View of Cartesian Product Of Interval Neutrosophic Automata

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Research Article

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Cartesian Product Of Interval Neutrosophic Automata

V. Karthikeyan 𝟏 R. Karuppaiya 𝟐

1_{Department of Mathematics, Government College of Engineering,Dharmapuri, Tamil Nadu, India.} 2_{Department 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 TN, Indeterminacy membership IN, and Falsity

membership FN, where TN, IN, and FN are real standard and non-standard subsets of] 0−, 1+[. Interval-valued

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 TN, an

indeterminacy membership IN and a falsity membership FN, where TN, IN, and FN are real standard or

non-standard subsets of] 0−,1+_{[. That is}

𝑁 = {〈 𝑥, 𝑇𝑁(𝑥), 𝐼𝑁(𝑥), 𝐹𝑁(𝑥) 〉, 𝑥 ∈ 𝑈, 𝑇𝑁, 𝐼𝑁, 𝐹𝑁∈ ] 0−, 1+[ } and 0− ≤ 𝑠𝑢𝑝 𝑇𝑁(𝑥) + 𝑠𝑢𝑝 𝐼𝑁(𝑥) + 𝑠𝑢𝑝 𝐹𝑁(𝑥) ≤ 3+. We need to take the interval [0,1] for instead of ] 0−, 1+[. .𝟐. 𝟐 Definition [𝟕]

An interval neutrosophic set (𝐼𝑁𝑆 for short) is 𝑁 = {〈𝛼𝑁(𝑥), 𝛽𝑁(𝑥), 𝛾𝑁(𝑥)〉 |𝑥 ∈ 𝑈}

= {〈𝑥, [𝑖𝑛𝑓 𝛼𝑁(𝑥), 𝑠𝑢𝑝 𝛼𝑁(𝑥)], [𝑖𝑛𝑓 𝛽𝑁(𝑥), 𝑠𝑢𝑝 𝛽𝑁(𝑥)], [𝑖𝑛𝑓 𝛾𝑁(𝑥), 𝑠𝑢𝑝 𝛾𝑁(𝑥)]〉}, 𝑥 ∈ 𝑈, where 𝛼𝑁(𝑥), 𝛽𝑁(𝑥), and 𝛾𝑁(𝑥) representing the truth-membership, indeterminacy-membership and falsity membership for each 𝑥 ∈ 𝑈. 𝛼𝑁(𝑥), 𝛽𝑁(𝑥), 𝛾𝑁(𝑥) ⊆ [0,1] and the condition that 0 ≤ 𝑠𝑢𝑝 𝛼𝑁(𝑥) + 𝑠𝑢𝑝 𝛽𝑁(𝑥) + 𝑠𝑢𝑝 𝛾𝑁(𝑥) ≤ 3.

𝟐. 𝟑 Definition [𝟕]

An 𝐼𝑁𝑆 𝑁 is empty if 𝑖𝑛𝑓 𝛼𝑁(𝑥) = 𝑠𝑢𝑝 𝛼𝑁(𝑥) = 0, 𝑖𝑛𝑓 𝛽𝑁(𝑥) = 𝑠𝑢𝑝 𝛽𝑁(𝑥) 1, 𝑖𝑛𝑓 𝛾𝑁(𝑥) = 𝑠𝑢𝑝 𝛾𝑁(𝑥) = 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 = {〈𝛼𝑁(𝑥), 𝛽𝑁(𝑥), 𝛾𝑁(𝑥)〉} is an 𝐼𝑁𝑆 in 𝑄 ×

