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Corrigendum to “The distributivity condition for uninorms and t -operators” [Fuzzy Sets and Systems, 128 (2002) 209–225]
Fuzzy Sets and Systems 153 (2005) 297 – 299 www.elsevier.com/locate/fss
Corrigendum
Corrigendum to “The distributivity condition for uninorms and t-operators” [Fuzzy Sets and Systems, 128 (2002) 209–225]夡 M. Mas, G. Mayor, J. Torrens∗ Dpt. de Ciències Matemàtiques i Informàtica, Universitat de les Illes Balears, Crta. de Valldemossa, km. 7.5, 07122 Palma de Mallorca, Spain Received 28 September 2004; received in revised form 22 December 2004; accepted 28 January 2005
Abstract In this paper, the study of the distributivity equation involving uninorms in U given in (Fuzzy Sets and Systems, 128 (2002) 209–225) is revised. Two wrong propositions in the mentioned reference are corrected and their right versions are proved. © 2005 Elsevier B.V. All rights reserved. Keywords: Distributivity; Functional equation; Uninorm; t-norm; t-conorm; t-operator
The distributivity condition for two uninorms from U was studied in Section 6 of [1]. This condition can be written as U1 (x, U2 (y, z)) = U2 (U1 (x, y), U1 (x, z))
for all
x, y, z ∈ [0, 1],
(1)
where U1 and U2 are uninorms from U . In that paper the authors consider two cases, first when U1 lies in UMax and second when U1 lies in UMin . In both cases similar results are obtained. Unfortunately, the result stated in Proposition 6.2 for uninorms from UMax is wrong and consequently the corresponding proposition for uninorms from UMin , Proposition 6.6, also fails. In the same way, Figs. 9 and 11 in [1] representing these results need to be changed. 夡
DOI of original article: 10.1016/S0165-0114(01)00123-3.
∗ Corresponding author.
E-mail addresses: [email protected] (M. Mas), [email protected] (G. Mayor), [email protected] (J. Torrens). 0165-0114/$ - see front matter © 2005 Elsevier B.V. All rights reserved. doi:10.1016/j.fss.2005.01.016
298
M. Mas et al. / Fuzzy Sets and Systems 153 (2005) 297 – 299 1
1
S
Max
Max
Max
Min
Max
e
e T
0
Max
e
1
0
e
1
Fig. 1. Two distributive uninorms U1 (left) and U2 (right) from UMax with 0 = e2 = e1 = e.
1
1
S
Min
Max
Min
Min
e
e T
0
Min
Min
e
1
0
e
1
Fig. 2. Two distributive uninorms U1 (left) and U2 (right) from UMin with 1 = e2 = e1 = e.
In Propositions 6.2 and 6.6 we present the corrected versions. Differences between old and new propositions are in the set of solutions. Uninorms in Proposition 6.2 (and Proposition 6.6) with 0 < e2 < e1 (and e1 < e2 < 1), where e1 and e2 are the neutral elements of U1 and U2 , respectively, are not actually solutions of Eq. (1). Note that a necessary and sufficient condition is claimed in Proposition 6.2 in [1]. In fact, this condition is necessary but not sufficient since many uninorms stated in this proposition are not distributive in general as it has been mentioned above. Thus, the corrected version is as follows. Proposition 6.2. Let U1 = (e1 , T , S) and U2 = (e2 , T , S ) be uninorms from UMax and e2 e1 . Then U1 and U2 satisfy Eq. (1) if, and only if, one of the following cases holds:
M. Mas et al. / Fuzzy Sets and Systems 153 (2005) 297 – 299
299
• e2 = 0 and U2 is the t-conorm maximum. • 0 = e2 = e1 and U2 is idempotent. Proof. The first two steps in the proof given in [1] are correct. Thus, U2 must be idempotent when U1 and U2 satisfy (1). Now, we can distinguish two cases: • e2 = 0. In this case, if U1 and U2 satisfy (1), U2 is an idempotent t-conorm and so U2 = max. Conversely, it is clear that any uninorm U1 is distributive with the maximum. • 0 < e2 . Let U1 and U2 be distributive and suppose that e2 < e1 . In this case, we have U2 (e1 , 0) = e1 . For 0 < x < e2 < e1 it follows from (1) that x = U1 (x, e1 ) = U1 (x, U2 (e1 , 0)) = U2 (U1 (x, e1 ), U1 (x, 0)) = U2 (x, 0) = 0 which is a contradiction. Thus e1 = e2 and we already know that U2 must be idempotent. Conversely, it is a straightforward computation to prove that U1 and U2 are distributive whenever e1 = e2 and U2 is idempotent. Similarly, Proposition 6.6 in [1] must be corrected as follows: Proposition 6.6. Let U1 = (e1 , T , S) and U2 = (e2 , T , S ) be uninorms from UMin and e1 e2 . Then U1 and U2 satisfy Eq. (1) if, and only if, one of the following cases holds: • e2 = 1 and U2 is the t-norm minimum. • 1 = e2 = e1 and U2 is idempotent. Finally, Figs. 9 and 11 in [1] must be replaced by Figs. 1 and 2 above, where the non-trivial solutions of Eq. (1) given in Propositions 6.2 and 6.6, respectively, can be viewed. Acknowledgements The authors are grateful to Fuzzy Sets and Systems for publishing this corrigendum and they actually apologize for not having found this error before the publication of the article. References [1] M. Mas, G. Mayor, J. Torrens, The distributivity condition for uninorms and t-operators, Fuzzy Sets and Systems 128 (2002) 209–225.
