Turkish Journal of Computer and Mathematics Education Vol.12 No.12 (2021), 3708-3712 Research Article Cartesian Product Of Interval Neutrosophic Automata V. Karthikeyan 𝟏 R. Karuppaiya 𝟐 1Department of Mathematics, Government College of Engineering,Dharmapuri, Tamil Nadu, India. 2Department of Mathematics, Annamalai University,Chidambaram, Tamil Nadu, India. Article History: Received: 11 January 2021; Revised: 12 February 2021; Accepted: 27 March 2021; Published online: 23 May 2021 Abstract We introduce Cartesian product of interval neutrosophic automata and prove that Cartesian product of cyclic interval neutrosophic automata is cyclic Key words: Cyclic, Cartesian product. AMS Mathematics subject classification: 03D05,20M35,18 B20,68Q45, 68Q70,94 A45 1 Introduction The neutrosophic set was introduced by Florentin Smarandache in 1999 [6]. The neutrosophic set is the generalization of classical sets, fuzzy set [11] and so on. The fuzzy set was introduced by Zadeh in 1965[11]. Bipolar fuzzy set, YinYang bipolar fuzzy set, NPN fuzzy set were introduced by W. R. Zhang in [8, 9, 10]. A neutrosophic set N is classified by a Truth membership T , Indeterminacy membership I , and Falsity N N membership F , where T , I , and F are real standard and non-standard subsets of] 0−, 1+[. Interval-valued N N N N neutrosophic sets was introduced by Wang etal.,[7]. The concept of interval neutrosophic finite state machine was introduced by Tahir Mahmood [5]. Generalized products of directable fuzzy automata are discussed in [1]. Retrievability, subsystems, and strong subsystems of INA were introduced in the papers [2, 3, 4]. In this paper, we introduce Cartesian product of interval neutrosophic automata and prove that Cartesian product of cyclic interval neutrosophic automata is cyclic. 