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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∗_{= {〈𝛼} 𝑁∗(𝑥), 𝛽𝑁∗(𝑥), 𝛾𝑁∗(𝑥)〉} in 𝑄 × 𝛴∗× 𝑄 by 𝛼𝑁∗(𝑞𝑖, 𝜖, 𝑞𝑗) = { [1, 1] 𝑖𝑓 𝑞𝑖= 𝑞𝑗 [0, 0] 𝑖𝑓 𝑞𝑖 ≠ 𝑞𝑗 𝛽𝑁∗(𝑞𝑖, 𝜖, 𝑞𝑗) = { [0, 0] 𝑖𝑓 𝑞𝑖= 𝑞𝑗 [1, 1] 𝑖𝑓 𝑞𝑖 ≠ 𝑞𝑗 𝛾𝑁∗(𝑞𝑖, 𝜖, 𝑞𝑗) = { [0, 0] 𝑖𝑓 𝑞𝑖= 𝑞𝑗 [1, 1] 𝑖𝑓 𝑞𝑖 ≠ 𝑞𝑗 𝛼𝑁∗(𝑞𝑖, 𝑤, 𝑞𝑗) = 𝛼𝑁∗(𝑞𝑖, 𝑥𝑦, 𝑞𝑗) = ∨𝑞_𝑟 ∈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 𝑀𝑖= (𝑄𝑖, Σ𝑖, 𝑁𝑖), 𝑖 = 1, 2 be interval neutrosophic automata and let Σ1∩ Σ2= ∅. Let 𝑀1 × 𝑀2= (𝑄1 × 𝑄2, Σ1 ∪ Σ2, 𝑁1 × 𝑁2), where (α1× α2) ((𝑞𝑖, 𝑞𝑗), 𝑎, (𝑞𝑘, 𝑞𝑙)) = { 𝛼1(𝑞𝑖, 𝑎, 𝑞𝑘) > [0,0] 𝑖𝑓 𝑎 ∈ Σ1 𝑎𝑛𝑑 𝑞𝑗= 𝑞𝑙 𝛼2(𝑞𝑖, 𝑎, 𝑞𝑘) > [0,0] 𝑖𝑓 𝑎 ∈ Σ2 𝑎𝑛𝑑 𝑞𝑖= 𝑞𝑘 0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 (𝛽1× 𝛽2) ((𝑞𝑖, 𝑞𝑗), 𝑎, (𝑞𝑘, 𝑞𝑙)) = { 𝛽1(𝑞𝑖, 𝑎, 𝑞𝑘) < [1,1] 𝑖𝑓 𝑎 ∈ Σ1 𝑎𝑛𝑑 𝑞𝑗= 𝑞𝑙 𝛽2(𝑞𝑖, 𝑎, 𝑞𝑘) < [1,1] 𝑖𝑓 𝑎 ∈ Σ2 𝑎𝑛𝑑 𝑞𝑖= 𝑞𝑘 0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 ( 𝛾1× 𝛾2) ((𝑞𝑖, 𝑞𝑗), 𝑎, (𝑞𝑘, 𝑞𝑙)) = { 𝛾1(𝑞𝑖, 𝑎, 𝑞𝑘) < [1,1] 𝑖𝑓 𝑎 ∈ Σ1 𝑎𝑛𝑑 𝑞𝑗= 𝑞𝑙 𝛾2(𝑞𝑖, 𝑎, 𝑞𝑘) < [1,1] 𝑖𝑓 𝑎 ∈ Σ2 𝑎𝑛𝑑 𝑞𝑖= 𝑞𝑘 0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

∀(𝑞𝑖, 𝑞𝑗), (𝑞𝑘, 𝑞𝑙) ∈ 𝑄1× 𝑄2, 𝑎 ∈ Σ1 ∪ Σ2. Then 𝑀1 × 𝑀2 is called the Cartesian product of interval neutrosophic automata.

4.2 Deﬁnition

Let 𝑀 = (𝑄, 𝛴, 𝑁) be an INA. 𝑀 is cyclic if ∃ 𝑞𝑖 ∈ Q such that Q = S(𝑞𝑖).

4.3 Deﬁnition [2]

Let 𝑀 = (𝑄, 𝛴, 𝑁) be INA. M is connected if ∀ 𝑞𝑗, 𝑞𝑖 and ∃ a ∈ Σ such that either αN (𝑞𝑖, 𝑎, 𝑞𝑗) > [0,0], βN (𝑞𝑖, 𝑎, 𝑞𝑗) < [1, 1], 𝛾𝑁(𝑞𝑖, 𝑎, 𝑞𝑗) < [1,1] or

αN (𝑞𝑗, 𝑎, 𝑞𝑖) > [0,0], βN (𝑞𝑗, 𝑎, 𝑞𝑖) < [1, 1], 𝛾𝑁(𝑞𝑗, 𝑎, 𝑞𝑖) < [1,1].

