Some new operators for Fermatean fuzzy matrices

1. Introduction

The concept of an intuitionistic fuzzy matrix (IFM) was introduced by Khan et al., [1] and simultaneously Im et al., [2] to generalize the concept of Thomason's [3] fuzzy matrix. Since the IFM was proposed, it has received a lot of attention in many fields, such as pattern recognition, medical diagnosis, and so on. But if the sum of the membership degree and the nonmembership degree is greater than 1, the IFM is no longer applicable. In [4] using IFM theory, we developed the Pythagorean fuzzy matrix (PFM) \(A=\left[\left\langle \zeta_{a_{ij}},\delta_{a_{ij}}\right\rangle\right]\), where the squared sum of its membership degree \(\zeta_{a_{ij}}\in [0,1]\) and nonmembership degree \(\delta_{a_{ij}}\in [0,1]\) is less than or equal to 1.

Since the PFM was brought up, it has been widely applied in different fields, such as investment decision making, service quality of domestic airline, collaborative-based recommender systems, and so on. Although the PFM generalizes the IFM, it cannot describe the following decision information. A panel of experts were invited to give their opinions about the feasibility of an investment plan, and they were divided into two independent groups to make a decision. One group considered the degree of the feasibility of the investment plan as 0.8, while the other group considered the nonmembership degree as 0.7. It was clearly seen that \(0.8+0.7>1,(0.8)^2+(0.7)^2>1\), and thus it could not be described by IFM and PFM.

After the IFM and PFM theory, many researchers attempted the important role in this theory, [5,6,7,8,9,10,11,12,13,14,15,16,17]. To describe such evaluation information, we have developed Fermatean fuzzy matrix (FFM) \(A=\left[\left\langle \zeta_{a_{ij}},\delta_{a_{ij}}\right\rangle\right]\), where represent the \(\zeta_{a_{ij}}\in [0,1]\) membership degree and \(\delta_{a_{ij}}\in [0,1]\) the non-membership degree for all \(i,j\), respectively, and \(0\leq \zeta^3_{a_{ij}}+\delta^3_{a_{ij}}\leq 1\) [18]. It was clearly seen that \(0.8+0.7>1,(0.8)^2+(0.7)^2>1,(0.8)^3+(0.7)^3\leq 1\). Then we defined some Fermatean fuzzy operators \(\boxplus_{F}, \boxtimes_{F}, @, \cap, \cup\) and properties are considered. In this paper we have developed some new operators for Fermatean fuzzy matrices and discussed several properties.

2. Preliminaries

In this section, some basic concepts related to the fuzzy matrix (FM), intuitionistic fuzzy matrix (IFM) and Pythagorean fuzzy matrix (PFM) have been given.

Definition 1. [3] A fuzzy matrix of \(A\) of order \(m\times n\) is defined as \(A=(a_{ij})\), where \(a_{ij}\in [0,1].\)

Definition 2. [1] An intuitionistic fuzzy matrix (IFM) is a pair \(A=\Big[\left\langle \zeta_{a_{ij}}, \delta_{a_{ij}}\right\rangle\Big]\) of a non negative real numbers \(\zeta_{a_{ij}}, \delta_{a_{ij}}\in [0,1]\) satisfying \(0\leq \zeta_{a_{ij}}+\delta_{a_{ij}}\leq 1\) for all \(i,j.\)

Definition 3. [4] A Pythagorean fuzzy matrix (PFM) is a pair \(A=\Big[\left\langle \zeta_{a_{ij}}, \delta_{a_{ij}}\right\rangle\Big]\) of non negative real numbers \(\zeta_{a_{ij}}, \delta_{a_{ij}}\in [0,1]\) satisfying the condition \(0\leq\zeta^2_{a_{ij}}+\delta^2_{a_{ij}}\leq 1\), for all \(i,j\). Where \(\zeta_{a_{ij}}\in[0,1]\) is called the degree of membership and \(\delta_{a_{ij}}\in[0,1]\) is called the degree of non-membership.