2 Preliminaries 2.1 Neutrosophic Set [6] Let 𝑈 be the universal set.. A neutrosophic set (NS) 𝑁 in 𝑈 is classified by a truth membership T , an N indeterminacy membership I and a falsity membership F , where T , I , and F are real standard or non- N N N N N standard subsets of] 0−,1+[. That is 𝑁 ={〈 𝑥,𝑇 (𝑥),𝐼 (𝑥),𝐹 (𝑥) 〉,𝑥 ∈𝑈, 𝑇 , 𝐼 , 𝐹 ∈ ] 0−,1+[ } and 0− ≤𝑠𝑢𝑝 𝑇 (𝑥)+ 𝑁 𝑁 𝑁 𝑁 𝑁 𝑁 𝑁 𝑠𝑢𝑝𝐼 (𝑥)+𝑠𝑢𝑝𝐹 (𝑥) ≤ 3+. We need to take the interval [0,1] for instead of ] 0−,1+[. 𝑁 𝑁 .𝟐.𝟐 Definition [𝟕] An interval neutrosophic set (𝐼𝑁𝑆 for short) is 𝑁 = {〈𝛼 (𝑥),𝛽 (𝑥),𝛾 (𝑥)〉 |𝑥 ∈ 𝑈} 𝑁 𝑁 𝑁 = {〈𝑥,[𝑖𝑛𝑓 𝛼 (𝑥),𝑠𝑢𝑝 𝛼 (𝑥)],[𝑖𝑛𝑓 𝛽 (𝑥),𝑠𝑢𝑝 𝛽 (𝑥)],[𝑖𝑛𝑓 𝛾 (𝑥),𝑠𝑢𝑝 𝛾 (𝑥)]〉},𝑥 ∈ 𝑈, where 𝛼 (𝑥), 𝑁 𝑁 𝑁 𝑁 𝑁 𝑁 𝑁 𝛽 (𝑥), and 𝛾 (𝑥) representing the truth-membership, indeterminacy-membership and falsity membership for 𝑁 𝑁 each 𝑥 ∈ 𝑈. 𝛼 (𝑥), 𝛽 (𝑥), 𝛾 (𝑥) ⊆ [0,1] and the condition that 0 ≤ 𝑠𝑢𝑝 𝛼 (𝑥) + 𝑠𝑢𝑝 𝛽 (𝑥) + 𝑁 𝑁 𝑁 𝑁 𝑁 𝑠𝑢𝑝 𝛾 (𝑥) ≤ 3. 𝑁 𝟐.𝟑 Definition [𝟕] An 𝐼𝑁𝑆 𝑁 is empty if 𝑖𝑛𝑓 𝛼 (𝑥) = 𝑠𝑢𝑝 𝛼 (𝑥) = 0,𝑖𝑛𝑓 𝛽 (𝑥) = 𝑠𝑢𝑝 𝛽 (𝑥) 1,𝑖𝑛𝑓 𝛾 (𝑥) = 𝑠𝑢𝑝 𝛾 (𝑥) = 𝑁 𝑁 𝑁 𝑁 𝑁 𝑁 1 for all 𝑥 ∈ 𝑈. 3 Interval Neutrosophic Automata 3.1 Definition [5] 𝑀 = (𝑄,𝛴,𝑁) is called interval neutrosophic automaton (𝐼𝑁𝐴 for short), where 𝑄 and 𝛴 are non-empty finite sets called the set of states and input symbols respectively, and 𝑁 = {〈𝛼 (𝑥),𝛽 (𝑥),𝛾 (𝑥)〉} is an 𝐼𝑁𝑆 in 𝑄× 𝑁 𝑁 𝑁 3708 Turkish Journal of Computer and Mathematics Education Vol.12 No.12 (2021), 3708-3712 Research Article 𝛴×𝑄. The set of all words of finite length of 𝛴 is denoted by Σ∗. The empty word is denoted by 𝜖 and the length of each 𝑥 ∈ Σ∗ is denoted by |𝑥|. 