Abstract
We extend to fuzzy similarity measures the study made for classical ones in a companion paper (Coletti and BouchonMeunier in Soft Comput 23:6827–6845, 2019). Using a classic method of measurement theory introduced by Tversky, we establish necessary and sufficient conditions for the existence of a particular class of fuzzy similarity measures, representing a binary relation among pairs of objects which expresses the idea of “no more similar than”. In this fuzzy context, the axioms are strictly dependent on the combination operators chosen to compute the union and the intersection.
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Funding
Giulianella Coletti: This work was partially supported by the Italian Ministry of Education, University and Research, funding of Research Projects of National Interest (PRIN 201011) under Grant 2010FP79LR_003 (Logical methods and information management) and by the Italian Ministry of Health under Grant J521I14001640001 (“Intelligent systems helping in decisions for the early alert and the dissuasion to the use of doping”).
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Appendices
Appendix 1
We recall the axioms introduced in Coletti and BouchonMeunier (2019a) for a comparative similarity in \({{\mathcal {X}}}^2\), and the relevant main results. For this we need to recall that, for any \(X\in {{\mathcal {X}}},\)\(I_X= \{i:x_i=1\}\) and X denotes the cardinality of \(I_X.\)
Basic Axioms

Axiom S0 [weak order]
\(\preceq \) is a weak order, i.e. it is a complete (or total), reflexive and transitive binary relation.

Axiom S1 [boundary conditions]
For every \( X, Y \in {{\mathcal {X}}}\), with \(X\ne Y\)
$$\begin{aligned} (X^c, X) \sim (Y^c, Y)\preceq (X, Y) \end{aligned}$$and, for \(X,Y\ne {\underline{0}}\)
$$\begin{aligned} (X^c, X) \prec (X, X) \sim (Y,Y) \end{aligned}$$ 
Axiom S2 [weak boundary conditions]
For every \( X, Y \in {{\mathcal {X}}}\), with \(X\cap Y={\underline{0}},\)
$$\begin{aligned} (X, Y) \sim (X_k^c, Y)\sim (X, Y_h^c) \end{aligned}$$for any \(k\in I_{X\cup (X^c\cap Y^c)}\) and \(h\in I_{Y\cup (X^c\cap Y^c)}\)
For every \( {\underline{0}}\ne X \in {{\mathcal {X}}}\), and for any k such that \(X_k^c\ne {\underline{0}}\)
$$\begin{aligned} (X, X) \sim (X_k^c, X_k^c) \end{aligned}$$ 
Axiom S4 [partial attribute uniformity]
For every \(X,Y\in {{\mathcal {X}}}\) and for every \(h,k\in \{1,\ldots ,p\},\)
$$\begin{aligned} (X, Y) \sim (X'_{h,k}, Y'_{h,k}). \end{aligned}$$where \(X'_{h,k}\) the element of \({{\mathcal {X}}}\) coinciding with X for \(i\ne h,k\) and such that \(x'_h =x_k\), \(x'_k =x_h\) and similarly for Y

Axiom S5 [distinctive attribute interchangeability]
For every \(X,Y\in {{\mathcal {X}}}\) and for every \(k\in I_{X{\varDelta } Y},\,\) we have
$$\begin{aligned} (X, Y) \sim (X^c_{k}, Y^c_{k}). \end{aligned}$$ 
Axiom S6 [attribute interchangeability]
For every \((X,Y)\in {{\mathcal {X}}}^2\) and for any \(k \in \{1,\ldots ,p\}\) we have
$$\begin{aligned} (X, Y) \sim (X_k^c, Y_k^c). \end{aligned}$$
Monotonicity Axioms

Axiom S7 \(_1\) [monotonicity]
For every \(X,Y, X',Y',X'', Y''\in {{\mathcal {X}}},\) if \(X\cap Y\supset X'\cap Y' = X''\cap Y''\ne {\underline{0}}\) and \({\underline{0}}\ne X{\varDelta } Y= X'{\varDelta } Y'\subset X''{\varDelta } Y'',\) then
$$\begin{aligned} (X'', Y'') \prec (X', Y')\prec (X, Y). \end{aligned}$$ 
Axiom S7 \(_2\) [weak monotonicity]
For every \(X,Y, X',Y',X'', Y''\in {{\mathcal {X}}},\) if \(X\cap Y\supset X'\cap Y' = X''\cap Y''\ne {\underline{0}},\,\)\({\underline{0}}\ne X\setminus Y= X'\setminus Y'\subseteq X''\setminus Y'',\,\)\({\underline{0}}\ne Y\setminus X= Y'\setminus X'\subseteq Y''\setminus X'',\,\) and at least one of the two following conditions is satisfied \(X'\setminus Y'\subset X''\setminus Y'',\,\)\(Y'\setminus X'\subset Y''\setminus X'',\,\) then
$$\begin{aligned} (X'', Y'') \prec (X', Y')\prec (X, Y). \end{aligned}$$ 
Axiom S7 \(_3\) [cumulative monotonicity]
For every \(X,Y, X',Y',X'', Y''\in {{\mathcal {X}}},\) if \(X{\varDelta } Y= X'{\varDelta } Y'\subset X''{\varDelta } Y'',\) then
$$\begin{aligned} (X'', Y'') \prec (X', Y')\sim (X, Y). \end{aligned}$$
Independence Axioms
Let \({\varGamma }\) be a set of comparisons on \({{\mathcal {X}}}^2,\) i.e.
\({\varGamma }=\{(X_i, Y_i)\preceq (Z_i, W_i)\}_{(X_i, Y_i),(Z_i, W_i)\in {{\mathcal {X}}}^2} \)
and let us consider the following sets:
\({{\mathcal {A}}}^{{\varGamma }} =\{X\cap Y\},\; {{\mathcal {B}}}^{{\varGamma }} =\{X{\varDelta } Y\},\; {{\mathcal {C}}}^{{\varGamma }} =\{(X^c\cap Y^c)\cup (X\cap Y)\}, {{\mathcal {D}}}_1^{{\varGamma }} = \{X\},\; {{\mathcal {D}}}_2^{{\varGamma }} = \{Y\},\)
with (X, Y) any element of \({{\mathcal {X}}}^2\) compared in at least one inequality present in \({\varGamma }\).

Axiom WkC [weakkth cancellation]
Let \(\preceq \) be a comparative similarity on \({{\mathcal {X}}}^2.\) Let us associate to each pair \((X,Y)\in {{\mathcal {X}}}^2\) the pair \((X\cap Y, X{\varDelta } Y).\) The relation \(\preceq \) satisfies the weak kth cancellation axiom if and only if for each \(h\le k\) and for every inequalities \((X_i,Y_i)\preceq (Z_i,W_i)\), \((i=1,\ldots ,h)\) we have the following: if for every \(A\in {{\mathcal {A}}}^{{\varGamma }}\) the number of pairs \((X_i, Y_i)\) with \(X_i\cap Y_i=A\) is the same as the number of pairs \((Z_i, W_i)\) with \(Z_i\cap W_i=A\), and similarly for every \(B\in {{\mathcal {B}}}^{{\varGamma }}\), then for every \(i=1,\ldots ,h\) we have
$$\begin{aligned} (X_i,Y_i)\sim (Z_i,W_i). \end{aligned}$$ 
Axiom WFC [weak finite cancellation]
A comparative similarity \(\preceq \) on \({{\mathcal {X}}}^2\) is said to satisfy the weak finite cancellation if and only if it satisfies the weak kth cancellation for each \(k \in {{\mathbb {N}}}\).