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Let 𝑀 = (𝑄, 𝛴, 𝑁) be INA. M is strongly connected if for every 𝑞𝑖, 𝑞𝑗 ∈ 𝑄, there exists u ∈ Σ∗ such that α𝑁∗(𝑞𝑖, 𝑢, 𝑞𝑗) > [0,0], β𝑁∗(𝑞𝑖, 𝑢, 𝑞𝑗) < [1, 1], γ𝑁∗(𝑞𝑖, 𝑢, 𝑞𝑗) < [1, 1].

Theorem 4.1 Let 𝑀𝑖= (𝑄𝑖, Σ𝑖, 𝑁𝑖), 𝑖 = 1, 2 be interval neutrosophic automata and let Σ1∩ Σ2= ∅. Let 𝑀1× 𝑀2= (𝑄1× 𝑄2 , Σ1∪ Σ2, 𝑁1× 𝑁2) be the Cartesian product of 𝑀1 and 𝑀2. Then ∀ 𝑥 ∈ Σ1∗ ∪ Σ2∗, 𝑥 ≠ 𝜖 (α1× α2)∗((𝑞𝑖, 𝑞𝑗), 𝑥, (𝑞𝑘, 𝑞𝑙)) = { 𝛼1(𝑞𝑖, 𝑥, 𝑞𝑘) > [0,0] 𝑖𝑓 𝑥 ∈ Σ1∗ 𝑎𝑛𝑑 𝑞𝑗 = 𝑞𝑙 𝛼2(𝑞𝑖, 𝑥, 𝑞𝑘) > [0,0] 𝑖𝑓 𝑥 ∈ Σ2∗𝑎𝑛𝑑 𝑞𝑖= 𝑞𝑘 0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 (𝛽1× 𝛽2)∗((𝑞𝑖, 𝑞𝑗), 𝑥, (𝑞𝑘, 𝑞𝑙)) = { 𝛽1(𝑞𝑖, 𝑥, 𝑞𝑘) < [1,1] 𝑖𝑓 𝑥 ∈ Σ1∗ 𝑎𝑛𝑑 𝑞𝑗= 𝑞𝑙 𝛽2(𝑞𝑖, 𝑥, 𝑞𝑘) < [1,1] 𝑖𝑓 𝑥 ∈ Σ2∗ 𝑎𝑛𝑑 𝑞𝑖= 𝑞𝑘 0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 (𝛾1× 𝛾2)∗((𝑞𝑖, 𝑞𝑗), 𝑥, (𝑞𝑘, 𝑞𝑙)) = { 𝛾1(𝑞𝑖, 𝑥, 𝑞𝑘) < [1,1] 𝑖𝑓 𝑥 ∈ Σ1∗𝑎𝑛𝑑 𝑞𝑗= 𝑞𝑙 𝛾2(𝑞𝑖, 𝑥, 𝑞𝑘) < [1,1] 𝑖𝑓 𝑥 ∈ Σ1∗𝑎𝑛𝑑 𝑞𝑖= 𝑞𝑘 0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 ∀(𝑞𝑖, 𝑞𝑗), (𝑞𝑘, 𝑞𝑙) ∈ 𝑄1× 𝑄2.

Proof . Let 𝑥 ∈ Σ1∗ ∪ Σ2∗, 𝑥 ≠ 𝜖 and let |𝑥| = 𝑚. Let 𝑥 ∈ Σ1∗. The result is trivial if 𝑚 = 1. Let the result is true ∀ 𝑦 ∈ Σ1∗, |𝑦| = 𝑚 − 1, 𝑚 > 1. Let 𝑥 = 𝑎𝑦 where 𝑎 ∈ Σ1, 𝑦 ∈ Σ1∗. Now,