3.2 Definition [5] 𝑀 = (𝑄,𝛴,𝑁) be an 𝐼𝑁𝐴. Define an 𝐼𝑁𝑆 𝑁∗ = {〈𝛼𝑁∗(𝑥), 𝛽𝑁∗(𝑥),𝛾𝑁∗(𝑥)〉} in 𝑄×𝛴∗×𝑄 by [1,1] 𝑖𝑓 𝑞 = 𝑞 𝑖 𝑗 𝛼𝑁∗(𝑞𝑖,𝜖,𝑞𝑗) = { [0,0] 𝑖𝑓 𝑞 ≠ 𝑞 𝑖 𝑗 [0,0] 𝑖𝑓 𝑞 = 𝑞 𝑖 𝑗 𝛽𝑁∗(𝑞𝑖,𝜖,𝑞𝑗) = { [1,1] 𝑖𝑓 𝑞 ≠ 𝑞 𝑖 𝑗 [0,0] 𝑖𝑓 𝑞 = 𝑞 𝑖 𝑗 𝛾𝑁∗(𝑞𝑖,𝜖,𝑞𝑗) = { [1,1] 𝑖𝑓 𝑞 ≠ 𝑞 𝑖 𝑗 𝛼𝑁∗(𝑞𝑖,𝑤,𝑞𝑗) = 𝛼𝑁∗(𝑞𝑖, 𝑥𝑦, 𝑞𝑗) = ∨𝑞𝑟 ∈𝑄[ 𝛼𝑁∗(𝑞𝑖, 𝑥, 𝑞𝑟)∧ 𝛼𝑁∗(𝑞𝑟, 𝑦, 𝑞𝑗)] 𝛽𝑁∗(𝑞𝑖,𝑤,𝑞𝑗) = 𝛽𝑁∗(𝑞𝑖, 𝑥𝑦, 𝑞𝑗) = ∧𝑞𝑟 ∈𝑄[ 𝛽𝑁∗(𝑞𝑖, 𝑥, 𝑞𝑟)∨ 𝛽𝑁∗(𝑞𝑟, 𝑦, 𝑞𝑗)] 𝛾𝑁∗(𝑞𝑖,𝑤,𝑞𝑗) = 𝛾𝑁∗(𝑞𝑖, 𝑥𝑦, 𝑞𝑗) = ∧𝑞𝑟 ∈𝑄[ 𝛾𝑁∗(𝑞𝑖, 𝑥, 𝑞𝑟)∨ 𝛾𝑁∗(𝑞𝑟, 𝑦, 𝑞𝑗)] ∀ 𝑞𝑖, 𝑞𝑗 ∈𝑄,𝑤 =𝑥𝑦 ,𝑥 ∈ Σ∗ and 𝑦 ∈Σ. 4 Cartesian Composition of Interval Neutrosophic Automata 4.1 Definition Let 𝑀 =(𝑄, Σ ,𝑁),𝑖 =1,2 be interval neutrosophic automata and let Σ ∩ Σ =∅. Let 𝑀 × 𝑀 = 𝑖 𝑖 𝑖 𝑖 1 2 1 2 (𝑄 × 𝑄 , Σ ∪ Σ , 𝑁 × 𝑁 ), where 1 2 1 2 1 2 𝛼 (𝑞,𝑎,𝑞 )>[0,0] 𝑖𝑓 𝑎 ∈ Σ 𝑎𝑛𝑑 𝑞 = 𝑞 1 𝑖 𝑘 1 𝑗 𝑙 (α1×α2)((𝑞𝑖,𝑞𝑗),𝑎,(𝑞𝑘,𝑞𝑙))= {𝛼2(𝑞𝑖,𝑎,𝑞𝑘)>[0,0] 𝑖𝑓 𝑎 ∈ Σ2 𝑎𝑛𝑑 𝑞𝑖 = 𝑞𝑘 0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 𝛽 (𝑞,𝑎,𝑞 )<[1,1] 𝑖𝑓 𝑎 ∈ Σ 𝑎𝑛𝑑 𝑞 = 𝑞 1 𝑖 𝑘 1 𝑗 𝑙 (𝛽1×𝛽2)((𝑞𝑖,𝑞𝑗),𝑎,(𝑞𝑘,𝑞𝑙))= {𝛽2(𝑞𝑖,𝑎,𝑞𝑘)<[1,1] 𝑖𝑓 𝑎 ∈ Σ2 𝑎𝑛𝑑 𝑞𝑖 = 𝑞𝑘 0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 𝛾 (𝑞 ,𝑎,𝑞 )<[1,1] 𝑖𝑓 𝑎 ∈ Σ 𝑎𝑛𝑑 𝑞 = 𝑞 1 𝑖 𝑘 1 𝑗 𝑙 ( 𝛾1× 𝛾2)((𝑞𝑖,𝑞𝑗),𝑎,(𝑞𝑘,𝑞𝑙))= { 𝛾2(𝑞𝑖,𝑎,𝑞𝑘)<[1,1] 𝑖𝑓 𝑎 ∈ Σ2 𝑎𝑛𝑑 𝑞𝑖 = 𝑞𝑘 0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 ∀(𝑞,𝑞 ), (𝑞 ,𝑞) ∈ 𝑄 ×𝑄 , 𝑎 ∈ Σ ∪ Σ . Then 𝑀 × 𝑀 is called the Cartesian product of interval 𝑖 𝑗 𝑘 𝑙 1 2 1 2 1 2 neutrosophic automata. 