Axiom CkC [cumulative kth cancellation]
Let \(\preceq \) be a comparative similarity on \({{\mathcal {X}}}^2\) and let us associate to each pair \((X,Y)\in {{\mathcal {X}}}^2\) the triple \((X\cap Y, X, Y).\) The relation \(\preceq \) satisfies the cumulative kth cancellation axiom if and only if for each \(h\le k\) and for every inequalities \((X_i,Y_i)\preceq (Z_i,W_i)\), \((i=1,\ldots ,h)\) we have the following: if for every \(A\in {{\mathcal {A}}}^{{\varGamma }}\) the number of pairs \((X_i, Y_i)\) with \(X_i\cap Y_i=A\) is the same as the number of pairs \((Z_i, W_i)\) with \(Z_i\cap W_i=A\), and similarly for every \(X\in {{\mathcal {D}}}^{{\varGamma }}_1\) and for every \(Y\in {{\mathcal {D}}}^{{\varGamma }}_2\), then for every \(i=1,\ldots ,h\) we have
$$\begin{aligned} (X_i,Y_i)\sim (Z_i,W_i). \end{aligned}$$ 
Axiom CFC [cumulative finite cancellation]
A comparative similarity \(\preceq \) on \({{\mathcal {X}}}^2\) is said to satisfy the cumulative finite cancellation if and only if it satisfies the cumulative kth cancellation for each \(k \in {{\mathbb {N}}}\).
Theorem 17
Let \(\preceq \) be a comparative similarity on \({{\mathcal {X}}}^2\). The following equivalences hold:

1.
the following conditions (i) and (ii) are equivalent:

(i)
\(\preceq \) satisfies S0, S1, S2, S4, S5, S\(7_1,\)

(ii)
there exists a (unique up to increasing transformations) function \(S: {{\mathcal {X}}}^2\rightarrow [0,1]\) representing \(\preceq \) and such that:

(a)
\(S(X, Y) = {\varPhi }(X\cap Y, X{\varDelta } Y),\)

(b)
\({\varPhi }\) is increasing with respect to \(X\cap Y\) and decreasing with respect to \(X{\varDelta } Y,\)

(c)
\({\varPhi }(0, b)=0,\) and \({\varPhi }(a, 0)=1,\) for every a and for every \(b\ne 0.\)

(a)

(i)

2.
the following conditions (j) and (jj) are equivalent:

(j)
\(\preceq \) satisfies S0, S1, S2, S4, S\(7_2,\)

(jj)
There exists a (unique up to increasing transformations) function \(S: {{\mathcal {X}}}^2\rightarrow [0,1]\) representing \(\preceq \) and such that:

(a’)
\(S(X,Y) = {\varPsi }(X\cap Y, X\setminus Y, Y\setminus X),\)

(b’)
\({\varPsi }\) is increasing with respect to \(X\cap Y\) and decreasing with respect to \(X\setminus Y\) and to \(Y\setminus X,\)

(c’)
\({\varPsi }(0,b, b')=0,\) and \({\varPsi }(a, 0, 0)=1,\) for every a and for every \(b,b'\) with \(b+b'\ne 0.\)

(a’)

(j)

3.
the following conditions (k) and (kk) are equivalent:

(k)
\(\preceq \) satisfies S0, S1, S4, S6, S\(7_3,\)

(kk)
There exists a (unique up to increasing transformations) function \(S: {{\mathcal {X}}}^2\rightarrow [0,1]\) representing \(\preceq \) and such that:

(a)
\(S(X,Y) = {\varXi }(X\cap Y,X^c\cap Y^c, X{\varDelta } Y),\)

(b)
\({\varXi }\) is increasing with respect to \(X\cap Y\) and \(X^c\cap Y^c\) and decreasing with respect to \(X{\varDelta } Y,\)

(c)
\({\varXi }(0, 0, b)=0,\) and \({\varXi }(a, c, 0)=1,\) for every a, c and for every \(b\ne 0.\)

(a)

(k)
Theorem 18
For a comparative similarity \(\preceq \) on \({{\mathcal {X}}}^2\), the following statements hold:

(i)
\(\preceq \) satisfies S0, S1, S2, S4, S5, S\(7_1\), WFC \(\; \Longleftrightarrow \) there exist two nonnegative increasing realvalued functions f, g, with \(f(0)= g(0) = 0\) such that function \(S_{f,g}:{{\mathcal {X}}}^2 \rightarrow [0,1],\) defined in Eq. (1) represents \(\preceq .\)

(ii)
\(\preceq \) satisfies S0, S1, S2, S4, S\(7_2\), CFC \(\; \Longleftrightarrow \) there exist two nonnegative increasing real valued functions f, g, with \(f(0)= g(0) = 0\) and \(h(x)=f( x) g(x)\) for every non negative number x, such that function \(T_{f,g}:{{\mathcal {X}}}^2 \rightarrow [0,1],\) defined in Eq. (2) represents \(\preceq .\)