(α𝑁₁× α𝑁₂)∗((𝑞𝑖, 𝑞𝑗), 𝑥, (𝑞𝑘, 𝑞𝑙)) = (α𝑁₁× α𝑁₂)∗((𝑞𝑖, 𝑞𝑗), 𝑎𝑦, (𝑞𝑘, 𝑞𝑙)) = ∨(𝑞𝑟, 𝑞𝑠) ∈ 𝑄1× 𝑄2{(α𝑁1× α𝑁2) ((𝑞𝑖, 𝑞𝑗), 𝑎, (𝑞𝑟, 𝑞𝑠)) ∧ (α1× α2) ∗_((𝑞 𝑟, 𝑞𝑠), 𝑦, (𝑞𝑘, 𝑞𝑙))} = ∨𝑞𝑟 ∈ 𝑄1{α𝑁1(𝑞𝑖, 𝑎, 𝑞𝑟) ∧ (α𝑁1× α𝑁2) ∗_((𝑞 𝑟, 𝑞𝑠), 𝑦, (𝑞𝑘, 𝑞𝑙))} = {∨𝑞_𝑟 ∈ 𝑄₁{α𝑁₁(𝑞𝑖, 𝑎, 𝑞𝑟) ∧ α𝑁₁∗(𝑞𝑟, 𝑦, 𝑞𝑘)} > [0,0] 𝑖𝑓 𝑞𝑗= 𝑞𝑙 0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 = {α𝑁1 ∗_(𝑞 𝑖, 𝑎𝑦, 𝑞𝑘) > [0,0] 𝑖𝑓 𝑞𝑗= 𝑞𝑙 0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 (β𝑁1× β𝑁2) ∗_((𝑞 𝑖, 𝑞𝑗), 𝑥, (𝑞𝑘, 𝑞𝑙)) = (β𝑁1× β𝑁2) ∗_((𝑞 𝑖, 𝑞𝑗), 𝑎𝑦, (𝑞𝑘, 𝑞𝑙)) = ∧(𝑞𝑟, 𝑞𝑠) ∈ 𝑄1× 𝑄2{(β𝑁1× β𝑁2) ((𝑞𝑖, 𝑞𝑗), 𝑎, (𝑞𝑟, 𝑞𝑠)) ∨ (β𝑁1× β𝑁2) ∗_((𝑞 𝑟, 𝑞𝑠), 𝑦, (𝑞𝑘, 𝑞𝑙))} = ∧𝑞_𝑟 ∈ 𝑄₁{β𝑁₁(𝑞𝑖, 𝑎, 𝑞𝑟) ∨ (β𝑁₁× β𝑁₂)∗((𝑞𝑟, 𝑞𝑠), 𝑦, (𝑞𝑘, 𝑞𝑙))} = {∧𝑞_𝑟 ∈ 𝑄₁{β𝑁₁(𝑞𝑖, 𝑎, 𝑞𝑟) ∨ β𝑁₁∗(𝑞𝑟, 𝑦, 𝑞𝑘)} < [1,1] 𝑖𝑓 𝑞𝑗= 𝑞𝑙 0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 = {β𝑁1 ∗_(𝑞 𝑖, 𝑎𝑦, 𝑞𝑘) < [1,1] 𝑖𝑓 𝑞𝑗= 𝑞𝑙 0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 (γ𝑁1× γ𝑁2) ∗_((𝑞 𝑖, 𝑞𝑗), 𝑥, (𝑞𝑘, 𝑞𝑙)) = (γ𝑁1× γ𝑁2) ∗_((𝑞 𝑖, 𝑞𝑗), 𝑎𝑦, (𝑞𝑘, 𝑞𝑙)) = ∧(𝑞_𝑟, 𝑞_𝑠) ∈ 𝑄₁× 𝑄₂{(γ𝑁₁× γ𝑁₂) ((𝑞𝑖, 𝑞𝑗), 𝑎, (𝑞𝑟, 𝑞𝑠)) ∨ (γ𝑁₁× γ𝑁₂)∗((𝑞𝑟, 𝑞𝑠), 𝑦, (𝑞𝑘, 𝑞𝑙))} = ∧𝑞𝑟 ∈ 𝑄1{γ𝑁1(𝑞𝑖, 𝑎, 𝑞𝑟) ∨ (γ𝑁1× γ𝑁2) ∗_((𝑞 𝑟, 𝑞𝑠), 𝑦, (𝑞𝑘, 𝑞𝑙))} = {∧𝑞𝑟 ∈ 𝑄1{γ𝑁1(𝑞𝑖, 𝑎, 𝑞𝑟) ∨ γ𝑁1 ∗_(𝑞 𝑟, 𝑦, 𝑞𝑘)} < [1,1] 𝑖𝑓 𝑞𝑗= 𝑞𝑙 0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 = {γ𝑁1 ∗_(𝑞 𝑖, 𝑎𝑦, 𝑞𝑘) < [1,1] 𝑖𝑓 𝑞𝑗= 𝑞𝑙 0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

The result is follows by induction. The Proof is similar if 𝑦 ∈ Σ2∗.