4.2 Definition Let 𝑀 = (𝑄,𝛴,𝑁) be an INA. 𝑀 is cyclic if ∃ 𝑞 ∈ Q such that Q = S(𝑞). 𝑖 𝑖 4.3 Definition [2] Let 𝑀 = (𝑄,𝛴,𝑁) be INA. M is connected if ∀ 𝑞 ,𝑞 and ∃ a ∈ Σ such that either α (𝑞,𝑎,𝑞 )>[0,0], 𝑗 𝑖 N 𝑖 𝑗 β (𝑞,𝑎,𝑞 )<[1,1], 𝛾 (𝑞,𝑎,𝑞 )<[1,1] or N 𝑖 𝑗 𝑁 𝑖 𝑗 α (𝑞 ,𝑎,𝑞)>[0,0], β (𝑞 ,𝑎,𝑞 )<[1,1], 𝛾 (𝑞 ,𝑎,𝑞)<[1,1]. N 𝑗 𝑖 N 𝑗 𝑖 𝑁 𝑗 𝑖 4.4 Definition [2] 3709 Turkish Journal of Computer and Mathematics Education Vol.12 No.12 (2021), 3708-3712 Research Article Let 𝑀 = (𝑄,𝛴,𝑁) be INA. M is strongly connected if for every 𝑞,𝑞 ∈ 𝑄, there exists u ∈ Σ∗ such that 𝑖 𝑗 α𝑁∗(𝑞𝑖,𝑢,𝑞𝑗)>[0,0],β𝑁∗(𝑞𝑖,𝑢,𝑞𝑗)<[1,1],γ𝑁∗(𝑞𝑖,𝑢,𝑞𝑗)<[1,1]. Theorem 4.1 Let 𝑀 =(𝑄,Σ ,𝑁),𝑖 =1,2 be interval neutrosophic automata and let Σ ∩ Σ = ∅. Let 𝑖 𝑖 𝑖 𝑖 1 2 𝑀 × 𝑀 = (𝑄 ×𝑄 ,Σ ∪ Σ ,𝑁 ×𝑁 ) be the Cartesian product of 𝑀 and 𝑀 . Then ∀ 𝑥 ∈Σ ∗ ∪ Σ ∗,𝑥 ≠ 1 2 1 2 1 2 1 2 1 2 1 2 𝜖 𝛼 (𝑞 ,𝑥,𝑞 )>[0,0] 𝑖𝑓 𝑥 ∈Σ ∗ 𝑎𝑛𝑑 𝑞 = 𝑞 1 𝑖 𝑘 1 𝑗 𝑙 (α1×α2)∗((𝑞𝑖,𝑞𝑗),𝑥,(𝑞𝑘,𝑞𝑙))= {𝛼2(𝑞𝑖,𝑥,𝑞𝑘)>[0,0] 𝑖𝑓 𝑥 ∈Σ2∗𝑎𝑛𝑑 𝑞𝑖 = 𝑞𝑘 0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 𝛽 (𝑞,𝑥,𝑞 )<[1,1] 𝑖𝑓 𝑥 ∈ Σ ∗ 𝑎𝑛𝑑 𝑞 = 𝑞 1 𝑖 𝑘 1 𝑗 𝑙 (𝛽1×𝛽2)∗((𝑞𝑖,𝑞𝑗),𝑥,(𝑞𝑘,𝑞𝑙))= {𝛽2(𝑞𝑖,𝑥,𝑞𝑘)<[1,1] 𝑖𝑓 𝑥 ∈ Σ2∗ 𝑎𝑛𝑑 𝑞𝑖 = 𝑞𝑘 0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 𝛾 (𝑞 ,𝑥,𝑞 )<[1,1] 𝑖𝑓 𝑥 ∈ Σ ∗𝑎𝑛𝑑 𝑞 = 𝑞 1 𝑖 𝑘 1 𝑗 𝑙 (𝛾1× 𝛾2)∗((𝑞𝑖,𝑞𝑗),𝑥,(𝑞𝑘,𝑞𝑙))= { 𝛾2(𝑞𝑖,𝑥,𝑞𝑘)<[1,1] 𝑖𝑓 𝑥 ∈ Σ1∗𝑎𝑛𝑑 𝑞𝑖 = 𝑞𝑘 0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 ∀(𝑞,𝑞 ), (𝑞 ,𝑞) ∈ 𝑄 ×𝑄 . 