(iii)
\(\preceq \) satisfies S0, S1, S4, S6, S\(7_3\), TWCF \(\; \Longleftrightarrow \) there exist two nonnegative increasing realvalued functions f, g with \(f(0)= g(0) = 0\) such that function \(H_{f,g}:{{\mathcal {X}}}^2 \rightarrow [0,1],\) defined in Eq. (3) represents \(\preceq .\)
Appendix 2
Proof of Lemma 1
First of all we notice that when the intersection is computed by \(\min \) and the union by \(\max \), we only obtain \(m(X\cap Y)=0\) in the case where the relevant supports \(s_X\) and \(s_Y\) are disjoint. Moreover we can have \(m(X{\varDelta } Y)=0\) if and only if X and Y are in \({{\mathcal {X}}}.\)
Therefore if either condition 2 or condition 3 holds, then the thesis follows immediately by applying Axioms FS1\(_G\) and FS0 a finite number of times.
By axiom FS2 all the pairs only differing through a permutation of indices are equivalent, so we can consider any element among them only differing from (X, Y) through a permutation and similarly for \((X', Y').\) First of all, we prove that (X, Y) and \((X', Y')\) cannot differ by only one element (\(x_k\ne x_k'\) and all the other elements of X and \(X'\) are equal to all the elements of Y and \(Y'\)). In fact, if \(x_k\ne x_k',\) then, by hypothesis \(x_k\) and \(x_k'\) are not minimum, so the minimum is, respectively, \(y_k\) and \(y_k'\) (which are supposed to be equal). Now, if \(x_k \le 1y_k\) and \(x_k'\le 1y_k'\) we have \(m(X\setminus Y) \ne m(X'\setminus Y')\,\) and if \(x_k \ge 1y_k\) and \(x_k'\ge 1y_k',\)\(m(Y\setminus X) \ne m(Y'\setminus X').\,\) Let us consider the case \(x_k < 1y_k\) and \(x_k'\ge 1y_k'.\) In this condition we have \(m(X\setminus Y) = m(X'\setminus Y')\,\) if and only if \(1x_k' = y_k = y_k'\) which implies \( m(Y\setminus X) \ne m (Y'\setminus X'),\) contrary to the hypothesis. By similar considerations, we obtain the assertion for the other two situations and for the case \(y_k\ne y_k'.\)
We now prove that we cannot have:
\(x_k\ne x_k',\,y_h\ne y_h'\) and \(x_h = x_h',\,y_k = y_k'.\)
Let us consider:
\(a=\sum _{r\ne k,h}\min \{x_r, y_r\} = \sum _{r\ne k,h}\min \{x_r', y_r'\}; \,\)
\(b=\sum _{r\ne k,h}\min \{x_r, 1 y_r\} = \sum _{r\ne k,h}\min \{x_r', 1  y_r'\}.\,\)
\(c=\sum _{r\ne k,h}\min \{1x_r, y_r\}= \sum _{r\ne k,h}\min \{1x_r', y_r'\}; \)
so \(m(X\cap Y)= a + \min (x_h,y_h) + \min (x_k,y_k)\) and \(m(X'\cap Y')= a + \min (x_h',y_h') + \min (x_k',y_k');\)
\(m(X\setminus Y)= b + \min (x_h,1y_h) + \min (x_k,1y_k)\) and \(m(X'\setminus Y')= b +\min (x_h',1y_h') + \min (x_k',1  y_k')\)
\(m(Y\setminus X)= c + \min (1x_h,y_h) + \min (1x_k,y_k)\) and \(m(Y'\setminus X') = c + \min (1x_h',y_h') + \min (1x_k',y_k').\)
It is immediate to see that \( m(X\cap Y) \ne m(X'\cap Y'),\,\) when one of the following situations occurs, where \(r=h,k\):
\(x_r\le y_r,\,\) and \(x_r'\le y_r',\,\) (or \(x_r\ge y_r,\,\) and \(x_r'\ge y_r'\) )
\(x_r\le y_r,\,\) and \(x_h'\le y_h',\,x_k'> y_k',\) (or \(x_h'> y_h',\,x_k'\le y_k'\))
We must prove that we cannot have \(x_r< y_r,\,\) and \(x_r'> y_r'.\,\)
Consider first the case \(x_r\le 1y_r\) and \(x_r'\le 1y_r'.\) In this case \( m(X\cap Y)= m(X'\cap Y'),\,\) if and only if \(x_h+x_k = y_h'+y_k\) and \(m(X\setminus Y)= m(X'\setminus Y')\) if and only if \(x_h + x_k = 2(y_h'+ y_k)\) so \(x_h + x_k = 1.\) On the other hand \(m(X\setminus Y)=m(X'\setminus Y')\) implies \(y_h+y_k= 2(x_h'+x_k)\) and since \(x_h'\ge 1y_h\ge x_h,\,\) and \(x_h'\ne x_h,\) we have \(y_h+y_k < 1 =x_h+x_k,\,\) contrary to the hypothesis.
The proof is similar for different inequalities between \(x_r\) and \(1y_r\) and \(x_r'\) and \(1y_r'\) and obviously for \(x_r> y_r,\,\) and \(x_r'< y_r'.\,\) Then we have to prove the assertion for \(x_h<y_h,\,x_k > y_k\) and \(x_h< 1 y_h,\,x_k < 1 y_k.\)
In this case we can only have either \(x_h<x_h',\; y_k'<y_k\) or \(x_h'< x_h,\; y_k,y_k'.\,\) In the first case we must have \(x_h'=x_h +\epsilon \) and \(y_k'=y_k\epsilon \), with \(\epsilon >0,\) while in the second case we have \(x_h'= x_h \alpha \) and \(y_k'=y_k +\alpha \) with \(\alpha >0.\,\) On the other hand, we have \(m(X\setminus Y)= b + x_h =x_k\) and \(m(Y\setminus X) = c + y_h =y_k.\,\) If \(x_r'< 1y_r'\)\((r=h,k),\) then we have in both above cases \(m(X'\setminus Y')= b+x_h'+x_k'=b+x_h'+x_k\ne b+x_h+x_k;\) if \(x_h'< 1y_h'\) and \(x_k'> 1y_k',\) then we have \(m(X'\setminus Y') = m(X\setminus Y)\) if and only if \(x_h'y_k'+1= x_h+x_k.\)
In the case where \(x_h'=x_h+\epsilon ,\) we obtain \(x_h+\epsilon y_k+\epsilon +1= x_h + x_k\) and so \(2\epsilon = x_k+y_k1<0,\) contrary to the hypothesis. In the case where \(x_h'= x_h \alpha ,\) we must have \(x_h \alpha y_k \alpha + 1 = x_h + x_k\) and so \(2\alpha = 1y_kx_k >0.\) For this \(\alpha \,\) we have \(m(Y'\setminus X')=c+y_h'+1x_k'= c+y_h+1x_k\) and then \(m(Y'\setminus X')= m(Y\setminus X)\) if and only if \(y_h+1x_k=y_h+y_k\), which implies \(\alpha =0\) contrary to the hypothesis. The proof is similar for the other inequalities. To conclude, if the hypotheses of the theorem are satisfied, then we have at least two indices h, k such that either \(x_r\ne x_r'\) (\(r=h,k\)) or \(y_r\ne y_r'\) (\(r=h,k\)). We analyse the different cases.
– First consider the case \(x_r'>x_r\) and \(y_r=y_r'\) for \(r=h,k.\) In this case \( m(X\cap Y)= m(X'\cap Y'),\,\) implies the absurd \(y_r \le x_r <x_r'\le y_r'=y_r.\)