Theorem 4.2 Let 𝑀𝑖= (𝑄𝑖, Σ𝑖, 𝑁𝑖), 𝑖 = 1, 2 be INA and let Σ1∩ Σ2= ∅. Then ∀ 𝑥 ∈ Σ1∗, 𝑦 ∈ Σ2∗, (α𝑁1× α𝑁2) ∗_((𝑝 𝑖, 𝑝𝑗), 𝑥𝑦, (𝑞𝑖, 𝑞𝑗)) = α𝑁1 ∗_(𝑝 𝑖, 𝑥, 𝑞𝑖) ∧ α𝑁2 ∗_(𝑝 𝑗, 𝑦, 𝑞𝑗) = (α𝑁1× α𝑁2) ∗_((𝑝 𝑖, 𝑝𝑗), 𝑦𝑥, (𝑞𝑖, 𝑞𝑗)) (β𝑁1× β𝑁2) ∗_((𝑝 𝑖, 𝑝𝑗), 𝑥𝑦, (𝑞𝑖, 𝑞𝑗)) = β𝑁1 ∗_(𝑝 𝑖, 𝑥, 𝑞𝑖) ∨ β𝑁2 ∗_(𝑝 𝑗, 𝑦, 𝑞𝑗) = (β𝑁1× β𝑁2) ∗_((𝑝 𝑖, 𝑝𝑗), 𝑦𝑥, (𝑞𝑖, 𝑞𝑗)) (γ𝑁1× 𝛾𝑁2) ∗_((𝑝 𝑖, 𝑝𝑗), 𝑥𝑦, (𝑞𝑖, 𝑞𝑗)) = γ𝑁1 ∗_(𝑝 𝑖, 𝑥, 𝑞𝑖) ∨ γ𝑁2 ∗_(𝑝 𝑗, 𝑦, 𝑞𝑗) = (γ𝑁₁× γ𝑁₂)∗((𝑝𝑖, 𝑝𝑗), 𝑦𝑥, (𝑞𝑖, 𝑞𝑗)), (𝑝𝑖, 𝑝𝑗), (𝑞𝑖, 𝑞𝑗) ∈ 𝑄1× 𝑄2. Proof .

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Research Article

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Let ∈ Σ1∗, 𝑦 ∈ Σ2∗, (𝑝𝑖, 𝑝𝑗), (𝑞𝑖, 𝑞𝑗) ∈ 𝑄1× 𝑄2. If 𝑥 = 𝜖 = 𝑦, then 𝑥𝑦 = 𝜖. Suppose (𝑝𝑖, 𝑝𝑗) = (𝑞𝑖, 𝑞𝑗). Then 𝑝𝑖= 𝑞𝑖 and 𝑝𝑗= 𝑞𝑗. Hence

(α𝑁₁× α𝑁₂)∗((𝑝𝑖, 𝑝𝑗), 𝑥𝑦, (𝑞𝑖, 𝑞𝑗)) = [1,1] = [1,1] ∧ [1,1] = α𝑁₁∗(𝑝𝑖, 𝑥, 𝑞𝑖) ∧ α𝑁₂∗(𝑝𝑗, 𝑦, 𝑞𝑗) (β𝑁₁× β𝑁₂)∗((𝑝𝑖, 𝑝𝑗), 𝑥𝑦, (𝑞𝑖, 𝑞𝑗)) = [0, 0] = [0, 0] ∨ [0, 0] = β𝑁₁∗(𝑝𝑖, 𝑥, 𝑞𝑖) ∨ β𝑁₁∗(𝑝𝑖, 𝑥, 𝑞𝑖) (γ𝑁₁× 𝛾𝑁₂)∗((𝑝𝑖, 𝑝𝑗), 𝑥𝑦, (𝑞𝑖, 𝑞𝑗)) = [0,0] = [0,0] ∨ [0,0] = γ𝑁₁∗(𝑝𝑖, 𝑥, 𝑞𝑖) ∨ γ𝑁₂∗(𝑝𝑗, 𝑦, 𝑞𝑗) If (𝑝𝑖, 𝑝𝑗) ≠ (𝑞𝑖, 𝑞𝑗), then either 𝑝𝑖 ≠ 𝑞𝑖 or 𝑝𝑗≠ 𝑞𝑗. Thus, α𝑁1 ∗_(𝑝 𝑖, 𝑥, 𝑞𝑖) ∧ α𝑁2 ∗_(𝑝 𝑗, 𝑦, 𝑞𝑗) = [0,0], β𝑁1 ∗_(𝑝 𝑖, 𝑥, 𝑞𝑖) ∨ β𝑁1 ∗_(𝑝 𝑖, 𝑥, 𝑞𝑖) = [1,1], γ𝑁₁∗(𝑝𝑖, 𝑥, 𝑞𝑖) ∨ γ𝑁₂∗(𝑝𝑗, 𝑦, 𝑞𝑗) = [1,1]. Hence (α𝑁₁× α𝑁₂)∗((𝑝𝑖, 𝑝𝑗), 𝑥𝑦, (𝑞𝑖, 𝑞𝑗)) = [0,0] = α𝑁₁∗(𝑝𝑖, 𝑥, 𝑞𝑖) ∧ α𝑁₂∗(𝑝𝑗, 𝑦, 𝑞𝑗) (β𝑁₁× β𝑁₂)∗((𝑝𝑖, 𝑝𝑗), 𝑥𝑦, (𝑞𝑖, 𝑞𝑗)) = [1,1] = β𝑁₁∗(𝑝𝑖, 𝑥, 𝑞𝑖) ∨ β𝑁₁∗(𝑝𝑖, 𝑥, 𝑞𝑖) (γ𝑁₁× 𝛾𝑁₂)∗((𝑝𝑖, 𝑝𝑗), 𝑥𝑦, (𝑞𝑖, 𝑞𝑗)) = [1,1] = γ𝑁₁∗(𝑝𝑖, 𝑥, 𝑞𝑖) ∨ γ𝑁₂∗(𝑝𝑗, 𝑦, 𝑞𝑗)