𝑖 𝑗 𝑘 𝑙 1 2 Proof . Let 𝑥 ∈Σ ∗ ∪ Σ ∗,𝑥 ≠ 𝜖 and let |𝑥|=𝑚. Let 𝑥 ∈Σ ∗. The result is trivial if 𝑚=1. Let the 1 2 1 result is true ∀ 𝑦 ∈ Σ ∗, |𝑦|=𝑚−1,𝑚 >1. Let 𝑥 =𝑎𝑦 where 𝑎 ∈ Σ ,𝑦 ∈ Σ ∗. Now, 1 1 1 (α ×α )∗((𝑞,𝑞 ),𝑥,(𝑞 ,𝑞))=(α ×α )∗((𝑞,𝑞 ),𝑎𝑦,(𝑞 ,𝑞)) 𝑁1 𝑁2 𝑖 𝑗 𝑘 𝑙 𝑁1 𝑁2 𝑖 𝑗 𝑘 𝑙 = ∨ {(α ×α )((𝑞,𝑞 ),𝑎,(𝑞 ,𝑞 ))∧(α ×α )∗((𝑞 ,𝑞 ),𝑦,(𝑞 ,𝑞))} (𝑞𝑟, 𝑞𝑠) ∈ 𝑄1× 𝑄2 𝑁1 𝑁2 𝑖 𝑗 𝑟 𝑠 1 2 𝑟 𝑠 𝑘 𝑙 = ∨ {α (𝑞, 𝑎, 𝑞 ) ∧ (α ×α )∗((𝑞 ,𝑞 ),𝑦,(𝑞 ,𝑞))} 𝑞𝑟 ∈ 𝑄1 𝑁1 𝑖 𝑟 𝑁1 𝑁2 𝑟 𝑠 𝑘 𝑙 ∨ {α (𝑞, 𝑎, 𝑞 ) ∧ α ∗(𝑞 ,𝑦,𝑞 )}>[0,0] 𝑖𝑓 𝑞 = 𝑞 = { 𝑞𝑟 ∈ 𝑄1 𝑁1 𝑖 𝑟 𝑁1 𝑟 𝑘 𝑗 𝑙 0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 α ∗(𝑞 ,𝑎𝑦,𝑞 )>[0,0] 𝑖𝑓 𝑞 = 𝑞 = { 𝑁1 𝑖 𝑘 𝑗 𝑙 0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 (β ×β )∗((𝑞,𝑞 ),𝑥,(𝑞 ,𝑞))=(β ×β )∗((𝑞,𝑞 ),𝑎𝑦,(𝑞 ,𝑞)) 𝑁1 𝑁2 𝑖 𝑗 𝑘 𝑙 𝑁1 𝑁2 𝑖 𝑗 𝑘 𝑙 = ∧ {(β ×β )((𝑞,𝑞 ),𝑎,(𝑞 ,𝑞 ))∨(β ×β )∗((𝑞 ,𝑞 ),𝑦,(𝑞 ,𝑞 ))} (𝑞𝑟, 𝑞𝑠) ∈ 𝑄1× 𝑄2 𝑁1 𝑁2 𝑖 𝑗 𝑟 𝑠 𝑁1 𝑁2 𝑟 𝑠 𝑘 𝑙 = ∧ {β (𝑞, 𝑎, 𝑞 ) ∨ (β ×β )∗((𝑞 ,𝑞 ),𝑦,(𝑞 ,𝑞))} 𝑞𝑟 ∈ 𝑄1 𝑁1 𝑖 𝑟 𝑁1 𝑁2 𝑟 𝑠 𝑘 𝑙 ∧ {β (𝑞 , 𝑎, 𝑞 ) ∨ β ∗(𝑞 ,𝑦,𝑞 )}<[1,1] 𝑖𝑓 𝑞 = 𝑞 = { 𝑞𝑟 ∈ 𝑄1 𝑁1 𝑖 𝑟 𝑁1 𝑟 𝑘 𝑗 𝑙 0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 β ∗(𝑞,𝑎𝑦,𝑞 )<[1,1] 𝑖𝑓 𝑞 = 𝑞 = { 𝑁1 𝑖 𝑘 𝑗 𝑙 0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 (γ ×γ )∗((𝑞,𝑞 ),𝑥,(𝑞 ,𝑞))=(γ ×γ )∗((𝑞 ,𝑞 ),𝑎𝑦,(𝑞 ,𝑞)) 𝑁1 𝑁2 𝑖 𝑗 𝑘 𝑙 𝑁1 𝑁2 𝑖 𝑗 𝑘 𝑙 = ∧ {(γ ×γ )((𝑞,𝑞 ),𝑎,(𝑞 ,𝑞 ))∨(γ ×γ )∗((𝑞 ,𝑞 ),𝑦,(𝑞 ,𝑞 ))} (𝑞𝑟, 𝑞𝑠) ∈ 𝑄1× 𝑄2 𝑁1 𝑁2 𝑖 𝑗 𝑟 𝑠 𝑁1 𝑁2 𝑟 𝑠 𝑘 𝑙 = ∧ {γ (𝑞, 𝑎, 𝑞 ) ∨ (γ ×γ )∗((𝑞 ,𝑞 ),𝑦,(𝑞 ,𝑞))} 𝑞𝑟 ∈ 𝑄1 𝑁1 𝑖 𝑟 𝑁1 𝑁2 𝑟 𝑠 𝑘 𝑙 ∧ {γ (𝑞, 𝑎, 𝑞 ) ∨ γ ∗(𝑞 ,𝑦,𝑞 )}<[1,1] 𝑖𝑓 𝑞 = 𝑞 = { 𝑞𝑟 ∈ 𝑄1 𝑁1 𝑖 𝑟 𝑁1 𝑟 𝑘 𝑗 𝑙 0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 γ ∗(𝑞,𝑎𝑦,𝑞 )<[1,1] 𝑖𝑓 𝑞 = 𝑞 = { 𝑁1 𝑖 𝑘 𝑗 𝑙 0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 The result is follows by induction. The Proof is similar if 𝑦 ∈ Σ ∗. 