– Consider now the case \(x_h'< x_h,\, x_k'<x_k,\,\)\(y_r=y_r',\,\)\(x_r\le y_r,\,\)\(x_r'\le y_r',\,\)\(1x_r\le y_r\) and \(1x_r'\le y_r'\) for \(r=h,k.\) In this case, by putting \(\epsilon _h= x_h x_h' \) and \(\epsilon _k= x_k' x_k\) we have \( m(X\cap Y) = a +x_h + x_k\) and \(m(X'\cap Y') = a +x_h \epsilon _h +x_k +\epsilon _k,\) so \(m(X\cap Y) = m(X'\cap Y')\) if and only if \(\epsilon _h=\epsilon _k=\epsilon .\) Under this equality, we have moreover \(m(X\setminus Y)= b+ 2y_hy_k = m(X'\setminus Y')\) and \(m(Y\setminus X)= c+ 2x_hx_k = m(Y'\setminus X').\) Now, by simple computation, we have \(\epsilon \le y_hx_h,\,\)\(\epsilon \le y_k+x_k1\) and obviously \(\epsilon \le x_k,\) we are in one of the hypotheses of Axiom FS4 and so
– For the case \(x_r'=x_r,\, x_k'<x_k,\,\)\(y_h<y_h',\,\)\(y_k>y_k',\,\)\(x_r\le y_r,\,\)\(x_r'\le y_r',\,\)\(1y_r\le x_r\) and \(1x_r'\le y_r'\) for \(r=h,k,\) by following the same line as for the previous point and applying again Axiom FS4, we obtain:
where \(\alpha = y_h' y_h= y_k'y_k.\)
– Finally consider the case where \(x_h<x_h',\,\)\(y_k>y_k'\)\(y_h<y_h'\) and \(y_k>y_k'.\) the equivalence between (X, Y) and \((X', Y')\) can be proved by using previous results and transitivity. In fact by putting \(X'= (X_h^{\epsilon })_k^\epsilon \) and \(Y'= (Y_h^{\alpha })_k^\alpha \) we obtain
We note that the indices in which Y is different from \(Y'\) are not necessarily h and k but can be any two others.
The proof of the other cases, involving different inequalities are very similar to the one of the previous case.
Consider now the situation in which q (with \(2 < q\le p\)) elements of X are different from those of \(X'.\) Let \(\epsilon _i = x_i'x_i\) (note that \(\epsilon _i\in [1,1]\) and \(\sum _i \epsilon _i=0).\) By axiom FS4 one has \((X',Y')\sim ({{\overline{X}}}, {{\overline{Y}}}),\) where \(({{\overline{X}}}, {{\overline{Y}}})\) is obtained from \((X',Y')\) by a permutation of indices in such a way that the first elements of \({{\overline{X}}}\) are the \(x_i'\ne x_i\) and moreover they are ordered in such a way the \(\epsilon _i\) are increasing.
If \(\epsilon _1>0\) then consider \(\Bigr ( ({\overline{X}}_1^{\epsilon _1})_{r_1}^{\epsilon _1}, {{\overline{Y}}}\Bigr )\sim ({{\overline{X}}}, {{\overline{Y}}})\,\) where \(r_1\) is the first index with \(\epsilon _i<0.\)
If \(\epsilon _1< 0\) then consider \(\Bigr ( ({{\overline{X}}}_1^{\epsilon _1})_{s_1}^{\epsilon _1}, {{\overline{Y}}}\Bigr )\sim ({{\overline{X}}}, {{\overline{Y}}}),\,\) where \(s_1\) is the first index with \(\epsilon _i>0.\)
We note that \(({{\overline{X}}}_1)=\Bigr ( ({{\overline{X}}}_1^{\epsilon _1})_{s_1}\Bigr )\) (or \(({{\overline{X}}}_1)= \Bigr ( ({{\overline{X}}}_1^{\epsilon _1})_{r_1}^{\epsilon _1}\Bigr )\)) differs from \(X'\) for \(q1\) elements. Consider now \( \min \{\epsilon _2, \epsilon _{r_1}+\epsilon _1\}\) (or \( \min \{\epsilon _2, \epsilon _{s_1}\epsilon _1\}\)) and put \(\eta _2\) equal to the value with minimum modulus. Repeat for \(\eta _2\) the same procedure as for \(\epsilon _1,\) starting from \(({{\overline{X}}}_1, {{\overline{Y}}})\) and obtaining \(({{\overline{X}}}_2, {{\overline{Y}}})\sim ({{\overline{X}}}, {{\overline{Y}}}).\) We note that \(({{\overline{X}}}_2)\) differs from \(X'\) at most for \(q2\) indices. By repeating the procedure m times, with \(m\le q,\) we obtain \(({{\overline{X}}}_m, {{\overline{Y}}})\sim ({{\overline{X}}}, {{\overline{Y}}})\sim (X,Y).\) We repeat a similar procedure for the indices for which \(y_i\ne y_i'.\) If X differs from \(X'\) for q indices and Y differs from \(Y'\) for \(q'\) indices, then we prove the equivalence, yielding \(({{\overline{X}}}_m, {{\overline{Y}}})\sim ({{\overline{X}}}_m, {{\overline{Y}}}')\sim ({{\overline{X}}}, {{\overline{Y}}}')\sim ({\underline{X}},{\underline{Y}})\sim ({\underline{X}},{\underline{Y}}_k)\sim (X,Y),\) where \(({\underline{X}},{\underline{Y}})\) is a permutation of \((X,Y')\) and k is the number of steps to arrive from \({\underline{X}}\) to Y.
For the other inequalities, the proof is similar. \(\square \)
Proof of Lemma 2
First of all we notice that when the intersection is computed by \(\odot _L\) and union by \(\oplus _L\), we obtain \(m(X\cap Y)=0\) in the case where the relevant supports \(s_X\) and \(s_Y\) are disjoint, and also when \(X,Y\in {{\mathcal {Y}}}\setminus {{\mathcal {Y}}}_L\). The condition \(m(X{\varDelta } Y)=0\) is satisfied if and only if \(X =Y.\)
Therefore if either condition 2. or condition 3. holds, then the thesis follows immediately by applying Axioms FS1\(_L\) and FS0 a finite number of times.
By axiom FS2, all the pairs only differing through a permutation of indices are equivalent, so we can consider any element among them only differing from (X, Y) through a permutation and similarly for \((X', Y').\)
Due to the use of \(\odot _L\) to compute the intersection, for the elements \(z_k\in Z=X\cap Y\) we have \(z_k=0\) or \(z_k=x_k+y_k1>0,\) while for \(v_k\in V=X\setminus Y\) and \(w_k\in W= Y\setminus X,\) one is equal to 0 and the other equal to \(x_ky_k.\) So (X, Y) and \((X',Y')\) cannot satisfy the hypothesis if they only differ through one element \(x_k\) and \(x_k'\) (or \(y_k\) and \(y_k'\) ). On the contrary, the pairs (X, Y) and \((X',Y')\) satisfy hypotheses of the theorem when \(x_k\ne y_k\) and \(y_k\ne y_k'\) if \(z_k=z_k'=0\) and either (a) or (b) holds