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, (α𝑁1× α𝑁2) ∗_((𝑝 𝑖, 𝑝𝑗), 𝑥𝑦, (𝑞𝑖, 𝑞𝑗)) = ∨(𝑟_𝑖, 𝑟_𝑗) ∈ 𝑄1× 𝑄2{(α𝑁1× α𝑁2) ∗_((𝑝 𝑖, 𝑝𝑗), 𝑥, (𝑟𝑖, 𝑟𝑗)) ∧ (α𝑁1× α𝑁2) ∗_((𝑟 𝑖, 𝑟𝑗), 𝑦, (𝑞𝑖, 𝑞𝑗))} = ∨𝑟_𝑖 ∈ 𝑄1{(α𝑁1× α𝑁2) ∗_((𝑝 𝑖, 𝑝𝑗), 𝑥, (𝑟𝑖, 𝑝𝑗)) ∧ (α𝑁1× α𝑁2) ∗_((𝑟 𝑖, 𝑝𝑗), 𝑦, (𝑞𝑖, 𝑞𝑗))} = α𝑁1 ∗_(𝑝 𝑖, 𝑥, 𝑞𝑖) ∧ α𝑁2 ∗_(𝑝 𝑗, 𝑦, 𝑞𝑗) (β𝑁1× β𝑁2) ∗_((𝑝 𝑖, 𝑝𝑗), 𝑥𝑦, (𝑞𝑖, 𝑞𝑗)) = ∧(𝑟𝑖, 𝑟𝑗) ∈ 𝑄1× 𝑄2{(β𝑁1× β𝑁2) ∗_((𝑝 𝑖, 𝑝𝑗), 𝑥, (𝑟𝑖, 𝑟𝑗)) ∨ (β𝑁1× β𝑁2) ∗_((𝑟 𝑖, 𝑟𝑗), 𝑦, (𝑞𝑖, 𝑞𝑗))} = ∧𝑟𝑖 ∈ 𝑄1{(β𝑁1× β𝑁2) ∗_((𝑝 𝑖, 𝑝𝑗), 𝑥, (𝑟𝑖, 𝑝𝑗)) ∨ (β𝑁1× 𝛽𝑁2) ∗_((𝑟 𝑖, 𝑝𝑗), 𝑦, (𝑞𝑖, 𝑞𝑗))} = 𝛽𝑁1 ∗_(𝑝 𝑖, 𝑥, 𝑞𝑖) ∨ 𝛽𝑁2 ∗ (𝑝𝑗, 𝑦, 𝑞𝑗) (γ𝑁₁× 𝛾𝑁₂)∗((𝑝𝑖, 𝑝𝑗), 𝑥𝑦, (𝑞𝑖, 𝑞𝑗)) = ∧(𝑟_𝑖, 𝑟_𝑗) ∈ 𝑄₁× 𝑄₂{(γ𝑁₁× γ𝑁₂)∗((𝑝𝑖, 𝑝𝑗), 𝑥, (𝑟𝑖, 𝑟𝑗)) ∨ (γ𝑁₁× γ𝑁₂)∗((𝑟𝑖, 𝑟𝑗), 𝑦, (𝑞𝑖, 𝑞𝑗))} = ∧𝑟_𝑖 ∈ 𝑄₁{(γ𝑁₁× γ𝑁₂)∗((𝑝𝑖, 𝑝𝑗), 𝑥, (𝑟𝑖, 𝑝𝑗)) ∨ (γ𝑁₁× γ𝑁₂)∗((𝑟𝑖, 𝑝𝑗), 𝑦, (𝑞𝑖, 𝑞𝑗))} = γ𝑁1 ∗_(𝑝 𝑖, 𝑥, 𝑞𝑖) ∨ γ𝑁2 ∗_(𝑝 𝑗, 𝑦, 𝑞𝑗) Similarly (α𝑁1× α𝑁2) ∗_((𝑝 𝑖, 𝑝𝑗), 𝑦𝑥, (𝑞𝑖, 𝑞𝑗)) = α𝑁2 ∗_(𝑝 𝑗, 𝑦, 𝑞𝑗) ∧ α𝑁1 ∗_(𝑝 𝑖, 𝑥, 𝑞𝑖) (β𝑁1× β𝑁2) ∗_((𝑝 𝑖, 𝑝𝑗), 𝑦𝑥, (𝑞𝑖, 𝑞𝑗)) = 𝛽𝑁2 ∗_(𝑝 𝑗, 𝑦, 𝑞𝑗) ∨ 𝛽𝑁1 ∗_(𝑝 𝑖, 𝑥, 𝑞𝑖) (γ𝑁1× 𝛾𝑁2) ∗_((𝑝 𝑖, 𝑝𝑗), 𝑦𝑥, (𝑞𝑖, 𝑞𝑗)) = γ𝑁2 ∗_(𝑝 𝑗, 𝑦, 𝑞𝑗) ∨ γ𝑁1 ∗_(𝑝 𝑖, 𝑥, 𝑞𝑖).