Corrigendum
Corrigendum to “The distributivity condition for uninorms and t-operators” [Fuzzy Sets and Systems, 128 (2002) 209–225]夡 M. Mas, G. Mayor, J. Torrens∗ Dpt. de Ciències Matemàtiques i Informàtica, Universitat de les Illes Balears, Crta. de Valldemossa, km. 7.5, 07122 Palma de Mallorca, Spain Received 28 September 2004; received in revised form 22 December 2004; accepted 28 January 2005
Abstract In this paper, the study of the distributivity equation involving uninorms in U given in (Fuzzy Sets and Systems, 128 (2002) 209–225) is revised. Two wrong propositions in the mentioned reference are corrected and their right versions are proved. © 2005 Elsevier B.V. All rights reserved. Keywords: Distributivity; Functional equation; Uninorm; t-norm; t-conorm; t-operator
The distributivity condition for two uninorms from U was studied in Section 6 of [1]. This condition can be written as U1 (x, U2 (y, z)) = U2 (U1 (x, y), U1 (x, z))
for all
x, y, z ∈ [0, 1],
(1)
where U1 and U2 are uninorms from U . In that paper the authors consider two cases, first when U1 lies in UMax and second when U1 lies in UMin . In both cases similar results are obtained. Unfortunately, the result stated in Proposition 6.2 for uninorms from UMax is wrong and consequently the corresponding proposition for uninorms from UMin , Proposition 6.6, also fails. In the same way, Figs. 9 and 11 in [1] representing these results need to be changed. 夡
DOI of original article: 10.1016/S0165-0114(01)00123-3.
∗ Corresponding author.
E-mail addresses: [email protected] (M. Mas), [email protected] (G. Mayor), [email protected] (J. Torrens). 0165-0114/$ - see front matter © 2005 Elsevier B.V. All rights reserved. doi:10.1016/j.fss.2005.01.016
298
M. Mas et al. / Fuzzy Sets and Systems 153 (2005) 297 – 299 1
1
S
Max
Max
Max
Min
Max
e
e T
0
Max
e
1
0
e
1
Fig. 1. Two distributive uninorms U1 (left) and U2 (right) from UMax with 0 = e2 = e1 = e.
1
1
S
Min
Max
Min
Min
e
e T
0
Min
Min
e
1
0
e
1
Fig. 2. Two distributive uninorms U1 (left) and U2 (right) from UMin with 1 = e2 = e1 = e.
In Propositions 6.2 and 6.6 we present the corrected versions. Differences between old and new propositions are in the set of solutions. Uninorms in Proposition 6.2 (and Proposition 6.6) with 0 < e2 < e1 (and e1 < e2 < 1), where e1 and e2 are the neutral elements of U1 and U2 , respectively, are not actually solutions of Eq. (1). Note that a necessary and sufficient condition is claimed in Proposition 6.2 in [1]. In fact, this condition is necessary but not sufficient since many uninorms stated in this proposition are not distributive in general as it has been mentioned above. Thus, the corrected version is as follows. Proposition 6.2. Let U1 = (e1 , T , S) and U2 = (e2 , T , S ) be uninorms from UMax and e2 e1 . Then U1 and U2 satisfy Eq. (1) if, and only if, one of the following cases holds:
M. Mas et al. / Fuzzy Sets and Systems 153 (2005) 297 – 299
299
• e2 = 0 and U2 is the t-conorm maximum. • 0 = e2 = e1 and U2 is idempotent. Proof. The first two steps in the proof given in [1] are correct. Thus, U2 must be idempotent when U1 and U2 satisfy (1). Now, we can distinguish two cases: • e2 = 0. In this case, if U1 and U2 satisfy (1), U2 is an idempotent t-conorm and so U2 = max. Conversely, it is clear that any uninorm U1 is distributive with the maximum. • 0 < e2 . Let U1 and U2 be distributive and suppose that e2 < e1 . In this case, we have U2 (e1 , 0) = e1 . For 0 < x < e2 < e1 it follows from (1) that x = U1 (x, e1 ) = U1 (x, U2 (e1 , 0)) = U2 (U1 (x, e1 ), U1 (x, 0)) = U2 (x, 0) = 0 which is a contradiction. Thus e1 = e2 and we already know that U2 must be idempotent. Conversely, it is a straightforward computation to prove that U1 and U2 are distributive whenever e1 = e2 and U2 is idempotent. Similarly, Proposition 6.6 in [1] must be corrected as follows: Proposition 6.6. Let U1 = (e1 , T , S) and U2 = (e2 , T , S ) be uninorms from UMin and e1 e2 . Then U1 and U2 satisfy Eq. (1) if, and only if, one of the following cases holds: • e2 = 1 and U2 is the t-norm minimum. • 1 = e2 = e1 and U2 is idempotent. Finally, Figs. 9 and 11 in [1] must be replaced by Figs. 1 and 2 above, where the non-trivial solutions of Eq. (1) given in Propositions 6.2 and 6.6, respectively, can be viewed. Acknowledgements The authors are grateful to Fuzzy Sets and Systems for publishing this corrigendum and they actually apologize for not having found this error before the publication of the article. References [1] M. Mas, G. Mayor, J. Torrens, The distributivity condition for uninorms and t-operators, Fuzzy Sets and Systems 128 (2002) 209–225.