(a)
\(v_k=v_k'=0\) and \(y_kx_k=y_k'x_k'>0\)

(b)
\(w_k=w_k'=0\) and \(x_ky_k=x_k'y_k'>0.\)
If a) holds, then we have \(\epsilon = y_k'y_k = x_k'x_k > 0 \) and if b) holds, then \(\eta = y_ky_k' = x_kx_k' > 0.\) Then, in both cases, we are in the hypothesis of Axiom FS4\(_L\) and so \((X,Y)\sim (X',Y').\) It is easy to see that we cannot have the only two inequalities \( x_k\ne x_k'\) and \(y_h\ne y_h'\) (with \(h\ne k\)).
Let us consider now the case in which \(x_i\ne x_i'\) (\(i= h,k\)) or equivalently \(y_i\ne y_i'\) (\(i= h,k\)) and \(x_j = x_j',\, y_j= y_j'\) (\(j\ne h,k\)). If \(x_i\ne x_i'\) and \(y_i= y_i'\), then we can only have \(m(X\cap Y) = m (X'\cap Y')\) if either c) or d) holds:

(c)
\(x_i+y_i \le 1\) and \(x_i'+y_i' = x_i'+y_i\le 1\) (\(i= h,k\))

(d)
\(x_i+y_i > 1\) and \(x_i'+y_i'=x_i'+y_i > 1\) (\(i= h,k\)) and \(x_h +x_k = x_h' +x_k'.\)
Let us consider case (c) and put \(\epsilon _i =x_i'x_i\) (\(i= h,k\)). We necessarily have \(\epsilon _i \le 1 y_ix_i\) (\(i= h,k\)). It is easy to see that \(m(X\setminus Y)= m(X'\setminus Y'),\) and \(m(Y\setminus X)= m(Y'\setminus X')\) only in the case where either \(x_i\le y_i\) (\(i= h,k\)) or \(x_i > y_i\) (\(i= h,k\)). If \(x_i\le y_i,\) then \(m(X\setminus Y) = m (X'\setminus Y')\) follows directly from the condition on \(\epsilon _i\) and we have also \(m(Y\setminus X) = m (Y'\setminus X')\) if and only if \(x_ky_k+y_k+x_hy_h=x_k'y_k+y_k+x_h'y_h\) and so \(\epsilon _h= \epsilon _k= \epsilon > 0.\) If \(x_i> y_i\) then \(m(X\setminus Y) = m (X'\setminus Y')\) and \(m(Y\setminus X) = m (Y'\setminus X')\) if and only if \(y_kx_k+y_hx_h=y_kx_k'+y_hx_h'\) and so \(\epsilon _h= \epsilon _k= \eta >0.\)
In both cases we are under the hypotheses of FS4\(_L\) and so \((X,Y)\sim (X',Y').\)
The proof in the case (d) follows by similar considerations, as well as the case where \(x_i\ne x_i'\) for (\(i=1,\ldots ,q>2\)). \(\square \)
Proof of Lemma 3
Let us suppose that Axioms FS0, FS2, FS4,FS4\('\), FS5, FS5\('\) and FS5\(_G\) hold. By axiom FS2, all the pairs only differing through a permutation of indices are equivalent, so we can consider any element among them only differing from (X, Y) only through a permutation and similarly for \((X', Y').\) We only need to prove the thesis in the case where \(m(X\cap Y)>0,\, m(X{\varDelta } Y)>0\) and either \( m(X\setminus Y)\ne m(X'\setminus Y') \,\) or \( m(Y\setminus X)\ne m(Y'\setminus X'), \,\) since the other case is proved in Lemma 1. First consider the case in which (X, Y) only differs from \((X',Y'),\) through one pair \((x_h,y_h),\,(x_h',y_h')\) and suppose \(x_h<x_h'.\) In this case, by condition \( m(X\cap Y) = m(X'\cap Y') \,\) it follows that one of the next situations occurs:

(1)
\(y_h\le x_h,\,y_h' < x_h',\,\) with \(y_h = y_h'\)

(2)
\(x_h\le y_h,\,y_h'\le x_h',\,\) with \(x_h = y_h'\) (or \(x_h\ge y_h\)\(y_h'\le x_h',\,\) with \(y_h = x_h'\)).
Put: \(\epsilon =x_h'x_h\) (\(\epsilon \le 1x_h\)),
\(e=\sum _{r\ne h}min\{x_r, y_r\} = \sum _{r\ne h}min\{x_r', yr'\}; \,\)
\(d =\sum _{i\ne h}\max \Big \{\min \{1x_i, y_i\}, \min \{1y_i, x_i\}\Big \}= \sum _{i\ne h}\max \Big \{\min \{1x_i', y_i'\}, \min \{1y_i', x_i'\}\Big \}.\)
If condition (1) holds, to have \( m(Y{\varDelta } X) = m (Y'{\varDelta } X'),\,\) we must have
\( m(Y{\varDelta } X)= d +\max \Big \{\min \{1x_h, y_h\}, \min \{1y_h, x_h\}\Big \}= d+ \max \Big \{ \min \{1x_h\epsilon , y_h\}, \min \{1y_h, x_h+\epsilon \}\Big \}= m (Y'{\varDelta } X').\)
Now, if \(y_h\ge 1x_h\) then \( m(Y{\varDelta } X)= Q +\max \{1x_h, 1y_h\}= Q +1y_h,\) and so \(1x_h\epsilon \le 1y_h.\) Thus we are under the hypotheses of FS5 and so \((X,Y)\sim (X',Y').\)
On the contrary, if \(y_h< 1x_h\) then \( m(Y{\varDelta } X)= d +\max \{x_h, y_h\}\) and \( m(Y'{\varDelta } X')= d +\max \{x_h +\epsilon , y_h\}= d+ x_h +\epsilon ,\) when \(\epsilon \le 1y_hx_h,\) and so \( m(Y\cap X)\ne m(Y'\cap X'),\) contrary to the hypothesis. On the other hand, if \(\epsilon > 1y_hx_h,\) implies \( m(Y{\varDelta } X)= d +\max \{1x_h+\epsilon , 1y_h\}=d + 1y_h.\) In this case, to have \( m(Y{\varDelta } X)= m(Y'{\varDelta } X'),\) we must have \(1y_h=x_h,\) contrary to the hypothesis \(y_h < 1x_h.\)
Let us consider situation 2). In this case, under the hypothesis \(y_h\ge 1x_h\), one has \( m(Y{\varDelta } X)= d +\max \{1x_h, 1y_h\}=d+ 1x_h\) and \( m(Y'{\varDelta } X')= d +\max \{1x_h', 1 y_h\}.\) The maximum cannot be \(1x_h'\) since in this case \( m(Y{\varDelta } X)\ne m(Y'{\varDelta } X').\) On the other hand if the max is \(1y_h\) then \( m(Y{\varDelta } X)\ne m(Y'{\varDelta } X')\) if and only if \(1y_h=1x_h,\) but in this case \(y_h=x_h=y_h'\) and so \(x_h'< y_h'\) contrary to the hypothesis.
Consider now the case \(x_h'<x_h\) and put \(\eta = x_hx_h'>0.\) Then to have \( m(Y\cap X) = m (Y'\cap X'),\,\) one of the following situations occurs:

(1’)
\(y_h< x_h,\,y_h' \le x_h',\,\) with \(y_h = y_h',\)

(2’)
\(y_h < y_h,\,x_h'\le y_h',\,\) with \(x_h' = y_h.\)
If we are in (1’) then necessarily \(\eta \le x_hy_h\) and then \( m(Y\cap X) = e+ 1y_h\) and \( m(X'{\varDelta } Y') = d + \max \Bigr \{ \min \{1x_h+\eta , y_h\}, \min \{1y_h,x_h+\eta \}\Bigr \}.\) Consider now the two cases \(\eta \le x_h+y_h1\) and \(\eta > x_h+y_h1.\) In the first case \(m (X'{\varDelta } Y')= d + \max \{1x_h+\eta , 1y_h\}= d +y_h\) and so \(m (X'{\varDelta } Y')= m (X{\varDelta } Y).\) But in this case, we are in the hypotheses of FS5\(_G\) and so \((X,Y)\sim (X',Y').\)
If we are in the second case, then, to obtain
\(m (X'{\varDelta } Y') = m (X{\varDelta } Y)\) we must have \(1y_h=x_h'=x_h\eta \), contrary to the hypothesis.
Let us consider now condition (2’). By equality \(x_h' = y_h,\) and by using Axiom FS4, we obtain a situation already studied in case (1), when \(y_h'\le x_h,\) and the one already studied in case (1’), when \(y_h>x_h.\)
Consider now the case where \(x_h\ne x_h'\) and \(y_h\ne y_h'.\) By using Axiom FS4\('\), we obtain situations equal to those studied before. If we have more than one index in which X and \(X'\) (or Y and \(Y'\)) differ, then we can follow the strategy used in the proof of Lemma 2, obtaining the equivalence between (X, Y) and \((X',Y')\) by a number of steps less or equal to the number of indices in which the pairs differ, if at each step i, indicating by \(X^i\) the relevant transformation of X, we have \(m(X^i{\varDelta } Y) = m(X^{i1}{\varDelta } Y),\) and similarly for Y.
Suppose now to have two indices h, k such that \(x_h\ne x_h',\, y_k\ne y_k',\)\(m (Y\cap X)= m (Y'\cap X')\) and either
\(m (X'{\varDelta } Y)\ne m (X'{\varDelta } Y')= m (X{\varDelta } Y)\) or \( m (X{\varDelta } Y')\ne m (X'{\varDelta } Y')= m (X{\varDelta } Y).\) Then it is easy to see that we can only obtain this situation in the following cases:
(3) \(y_h> x_h,\,y_k > x_k,\, x_i\le 1y_i\) (or \(x_i\ge 1y_i\)) for \(i=h, k,\) and \(x_h'>x_h\ge y_h,\, x_k\le x_k' < y_k\,\) and \(x_h'\le 1y_h, x_k\le 1y_k'\,\) (or \(x_h'> 1y_h, x_k > 1y_k'\))
(4) \(x_h < y_h,\,y_k\le x_k,\,x_i\le 1 y_i\) (or \(x_i > 1y_i\)) for \(i=h, k,\) and \(x_h \le x_h'< y_h,\, y_k'<y_k\le x_k,\,\)\(x_h'\le 1y_h,\, x_k\le 1y_k'\) (or \(x_h'>1y_h,\, x_k > 1y_k').\)
Let us consider situation 3) with \(x_i\le 1y_i\) and put \(\epsilon =x_h'x_h.\) then we necessarily have \(\epsilon \le 1x_hy_h.\) In this case we have \( m(X\cap Y) = m (X'\cap Y') = \sum _{i=1}^{n} \min \{x_i, y_i\}.\)
If we denote
\(g=\sum _{r\ne k,h}\max \Bigr \{\min \{x_r, 1 y_r\}, \min \{y_r, 1 x_r\} \Bigr \},\)
we have:
\(m(X'{\varDelta } Y) = g+ x_h+y_k'\ne m (X{\varDelta } Y)= g+ x_h+y_k\ne m(X{\varDelta } Y')=g+x_h'+y_k.\)
To have \(m(X'{\varDelta } Y')=g+ x_h'+y_h'=g +x_h+y_h = m(X{\varDelta } Y)\) we must have \(y_k'= y_k\epsilon ,\) and from condition 3) \(\epsilon \le y_kx_k.\) Then we are in the hypothesis of Axiom FS5\('\) and so \((X,Y)\sim (X',Y').\)
Similarly for \(x_i\ge 1y_i\) and for condition 4).
Finally if we have more than 2 pairs in which (X, Y) and \((X',Y')\) differ then we follow the strategy analysed in the previous steps. \(\square \)
Proof of Lemma 4
The proof is similar to that of Lemma 2. So, also here, we only consider the pairs with \(m(X\cap Y)>0,\, m(X{\varDelta } Y)>0\) and either \( m(X\setminus Y)\ne m(X'\setminus Y') \,\) or \( m(Y\setminus X)\ne m(Y'\setminus X'), \,\) since the other case is proved in Lemma 1. We start by considering the case where (X, Y) only differs from \((X',Y'),\) through one pair \((x_h, y_h),\,(x_h',y_h')\) and we suppose \(x_h<x_h'.\) Then by condition \( m(X\cap Y) = m(X'\cap Y') \,\) it follows that one of the following situations occurs:

(1)
\(y_h < 1x_h,\,y_h'\le 1 x_h';\,\)

(2)
\(x_h + y_h=y_h'+x_h'>1\).
Put: \(\epsilon = x_h'x_h\) (\(\epsilon \le 1x_h\)).
Let us consider situation 1). In this case, we obtain \(x_h\odot _{L} y_h = x_hy_h\ne x_h'y_h'= x_h'\odot _{L}y_h'\) only when \(y_h'= y_h \epsilon ,\) which assures that we are in the hypothesis of FS4\(_L\) and so \((X,Y)\sim (X',Y').\)
Let us consider situation (2). It is immediate to see that we cannot have it. In fact, in this case, we can have \(m(X\cap Y)= m(X'\cap Y')\) if (and only if) \(y_h'=y_h\epsilon .\) But for these values one has \(x_h'y_h'=x_hy_h +2\epsilon \ne x_hy_h\) and so \(m(X{\varDelta } Y)\ne m(X'{\varDelta } Y),\) contrary to the hypothesis. The proof for the case \(x_h\ge x_h'.\) is analogous.
Suppose now that we have two indices h, k such that \(x_h\ne x_h',\, y_k\ne y_k',\)\(m (Y\cap X)= m (Y'\cap X')\) and either \(m (X'{\varDelta } Y)\ne m (X'{\varDelta } Y')= m (X{\varDelta } Y)\) or \( m (X{\varDelta } Y')\ne m (X'{\varDelta } Y')= m (X{\varDelta } Y).\) Then it is easy to see that we can only obtain this situation in the following cases:
(3) \(y_i \le 1= x_i,\, x_i'\le 1y_i'\) (\(i= h. k\)) and \(x_h  y_h+ x_ky_k= x_h'  y_h'+ x_k'y_k',\) with either \(x_h  y_h\ne x_h'  y_h'\) or \(x_ky_k\ne x_k'y_k',\)
(4) \(y_i> 1 x_i\) and \(y_i'> 1 x_i'\) (for \(i=h, k,\) ), \(x_h + y_h+ x_k+y_k= x_h' + y_h'+ x_k'+y_k'\) and \(x_h  y_h+ x_ky_k= x_h'  y_h'+ x_k'y_k',\) with at least one of the following inequalities hold \(x_h + y_h\ne x_h' + y_h',\,\)\(x_k+y_k\ne x_k'+y_k',\,\)\(x_h  y_h\ne x_h'  y_h',\,\)\(x_ky_k\ne x_k'y_k'.\)
Suppose \(y_k'>y_k\) and put \(\eta = y_k'y_k.\)
When we are in the situation (3) we are in the hypothesis of Axiom FS5\('\), while when we are in the situation (4) we are in the hypothesis of Axiom FS5\(_L.\) Then in both cases we obtain \((X,Y)\sim (X',Y').\)
If we have more than 2 pairs in which (X, Y) and \((X',Y')\) differ then we follow the strategy analysed in the previous steps. \(\square \)
Proof of Lemma 5
We first consider \((\min , \max )\) as operators and prove item a). Recalling that for every i the \(i\)th element of intersection is \(\min \{x_i, y_i\}\) and the corresponding element of the symmetrical difference \({\varDelta }\) is \(\max \{x_i, y_i\},\) for \(x_i+y_i\le 1\) and \(\max \{1x_i, 1 y_i\},\) for \(x_i+y_i\ge 1,\) we can obtain \((X^*,Y^*),\) by subtracting from each minimum element a positive quantity until we get the sum of the new minima equal to k. For every index i such that \(x_i+y_i\le 1\) this procedure does not influence the ith element of \({\varDelta }\). For any other index j, we can only act in the case where the quantity we would decrease is sufficiently large to arrive at least to \(1\min \{x_j, y_j\};\) in this case \(\max \{x_j^*,y_j^*\}=1\min \{x_j, y_j\}.\)
To construct the pair \((X^{**},Y^{**})\) in item b), first of all we notice that, for every i, the maximum value we can obtain in \(X'{\varDelta } Y'\) (and in \(X{\varDelta } Y\)) is \(\min \{x_i ,y_i\}\) (and this value is unique for all the pairs \(\{x_i,y_i)\) such that \(x_i+y_i\ge 1\)); so necessarily \(h\le \sum _i\min \{x_i ,y_i\}.\) To construct \((X^{**},Y^{**})\) it is sufficient to increase the \(\max \{x_i, y_i\}\) without exceeding the above limits.
To construct the pair \((X^o,Y^o)\) in item c), we can proceed in this way to obtain \((x_i^o,y_i^o)\): if \(x_i\le y_i,\) we can only act when \(\,x_i+y_i\le 1\,\) (in the opposite case the pair is uniquely identified), in this case we can increase \( y_i\) with the constraint \( 1y_i^o\ge x_i;\,\) in the same way, if \(x_i\ge y_i,\) we can decrease \(x_i\) with the constraint \( 1y_i^o\le x_i.\)
The proof of item (c\('\)) is analogous.
Consider now \((\odot _{L},\oplus _{L})\) as operators. Recalling that for every i the \(i\)th element of intersection is either \(x_i+y_i1,\) (if \(x_i+y_i\ge 1\)) or 0 (if \(x_i+y_i < 1\)), we only can act in the pairs \((x_i,y_i)\) with \(x_i+y_i\ge 1.\) Then to prove item a), it is sufficient to construct \((X^*,Y^*),\) by subtracting an equal amount to \(x_i\) and to \(y_i,\) maintaining the corresponding value \(x_iy_i,\) in \(X{\varDelta } Y.\)
To construct the pair \((X^{**},Y^{**})\) in item b), we can increase the value of \(x_iy_i,\) until \(x_iy_i=1\), for the pairs \((x_i,y_i)\) such that \(x_i+y_i\le 1\), and increase \(x_iy_i,\) maintaining the condition \(x^{**}_i+y^{**}_i=x_i+y_i.\)
Finally, to construct the pair \((X^{o},Y^{o})\) in item c), we only can act on the pairs \((x_i,y_i)\) with \(x_i\le y_i,\) increasing the value of \(y_ix_i,\) maintaining the sum constant. The proof of item (c\('\)) is analogous. \(\square \)
Proof of Theorem 3
We first prove that (i)\(\Rightarrow \) (ii). Axioms FS0 and Q are sufficient conditions for the existence of a function \(S :{{\mathcal {Y}}}^2\rightarrow [0,1]\) representing \(\preceq \) (Krantz et al. 1971). Axioms FS0, FS2, FS4, FS4\('\), FS5, FS5\('\) and FS5\(_G\) through Lemma 3 ensure that S only depends on \(m(X\cap Y)\) and \(m(X{\varDelta } Y)\). Axioms FS0, FS2, FS5, FS5\('\), FS5\(_G\) and S7\(_1\) through Lemma 6 ensure that S is increasing with respect to \(m(X\cap Y)\) and decreasing with respect to \(m(X{\varDelta } Y).\)
Finally Axioms FS0, FS1\(_G\), FS2, FS4 and FS4\('\), taking into account Lemma 1 and Corollary 1, ensure the validity of statement (c) of condition (ii).
We now prove \(ii)\Rightarrow i).\) Every binary relation \(\preceq \) representable by a real function satisfies Axiom FS0 and Q (Krantz et al. 1971). We prove statement a) of FS1\(_G\). Condition (c) in (ii) implies that \(S(X,X)=1,\) if and only if \(X\in {{\mathcal {X}}}.\) To prove that, for every \(X, Y\in {{\mathcal {X}}}\) one has \(S(X, Y) \le S(X, X),\) consider first that \(m (X\cap Y) = \sum _i min(x_i,y_i)\le \sum _i x_i= m(X\cap X).\) Moreover for every \(x_i\le 1x_i\), we have \(\max (x_i,y_i)\ge x_i,\) if \(x_i< 1y_i\) and \(\max (1x_i, 1y_i)\ge x_i,\) if \(x_i\ge 1y_i.\) Similarly for the case \(x_i > 1y_i.\) Therefore
Then condition (a) of axiom FS1\(_G\) follows from the representability of \(\preceq \) by S and properties of S in condition (a) and (b) of (ii).
To prove condition (b) of FS1\(_G,\) let us consider that, under the hypotheses indicated in the axiom, \(m(X\cap X')=m(Y\cap Y')=0,\,\)\(m(X\cap Y) > 0\) and \(m(X{\varDelta } X')=m(Y{\varDelta } Y')\ne 0.\) Then condition b) follows from the representability of \(\preceq \) by S and properties of S quoted in (ii).
To prove axiom FS2, let us consider that, if \(X\cap Y={\underline{0}},\) we also have \(X_k^c\cap Y = X\cap Y_h^c ={\underline{0}}\) for every \(k\in s_{X\cup (X^c\cap Y^c)}\) and \(h\in s_{Y\cup (X^c\cap Y^c)}.\) So, since \({\varPhi }(0,\cdot ) = 0\), we have \(S(X, Y) = S(X_k^c, Y)= S(X, Y_h^c) = 0.\) Similarly, if \(X{\varDelta } Y={\underline{0}},\) and \(X\cap Y\ne {\underline{0}}\), we have, for every k, \(X_k^c=Y_k^c\) and, if \(X_k^c \ne {\underline{0}},\) by condition (c) we obtain \(S(X,X)= S(X_k^c, X_k^c) =1.\) Then Axiom FS2 follows from the representability of \(\preceq \) by S.
To prove axiom S4, consider that the fuzzy cardinality does not depend on the specific elements, and S(X, Y) is a function of the cardinality of the intersection and the symmetrical difference of X and Y.
To prove axiom FS5, it is sufficient to consider that, for every \(k\in \sigma _{X{\varDelta } Y},\) (X, Y) and \((X_k^c, Y_k^c)\) are such that \(m(X\cap Y)= m(X_k^c\cap Y_k^c)\) and \(m(X{\varDelta } Y)= m(X_k^c{\varDelta } Y_k^c)\) and so \(S(X,Y)=S(X_k^c, Y_k^c),\) which implies \((X,Y)\sim (X_k^c, Y_k^c).\)
To prove axiom FS4, it is sufficient to consider that, for every \(\alpha ,\beta ,\varepsilon \) and \(\eta ,\) with ranges as in the hypotheses of the axiom, the fuzzy cardinality of the intersection and the symmetrical difference of all the considered pairs is the same. So the value of the relevant S is the same and then all the pairs are necessarily equivalent.
The proof of axioms FS5\('\) and FS5\(_G\) is obtained by the same considerations as for axiom FS4\('\).
To prove axiom S\(7_1,\) it is sufficient to consider that the fuzzy cardinality is increasing with respect to the partial order induced by the inclusion and the function \({\varPhi }\) is increasing with respect to the first variable \(m(X\cup Y)\) and decreasing with respect to the second variable \(m(X{\varDelta } Y).\)\(\square \)