Theorem 4.3 Let 𝑀𝑖= (𝑄𝑖, Σ𝑖, 𝑁𝑖), 𝑖 = 1, 2 be INA and let Σ1∩ Σ2= ∅. Cartesian product of 𝑀1 × 𝑀2 is cyclic iff 𝑀1 and 𝑀2 are cyclic.

Proof. Let 𝑀1 and 𝑀2 are cyclic. Then 𝑄1= 𝑆(𝑞𝑖) and 𝑄2= 𝑆(𝑝𝑗) for some 𝑞𝑖 ∈ 𝑄1, 𝑝𝑗 ∈ 𝑄2. Let (𝑞𝑘, 𝑝𝑙) ∈ 𝑄1 × 𝑄2. Then ∃ 𝑥 ∈ Σ1∗ and 𝑦 ∈ Σ2∗ such that

α𝑁₁∗(𝑞𝑖, 𝑥, 𝑞𝑘) > [0,0], β𝑁₁∗(𝑞𝑖, 𝑥, 𝑞𝑘) < [1,1], γ𝑁₁∗(𝑞𝑖, 𝑥, 𝑞𝑘) < [1, 1] and α𝑁₂∗(𝑝𝑗, 𝑦, 𝑝𝑙) > [0,0], β𝑁₂∗(𝑝𝑗, 𝑦, 𝑝𝑙) < [1,1], γ𝑁₂∗(𝑝𝑗, 𝑦, 𝑝𝑙) < [1, 1]. Thus