2 Theorem 4.2 Let 𝑀 =(𝑄,Σ ,𝑁),𝑖 =1,2 be INA and let Σ ∩ Σ = ∅. Then ∀ 𝑥 ∈ Σ ∗,𝑦 ∈ Σ ∗, 𝑖 𝑖 𝑖 𝑖 1 2 1 2 (α ×α )∗((𝑝,𝑝 ),𝑥𝑦,(𝑞 ,𝑞 ))= α ∗(𝑝,𝑥,𝑞) ∧ α ∗(𝑝 ,𝑦,𝑞 ) 𝑁1 𝑁2 𝑖 𝑗 𝑖 𝑗 𝑁1 𝑖 𝑖 𝑁2 𝑗 𝑗 = (α ×α )∗((𝑝,𝑝 ),𝑦𝑥,(𝑞,𝑞 )) 𝑁1 𝑁2 𝑖 𝑗 𝑖 𝑗 (β ×β )∗((𝑝,𝑝 ),𝑥𝑦,(𝑞,𝑞 ))= β ∗(𝑝,𝑥,𝑞) ∨ β ∗(𝑝 ,𝑦,𝑞 ) 𝑁1 𝑁2 𝑖 𝑗 𝑖 𝑗 𝑁1 𝑖 𝑖 𝑁2 𝑗 𝑗 = (β ×β )∗((𝑝,𝑝 ),𝑦𝑥,(𝑞,𝑞 )) 𝑁1 𝑁2 𝑖 𝑗 𝑖 𝑗 (γ ×𝛾 )∗((𝑝,𝑝 ),𝑥𝑦,(𝑞,𝑞 ))= γ ∗(𝑝,𝑥,𝑞) ∨ γ ∗(𝑝 ,𝑦,𝑞 ) 𝑁1 𝑁2 𝑖 𝑗 𝑖 𝑗 𝑁1 𝑖 𝑖 𝑁2 𝑗 𝑗 = (γ ×γ )∗((𝑝,𝑝 ),𝑦𝑥,(𝑞,𝑞 )), 𝑁1 𝑁2 𝑖 𝑗 𝑖 𝑗 (𝑝,𝑝 ),(𝑞,𝑞 )∈𝑄 ×𝑄 . 𝑖 𝑗 𝑖 𝑗 1 2 Proof . 3710 Turkish Journal of Computer and Mathematics Education Vol.12 No.12 (2021), 3708-3712 Research Article Let ∈ Σ ∗,𝑦 ∈ Σ ∗, (𝑝, 𝑝 ),(𝑞 , 𝑞 )∈𝑄 ×𝑄 . If 𝑥 = 𝜖 =𝑦, then 𝑥𝑦=𝜖. Suppose (𝑝, 𝑝 )= 1 2 𝑖 𝑗 𝑖 𝑗 1 2 𝑖 𝑗 (𝑞 , 𝑞 ). Then 𝑝 = 𝑞 and 𝑝 =𝑞 . Hence 𝑖 𝑗 𝑖 𝑖 𝑗 𝑗 (α ×α )∗((𝑝,𝑝 ),𝑥𝑦,(𝑞,𝑞 ))=[1,1]=[1,1] ∧[1,1]= α ∗(𝑝,𝑥,𝑞) ∧ α ∗(𝑝 ,𝑦,𝑞 ) 𝑁1 𝑁2 𝑖 𝑗 𝑖 𝑗 𝑁1 𝑖 𝑖 𝑁2 𝑗 𝑗 (β ×β )∗((𝑝,𝑝 ),𝑥𝑦,(𝑞,𝑞 ))=[0,0]=[0,0]∨[0,0]= β ∗(𝑝,𝑥,𝑞)∨β ∗(𝑝,𝑥,𝑞) 𝑁1 𝑁2 𝑖 𝑗 𝑖 𝑗 𝑁1 𝑖 𝑖 𝑁1 𝑖 𝑖 (γ ×𝛾 )∗((𝑝,𝑝 ),𝑥𝑦,(𝑞,𝑞 ))=[0,0]=[0,0]∨[0,0]= γ ∗(𝑝,𝑥,𝑞)∨ γ ∗(𝑝 ,𝑦,𝑞 ) 𝑁1 𝑁2 𝑖 𝑗 𝑖 𝑗 𝑁1 𝑖 𝑖 𝑁2 𝑗 𝑗 If (𝑝,𝑝 ) ≠ (𝑞,𝑞 ), then either 𝑝 ≠ 𝑞 or 𝑝 ≠𝑞 . 𝑖 𝑗 𝑖 𝑗 𝑖 𝑖 𝑗 𝑗 Thus, α ∗(𝑝,𝑥,𝑞) ∧ α ∗(𝑝 ,𝑦,𝑞 )=[0,0], β ∗(𝑝,𝑥,𝑞)∨β ∗(𝑝,𝑥,𝑞)=[1,1], 𝑁1 𝑖 𝑖 𝑁2 𝑗 𝑗 𝑁1 𝑖 𝑖 𝑁1 𝑖 𝑖 γ ∗(𝑝,𝑥,𝑞)∨ γ ∗(𝑝 ,𝑦,𝑞 )=[1,1]. 𝑁1 𝑖 𝑖 𝑁2 𝑗 𝑗 Hence (α ×α )∗((𝑝,𝑝 ),𝑥𝑦,(𝑞,𝑞 ))=[0,0] = α ∗(𝑝,𝑥,𝑞) ∧ α ∗(𝑝 ,𝑦,𝑞 ) 𝑁1 𝑁2 𝑖 𝑗 𝑖 𝑗 𝑁1 𝑖 𝑖 𝑁2 𝑗 𝑗 (β ×β )∗((𝑝,𝑝 ),𝑥𝑦,(𝑞,𝑞 ))=[1,1]= β ∗(𝑝,𝑥,𝑞)∨β ∗(𝑝,𝑥,𝑞) 𝑁1 𝑁2 𝑖 𝑗 𝑖 𝑗 𝑁1 𝑖 𝑖 𝑁1 𝑖 𝑖 (γ ×𝛾 )∗((𝑝,𝑝 ),𝑥𝑦,(𝑞,𝑞 ))=[1,1]= γ ∗(𝑝,𝑥,𝑞)∨ γ ∗(𝑝 ,𝑦,𝑞 ) 𝑁1 𝑁2 𝑖 𝑗 𝑖 𝑗 𝑁1 𝑖 𝑖 𝑁2 𝑗 𝑗 If x = 𝜖 and y ≠ 𝜖 or x ≠ 𝜖 and y = 𝜖, then the result follows by Theorem 4.1. Suppose x ≠ 𝜖 and y ≠ 𝜖. Now, (α ×α )∗((𝑝,𝑝 ),𝑥𝑦,(𝑞 ,𝑞 ))= ∨ {(α ×α )∗((𝑝,𝑝 ),𝑥,(𝑟,𝑟))∧ 𝑁1 𝑁2 𝑖 𝑗 𝑖 𝑗 (𝑟𝑖, 𝑟𝑗) ∈ 𝑄1× 𝑄2 𝑁1 𝑁2 𝑖 𝑗 𝑖 𝑗 (α ×α )∗((𝑟,𝑟),𝑦,(𝑞,𝑞 ))} 𝑁1 𝑁2 𝑖 𝑗 𝑖 𝑗 = ∨ {(α ×α )∗((𝑝,𝑝 ),𝑥,(𝑟,𝑝 ))∧ (α ×α )∗((𝑟,𝑝 ),𝑦,(𝑞 ,𝑞 ))} 𝑟𝑖 ∈ 𝑄1 𝑁1 𝑁2 𝑖 𝑗 𝑖 𝑗 𝑁1 𝑁2 𝑖 𝑗 𝑖 𝑗 = α ∗(𝑝,𝑥,𝑞 ) ∧ α ∗(𝑝 ,𝑦,𝑞 ) 𝑁1 𝑖 𝑖 𝑁2 𝑗 𝑗 (β ×β )∗((𝑝,𝑝 ),𝑥𝑦,(𝑞,𝑞 ))= ∧ {(β ×β )∗((𝑝,𝑝 ),𝑥,(𝑟,𝑟))∨ 𝑁1 𝑁2 𝑖 𝑗 𝑖 𝑗 (𝑟𝑖, 𝑟𝑗) ∈ 𝑄1× 𝑄2 𝑁1 𝑁2 𝑖 𝑗 𝑖 𝑗 (β ×β )∗((𝑟,𝑟),𝑦,(𝑞,𝑞 ))} 𝑁1 𝑁2 𝑖 𝑗 𝑖 𝑗 = ∧ {(β ×β )∗((𝑝,𝑝 ),𝑥,(𝑟,𝑝 ))∨ (β ×𝛽 )∗((𝑟,𝑝 ),𝑦,(𝑞,𝑞 ))} 𝑟𝑖 ∈ 𝑄1 𝑁1 𝑁2 𝑖 𝑗 𝑖 𝑗 𝑁1 𝑁2 𝑖 𝑗 𝑖 𝑗 = 𝛽 ∗(𝑝,𝑥,𝑞) ∨ 𝛽 ∗(𝑝 ,𝑦,𝑞 ) 𝑁1 𝑖 𝑖 𝑁2 𝑗 𝑗 (γ ×𝛾 )∗((𝑝,𝑝 ),𝑥𝑦,(𝑞,𝑞 ))= ∧ {(γ ×γ )∗((𝑝,𝑝 ),𝑥,(𝑟,𝑟))∨ 𝑁1 𝑁2 𝑖 𝑗 𝑖 𝑗 (𝑟𝑖, 𝑟𝑗) ∈ 𝑄1× 𝑄2 𝑁1 𝑁2 𝑖 𝑗 𝑖 𝑗 (γ ×γ )∗((𝑟,𝑟),𝑦,(𝑞 ,𝑞 ))} 𝑁1 𝑁2 𝑖 𝑗 𝑖 𝑗 = ∧ {(γ ×γ )∗((𝑝,𝑝 ),𝑥,(𝑟,𝑝 ))∨ (γ ×γ )∗((𝑟,𝑝 ),𝑦,(𝑞,𝑞 ))} 𝑟𝑖 ∈ 𝑄1 𝑁1 𝑁2 𝑖 𝑗 𝑖 𝑗 𝑁1 𝑁2 𝑖 𝑗 𝑖 𝑗 = γ ∗(𝑝,𝑥,𝑞) ∨ γ ∗(𝑝 ,𝑦,𝑞 ) 𝑁1 𝑖 𝑖 𝑁2 𝑗 𝑗 Similarly (α ×α )∗((𝑝,𝑝 ),𝑦𝑥,(𝑞 ,𝑞 ))= α ∗(𝑝 ,𝑦,𝑞 ) ∧ α ∗(𝑝,𝑥,𝑞 ) 𝑁1 𝑁2 𝑖 𝑗 𝑖 𝑗 𝑁2 𝑗 𝑗 𝑁1 𝑖 𝑖 (β ×β )∗((𝑝,𝑝 ),𝑦𝑥,(𝑞,𝑞 ))= 𝛽 ∗(𝑝 ,𝑦,𝑞 ) ∨ 𝛽 ∗(𝑝,𝑥,𝑞) 𝑁1 𝑁2 𝑖 𝑗 𝑖 𝑗 𝑁2 𝑗 𝑗 𝑁1 𝑖 𝑖 (γ ×𝛾 )∗((𝑝,𝑝 ),𝑦𝑥,(𝑞,𝑞 ))= γ ∗(𝑝 ,𝑦,𝑞 ) ∨ γ ∗(𝑝,𝑥,𝑞). 𝑁1 𝑁2 𝑖 𝑗 𝑖 𝑗 𝑁2 𝑗 𝑗 𝑁1 𝑖 𝑖 Theorem 4.3 Let 𝑀 =(𝑄,Σ ,𝑁),𝑖 =1,2 be INA and let Σ ∩ Σ = ∅. Cartesian product of 𝑀 ×𝑀 is 𝑖 𝑖 𝑖 𝑖 1 2 1 2 cyclic iff 𝑀 and 𝑀 are cyclic. 