Proof of Theorem 9
To prove that \(\preceq \) satisfies axioms FS0, FS1\(_G\), FS2, FS4\('\), FS5, FS5\('\), FS5\(_G\), S\(7_1\) and Q it is sufficient to prove that \(S_{f,g}\) is a function fulfilling properties of function \({\varPhi }\) in Theorem 3. In fact \(S_{f,g}(X,Y)= {\varPhi }\big (m(X\cap Y), m(X{\varDelta } Y)\big )\). The nonnegativity of f and g implies that the values taken by \(S_{f,g}\) are in [0, 1]. Condition \(f(0)=g(0)=0\) and the definition of \(S_{f,g}({\underline{0}},{\underline{0}})=0\) ensures that \({\varPhi }(0,b) = 0\) and \({\varPhi }(a,0) = 1,\) for every \(a\ne 0\). Moreover, taking into account that f and g are increasing functions, it is immediate to see that \(S_{f,g}\) is increasing with respect to \(m(X\cap Y)\) and decreasing with respect to \(m(X{\varDelta } Y).\)
Axiom FWI follows by a direct computation, taking into account Lemma 10 and the fact that f and g are strictly increasing.
We prove now that there exists a weak order \(\preceq '_*,\) extending \(\preceq '\) satisfying conditions I), NT), AC), RS) and T). For that we first note that \(\preceq '\) satisfies I) thanks to Lemma 10 and condition NT) immediately follows, since \({{\mathcal {Y}}}\) contains all the fuzzy subsets of \({{\mathcal {H}}}.\) Condition AC) is implied by the Archimedean property of real numbers, \({\varTheta }_i\subseteq {\mathbb {R}},\)\((i=1,2)\). Finally to prove conditions RS) and T) it is sufficient to refer to items (a) and (b) of Lemma 5. Then the relation \(\preceq '_*\) induced on \({\varTheta }'_*\) by function in (5) extends \(\preceq '.\) The proof that \(\preceq '_*\) satisfies I). NT), AC) and RS) is direct. \(\square \)
Proof of Theorem 13
The implication (ii)\(\Rightarrow \) (i) has been proved in Theorem 9. We prove (i)\(\Rightarrow \) (ii). First of all we note that, since \(\preceq \) satisfies conditions FS0, FS2, FS5, FS5\('\) and FS5\(_G\), thanks to Theorem 1, if a real function represents \(\preceq '\), then it represents \(\preceq .\)
Let us consider \(\preceq '_*\) in \({\varTheta }_1\times {\varTheta }_2\) and \({{\mathcal {C}}}= {\mathbb {R}}^* ={\mathbb {R}}\cup \{\infty , +\infty \}\) (that is the compactification of \({\mathbb {R}}\)) which is a totally ordered set, by extending in a natural way the usual order of \({\mathbb {R}}.\)
The implication \(ii)\Rightarrow i)\) has been proved in Theorem 10. We prove \(i)\Rightarrow ii).\) First of all we note that, since \(\preceq \) satisfies conditions FS0, FS2, FS5, FS5\('\) and FS5\(_G\), thanks to Theorem 1, if a real function represents \(\preceq '\), then it represents \(\preceq .\)
We first note that, by Axioms FS0, FS1\(_G\), FS2, FS4 and FS4\('\), through Corollary 1, we have \( ({\underline{0}},{\underline{0}})\sim (X^c,X)\) for every \(X\in {{\mathcal {X}}},\,\)\(X\ne {\underline{0}}\), so we can consider \(\preceq \) restricted to \({{\mathcal {X}}}^2\setminus \{({\underline{0}},{\underline{0}})\}\) and then extend the function representing \(\preceq ,\) by assigning to \(({\underline{0}},{\underline{0}})\) the same value assigned to the pairs \((X^c,X)\).
Now, by Theorem 7, there exist two functions \({\varPhi }_i: {\varTheta }_i \rightarrow {\mathbb {R}}^*\) (\(i=1,2\)) such that the function
represents \(\preceq '_*\) and so the restriction \(\preceq '\) in \({\varTheta }'.\) Therefore, by Theorem 1, \(G(X,Y) = {\varPhi }(m(X\cap Y), m(X{\varDelta } Y))\) represents \(\preceq .\) From Theorem 3, function \({\varPhi }_1\) is strictly increasing and function \({\varPhi }_2\) is strictly decreasing.
The conclusion of the proof can be obtained by the same consideration as in the proof of Theorem 5 in Coletti and BouchonMeunier (2019a), we report it for completeness.
By Axiom FS2, \({\varPhi }_2 (0)= +\infty \): in fact condition b) of Axiom FS1\(_G\) implies that for every \(X,Y\in {{\mathcal {X}}}\subset {{\mathcal {Y}}},\,\)\(X\ne {\underline{0}}\) one has \((X,X)\sim (Y,Y)\,\), and so, if \(x = X\cap X< Y\cap Y = y\), we must have \({\varPhi }_1(x) +{\varPhi }_2(0) = {\varPhi }_1(y) +{\varPhi }_2(0)\). So, since \({\varPhi }_1\) is strictly increasing, necessarily \({\varPhi }_2(0)= +\infty .\) With the same considerations, we have \({\varPhi }_1(0)= \infty .\) Now we put \({\varPhi }'_2= {\varPhi }_2\) and we consider the function \(\ln :[0, +\infty ]\rightarrow {\mathbb {R}}^*\) (defined by putting \(\ln (0)= \infty \) and \(\ln (+\infty ) = + \infty \)) and the function \(\exp : {\mathbb {R}}^* \rightarrow [0, +\infty ]\) (defined by putting \(\exp (\infty ) = 0\) and \(\exp (+\infty ) = + \infty \)). By using these functions, we can write for every \((X,Y)\ne ({\underline{0}},{\underline{0}}):\)
The function \(G: {{\mathcal {Y}}}^2\setminus \{({\underline{0}},{\underline{0}})\}\rightarrow [\infty , +\infty ]\) represents \(\preceq \) on \({{\mathcal {Y}}}^2\setminus \{({\underline{0}},{\underline{0}})\}\); taking into account the previous considerations, it can be extended to \({{\mathcal {Y}}}^2\) by putting:
Since \(G': {{\mathcal {Y}}}^2\ \rightarrow {\mathbb {R}}^*\) represents \(\preceq \) on \({{\mathcal {Y}}}^2\), then any of its strictly increasing transformations \(\varphi \) also represents \(\preceq \) on \({{\mathcal {Y}}}^2\). In particular \(\varphi (\exp (G'))\) represents \(\preceq \), with \(\exp :{\mathbb {R}}^* \rightarrow [0, +\infty ]\) and \(\varphi : [\infty , +\infty ]\rightarrow [0,1]\) defined as follows:
By putting \(\exp ({\varPhi }_1)=f\) and \(\exp ({\varPhi }_2') = g\) we obtain function \(S_{f,g}\) as defined in Eq. (5). \(\square \)
Proof of Theorem 14
The proof follows the same line as the proof of Theorem 13, by considering this time \({\varTheta }''_*\) and \(\preceq ''_*\) and taking into account Lemma 1. In this case, from Theorem 7, there exist three functions \({\varPhi }_1, {\varPhi }_2, {\varPhi }_3,\) taking values on \({\mathbb {R}}^*,\) such that:
\(G(X,Y) ={\varPhi }(m(X Y),m[(X Y)\cup (X\setminus Y)], m[(X Y)\cup (Y\setminus X)]) ={\varPhi }_1(m(X Y)) + {\varPhi }_2(m[(X Y)\cup (X\setminus Y)]) +{\varPhi }_3 (m[(X Y)\cup (Y\setminus X)])\)
(where \(XY=X\cap Y\)) represents \(\preceq '',\) and so represents \(\preceq .\) From Theorem 8, function \({\varPhi }_1\) is strictly increasing and functions \({\varPhi }_2, {\varPhi }_3\) are strictly decreasing.
The conclusion of the proof can be obtained by the same considerations as in the proof of Theorem 6 in Coletti and BouchonMeunier (2019a).
Among the possible triples \(\{{\varPhi }_i\}\), we can choose one of them such that \({\varPsi }_i(1) = 0\) for \(i=1,2,3.\) This is possible since, if \({\varPhi }={\varPhi }_1+{\varPhi }_2+{\varPhi }_3\) represents \(\preceq ,\)\({\varPhi }' = ({\varPhi }_1 k_1)+({\varPhi }_2 k_2)+({\varPhi }_3k_3)\) also represents \(\preceq .\) From Axiom FS1\(_G\), by following the same considerations as in the previous theorem, we have \({\varPhi }_1(0) = \infty \) and \({\varPhi }_2(0) = {\varPhi }_3 (0)= +\infty .\) Moreover, still from condition b) of Axiom FS1\(_G\), we have that, for every \(x, y > 0\), \({\varPhi }_1(x) + {\varPhi }_2(x) +{\varPhi }_3 (x) = {\varPhi }_1(y) + {\varPhi }_2(y) +{\varPhi }_3 (y).\) So, taking into account that \( {\varPhi }_1(1) = {\varPhi }_2(1) {\varPhi }_3 (1)=0,\) necessarily \( {\varPhi }_1(x) = {\varPhi }_2(x) {\varPhi }_3 (x).\) By putting \({\varPhi }_i'=  {\varPhi }_i\,\)\((i=2,3)\) and considering the functions \(\exp \) and \(\ln \) defined on \({\mathbb {R}}^*\) as in Theorem 13, we obtain the function \(T_{f,g}(X,Y)\) defined by Eq. (6), with \(\exp ({\varPhi }_1)=h,\,\exp ({\varPhi }_2') = f\) and \(\exp ({\varPhi }_3') = g\), with \(h(x)=f( x) g(x)\) for every x, as required. The considerations about the extension to \({{\mathcal {Y}}}^2\) of the function representing \(\preceq \) on \({{\mathcal {Y}}}^2\setminus \{({\underline{0}},{\underline{0}})\},\) made in the proof of Theorem 13, also hold in this context. \(\square \)
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Coletti, G., BouchonMeunier, B. A study of similarity measures through the paradigm of measurement theory: the fuzzy case. Soft Comput 24, 11223–11250 (2020). https://doi.org/10.1007/s00500020050549
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Keywords
 Fuzzy similarity measures
 Comparative similarity
 Boundary axioms
 Monotonicity axioms
 Independence axioms
 Representability by fuzzy similarity measures