(α𝑁₁× α𝑁₂)∗((𝑞𝑖, 𝑝𝑗), 𝑥𝑦, (𝑞𝑘, 𝑝𝑙)) = α𝑁₁∗(𝑞𝑖, 𝑥, 𝑞𝑘) ∧ α𝑁₂∗(𝑝𝑗, 𝑦, 𝑝𝑙) > [0,0] (β𝑁₁× β𝑁₂)∗((𝑞𝑖, 𝑝𝑗), 𝑥𝑦, (𝑞𝑘, 𝑝𝑙)) = β𝑁₁∗(𝑞𝑖, 𝑥, 𝑞𝑘) ∨ β𝑁₂∗(𝑝𝑗, 𝑦, 𝑝𝑙) < [1,1] (γ𝑁₁× 𝛾𝑁₂)∗ ((𝑞𝑖, 𝑝𝑗), 𝑥𝑦, (𝑞𝑘, 𝑝𝑙)) = γ𝑁₁∗(𝑞𝑖, 𝑥, 𝑞𝑘) ∨ γ𝑁₂∗(𝑝𝑗, 𝑦, 𝑝𝑙) < [1, 1]. Hence (𝑞𝑘, 𝑝𝑙) ∈ 𝑆 ((𝑞𝑖, 𝑝𝑗)). 𝑄1 × 𝑄2= 𝑆 ((𝑞𝑖, 𝑝𝑗)). Hence 𝑀1 × 𝑀2 is cyclic.

Conversely, let 𝑀1 × 𝑀2 is cyclic. Then 𝑄1 × 𝑄2= 𝑆 ((𝑞𝑖, 𝑝𝑗)) for some (𝑞𝑖, 𝑝𝑗) ∈ 𝑄1 × 𝑄2. Let 𝑞𝑘 ∈ 𝑄1 and 𝑝𝑙 ∈ 𝑄2. Then ∃ 𝑤 ∈ (Σ1∪ Σ2)∗ such that

(α𝑁1× α𝑁2) ∗_((𝑞 𝑖, 𝑝𝑗), 𝑤, (𝑞𝑘, 𝑝𝑙)) > [0,0], (β𝑁1× β𝑁2) ∗_((𝑞 𝑖, 𝑝𝑗), 𝑤, (𝑞𝑘, 𝑝𝑙)) < [1,1] and (γ𝑁1× 𝛾𝑁2) ∗_((𝑞

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Research Article

3712

α𝑁₁∗(𝑞𝑖, 𝑢, 𝑞𝑘) ∧ α𝑁₂∗(𝑝𝑗, 𝑣, 𝑝𝑙) = (α𝑁₁× α𝑁₂)∗((𝑞𝑖, 𝑝𝑗), 𝑤, (𝑞𝑘, 𝑝𝑙)) > [0,0] β𝑁1 ∗_(𝑞 𝑖, 𝑢, 𝑞𝑘) ∨ β𝑁2 ∗ (𝑝𝑗, 𝑣, 𝑝𝑙) = (β𝑁1× β𝑁2) ∗_((𝑞 𝑖, 𝑝𝑗), 𝑤, (𝑞𝑘, 𝑝𝑙)) < [1,1] γ𝑁1 ∗_(𝑞 𝑖, 𝑢, 𝑞𝑘) ∨ γ𝑁2 ∗_(𝑝 𝑗, 𝑣, 𝑝𝑙) = (γ𝑁1× 𝛾𝑁2) ∗_((𝑞 𝑖, 𝑝𝑗), 𝑤, (𝑞𝑘, 𝑝𝑙)) < [1, 1].

Hence ∃ 𝑢 ∈ Σ1∗ and 𝑣 ∈ Σ2∗ such that α𝑁₁∗(𝑞𝑖, 𝑢, 𝑞𝑘) > [0,0], β𝑁₁∗(𝑞𝑖, 𝑢, 𝑞𝑘) < [1,1], γ𝑁₁∗(𝑞𝑖, 𝑢, 𝑞𝑘) < [1,1] and α𝑁2 ∗_(𝑝 𝑗, 𝑣, 𝑝𝑙) > [0,0], β𝑁2 ∗ (𝑝𝑗, 𝑣, 𝑝𝑙) < [1,1], γ𝑁2 ∗_(𝑝 𝑗, 𝑣, 𝑝𝑙) < [1,1]. Thus 𝑞𝑘 ∈ 𝑆(𝑞𝑖) and 𝑝𝑙 ∈ 𝑆(𝑝𝑗). Hence 𝑄1∈ 𝑆(𝑞𝑖) and 𝑄2 ∈ 𝑆(𝑝𝑗). Therefore 𝑀1 × 𝑀2 is cyclic.

5 Conclusion

The purpose of this paper is to study the Cartesian product of INA. We prove that Cartesian product of cyclic of interval neutrosophic automata is cyclic.

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