1 2 Proof. Let 𝑀 and 𝑀 are cyclic. Then 𝑄 =𝑆(𝑞) and 𝑄 =𝑆(𝑝 ) for some 𝑞 ∈𝑄 ,𝑝 ∈ 𝑄 . Let 1 2 1 𝑖 2 𝑗 𝑖 1 𝑗 2 (𝑞 ,𝑝) ∈ 𝑄 × 𝑄 . Then ∃ 𝑥 ∈ Σ ∗ and 𝑦 ∈ Σ ∗ such that 𝑘 𝑙 1 2 1 2 α ∗(𝑞,𝑥,𝑞 )>[0,0],β ∗(𝑞 ,𝑥,𝑞 )<[1,1],γ ∗(𝑞,𝑥,𝑞 ) < [1, 1] and 𝑁1 𝑖 𝑘 𝑁1 𝑖 𝑘 𝑁1 𝑖 𝑘 α ∗(𝑝 ,𝑦,𝑝)>[0,0],β ∗(𝑝 ,𝑦,𝑝)<[1,1],γ ∗(𝑝 ,𝑦,𝑝) < [1, 1]. Thus 𝑁2 𝑗 𝑙 𝑁2 𝑗 𝑙 𝑁2 𝑗 𝑙 (α ×α )∗((𝑞,𝑝 ),𝑥𝑦,(𝑞 ,𝑝))= α ∗(𝑞,𝑥,𝑞 ) ∧ α ∗(𝑝 ,𝑦,𝑝)>[0,0] 𝑁1 𝑁2 𝑖 𝑗 𝑘 𝑙 𝑁1 𝑖 𝑘 𝑁2 𝑗 𝑙 (β ×β )∗((𝑞,𝑝 ),𝑥𝑦,(𝑞 ,𝑝))= β ∗(𝑞,𝑥,𝑞 ) ∨ β ∗(𝑝 ,𝑦,𝑝)<[1,1] 𝑁1 𝑁2 𝑖 𝑗 𝑘 𝑙 𝑁1 𝑖 𝑘 𝑁2 𝑗 𝑙 (γ ×𝛾 )∗ ((𝑞,𝑝 ),𝑥𝑦,(𝑞 ,𝑝))= γ ∗(𝑞,𝑥,𝑞 ) ∨ γ ∗(𝑝 ,𝑦,𝑝)<[1,1]. 𝑁1 𝑁2 𝑖 𝑗 𝑘 𝑙 𝑁1 𝑖 𝑘 𝑁2 𝑗 𝑙 Hence (𝑞 ,𝑝) ∈𝑆((𝑞,𝑝 )). 𝑄 × 𝑄 = 𝑆((𝑞,𝑝 )). Hence 𝑀 ×𝑀 is cyclic. 𝑘 𝑙 𝑖 𝑗 1 2 𝑖 𝑗 1 2 Conversely, let 𝑀 ×𝑀 is cyclic. Then 𝑄 × 𝑄 = 𝑆((𝑞 ,𝑝 )) for some (𝑞,𝑝 )∈ 𝑄 × 𝑄 . 1 2 1 2 𝑖 𝑗 𝑖 𝑗 1 2 Let 𝑞 ∈ 𝑄 and 𝑝 ∈ 𝑄 . Then ∃ 𝑤 ∈ (Σ ∪ Σ )∗ such that 𝑘 1 𝑙 2 1 2 (α ×α )∗((𝑞 ,𝑝 ),𝑤,(𝑞 ,𝑝))>[0,0], (β ×β )∗((𝑞,𝑝 ),𝑤,(𝑞 ,𝑝))<[1,1] and 𝑁1 𝑁2 𝑖 𝑗 𝑘 𝑙 𝑁1 𝑁2 𝑖 𝑗 𝑘 𝑙 (γ ×𝛾 )∗ ((𝑞,𝑝 ),𝑤,(𝑞 ,𝑝)) < [1, 1]. Then by the theorem 4.2 ∃ 𝑢 ∈Σ ∗ and 𝑣 ∈ Σ ∗ such that 𝑁1 𝑁2 𝑖 𝑗 𝑘 𝑙 1 2 3711 Turkish Journal of Computer and Mathematics Education Vol.12 No.12 (2021), 3708-3712 Research Article α ∗(𝑞,𝑢,𝑞 ) ∧ α ∗(𝑝 ,𝑣,𝑝)= (α ×α )∗((𝑞,𝑝 ),𝑤,(𝑞 ,𝑝)) >[0,0] 𝑁1 𝑖 𝑘 𝑁2 𝑗 𝑙 𝑁1 𝑁2 𝑖 𝑗 𝑘 𝑙 β ∗(𝑞,𝑢,𝑞 ) ∨ β ∗(𝑝 ,𝑣,𝑝)= (β ×β )∗((𝑞,𝑝 ),𝑤,(𝑞 ,𝑝))<[1,1] 𝑁1 𝑖 𝑘 𝑁2 𝑗 𝑙 𝑁1 𝑁2 𝑖 𝑗 𝑘 𝑙 γ ∗(𝑞,𝑢,𝑞 ) ∨ γ ∗(𝑝 ,𝑣,𝑝)= (γ ×𝛾 )∗ ((𝑞 ,𝑝 ),𝑤,(𝑞 ,𝑝)) < [1, 1]. 𝑁1 𝑖 𝑘 𝑁2 𝑗 𝑙 𝑁1 𝑁2 𝑖 𝑗 𝑘 𝑙 Hence ∃ 𝑢 ∈Σ ∗ and 𝑣 ∈ Σ ∗ such that α ∗(𝑞,𝑢,𝑞 ) >[0,0], β ∗(𝑞,𝑢,𝑞 )<[1,1], γ ∗(𝑞,𝑢,𝑞 ) <[1,1] 1 2 𝑁1 𝑖 𝑘 𝑁1 𝑖 𝑘 𝑁1 𝑖 𝑘 and α ∗(𝑝 ,𝑣,𝑝)>[0,0], β ∗(𝑝 ,𝑣,𝑝)<[1,1],γ ∗(𝑝 ,𝑣,𝑝)<[1,1]. Thus 𝑞 ∈𝑆(𝑞 ) and 𝑝 ∈𝑆(𝑝 ). 𝑁2 𝑗 𝑙 𝑁2 𝑗 𝑙 𝑁2 𝑗 𝑙 𝑘 𝑖 𝑙 𝑗 Hence 𝑄 ∈𝑆(𝑞) and 𝑄 ∈𝑆(𝑝 ). 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