. Gust, H., Krumnack, U., Kühnberger, K.-U., and Schwering, A. (2008). Analogical reasoning: a core of cognition.  Künstliche Intelligenz, 22(1), 8–12.
  • Hesse Hesse, M. B. (1966). Models and Analogies in Science. University of Notre Dame Press.
  • Hofstadter Hofstadter, D. (2001). Analogy as the core of cognition.  In Gentner, D., Holyoak, K. J., and Kokinov, B. K. (Eds.), The Analogical Mind: Perspectives from Cognitive Science, pp. 499–538. MIT Press/Bradford Book, Cambridge MA.
  • Hofstadter and Sander Hofstadter, D.,  and Sander, E. (2013). Surfaces and Essences. Analogy as the Fuel and Fire of Thinking. Basic Books, New York.
  • Krieger Krieger, M. H. (2003). Doing Mathematics: Convention, Subject, Calculation, Analogy. World Scientific, New Jersey.
  • Lepage Lepage, Y. (2003). De L’Analogie. Rendant Compte de la Commutation en Linguistique. Habilitation à diriger les recherches, Université Joseph Fourier, Grenoble.
  • Lim et al. Lim, S., Prade, H., and Richard, G. (2021). Classifying and completing word analogies by machine learning.  International Journal of Approximate Reasoning, 132, 1–25.
  • Miclet et al. Miclet, L., Bayoudh, S., and Delhay, A. (2008). Analogical dissimilarity: definition, algorithms and two experiments in machine learning.  Journal of Artificial Intelligence Research, 32, 793–824.
  • Miclet and Prade Miclet, L.,  and Prade, H. (2009). Handling analogical proportions in classical logic and fuzzy logics settings.  In Sossai, C.,  and Chemello, G. (Eds.), ECSQARU 2009, LNAI 5590, pp. 638–650. Springer-Verlag, Berlin/Heidelberg.
  • Pólya Pólya, G. (1954). Induction and Analogy in Mathematics, Vol. 1 of Mathematics and Plausible Reasoning. Princeton University Press, Princeton, New Jersey.
  • Prade and Richard Prade, H.,  and Richard, G. (2013). From analogical proportion to logical proportions.  Logica Universalis, 7, 441–505.
  • Prade and Richard Prade, H.,  and Richard, G. (2021). Analogical proportions: why they are useful in AI.  In Zhou, Z.-H. (Ed.), IJCAI 2021, pp. 4568–4576.
  • Winston Winston, P. H. (1980). Learning and reasoning by analogy.  Communications of the ACM, 23(12), 689–703.