Definition 4. [18] A Fermatean fuzzy matrix (FFM) is a pair \(A=\Big[\left\langle \zeta_{a_{ij}}, \delta_{a_{ij}}\right\rangle\Big]\) of non negative real numbers \(\zeta_{a_{ij}}, \delta_{a_{ij}}\in [0,1]\) satisfying the condition \(0\leq\zeta^3_{a_{ij}}+\delta^3_{a_{ij}}\leq 1\), for all \(i,j\). Where \(\zeta_{a_{ij}}\in[0,1]\) is called the degree of membership and \(\delta_{a_{ij}}\in[0,1]\) is called the degree of non-membership.

Definition 5. [18] Let \(F_{m \times n}\) denote the family of all \(FFMs\) for all \(i,j\), and \(A,B\in F_{m \times n}\) be given as \[A=\left[\left\langle\zeta_{a_{ij}},\delta_{a_{ij}}\right\rangle\right]\ \ \ \text{and}\ \ \ B=\left[\left\langle\zeta_{b_{ij}},\delta_{b_{ij}}\right\rangle\right]\] then

• (i) \(A\vee B=\left[\left\langle\max\left\{\zeta_{a_{ij}},\zeta_{b_{ij}}\right\},\min\left\{\delta_{a_{ij}},\delta_{b_{ij}}\right\}\right\rangle\right]\),
• (ii) \(A\wedge B=\left[\left\langle\min\left\{\zeta_{a_{ij}},\zeta_{b_{ij}}\right\},\max\left\{\delta_{a_{ij}},\delta_{b_{ij}}\right\}\right\rangle\right]\),
• (iii) \(A^C=\left[\left\langle(\delta_{a_{ij}}),(\zeta_{a_{ij}})\right\rangle\right]\),
• (iv) \(A\boxplus_{F} B=\left[\left\langle \sqrt[3]{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}-\zeta^3_{a_{ij}}\zeta^3_{b_{ij}}}, \delta_{a_{ij}}\delta_{b_{ij}}\right\rangle\right]\),
• (v) \( A\boxtimes_{F} B=\left[\left\langle \zeta_{a_{ij}}\zeta_{b_{ij}},\sqrt[3]{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}-\delta^3_{a_{ij}}\delta^3_{b_{ij}}}\right\rangle\right] \),
• (vi) \(A @ B=\left[\left\langle \sqrt[3]{\dfrac{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}}{2}},\sqrt[3]{\dfrac{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}}{2}}\right\rangle\right]\).

3. New operators for Fermatean fuzzy matrices

In this section, we define the new operators for Fermatean fuzzy matrices and investigates the several algebraic properties.

Definition 6. Let \(F_{m \times n}\) denote the family of all \(FFMs\) for all \(i,j\), and let \(A,B\in F_{m \times n}\) be given as

• (i) \(A\$ B=\left[\left\langle \sqrt[3]{\zeta_{a_{ij}}\zeta_{b_{ij}}},\sqrt[3]{\delta_{a_{ij}}\delta_{b_{ij}}}\right\rangle\right]\),
• (ii) \(A\# B=\left[\left\langle \dfrac{\sqrt[3]{2}\zeta_{a_{ij}}\zeta_{b_{ij}}}{\sqrt[3]{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}}}, \dfrac{\sqrt[3]{2}\delta_{a_{ij}}\delta_{b_{ij}}}{\sqrt[3]{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}}}\right\rangle\right]\).

For which we shall accept that if \(\zeta_{a_{ij}}=\zeta_{b_{ij}}=0\) then \(\dfrac{\zeta_{a_{ij}}\zeta_{b_{ij}}}{\zeta_{a_{ij}}+\zeta_{b_{ij}}}=0\) and if \(\delta_{a_{ij}}=\delta_{b_{ij}}=0\), then \(\dfrac{\delta_{a_{ij}}\delta_{b_{ij}}}{\delta_{a_{ij}}+\delta_{b_{ij}}}=0\).

• (iii) \(A\ast B =\left[\left\langle\sqrt[3]{\dfrac{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}}{2(\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}+1)}},\sqrt[3]{\dfrac{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}}{2(\delta^3_{a_{ij}}+\delta^3_{b_{ij}}+1)}} \right\rangle\right]\),
• (iv) \(A\rightarrow B=\left[\left\langle\max\left\{\delta_{a_{ij}},\zeta_{b_{ij}}\right\},\min\left\{\zeta_{a_{ij}},\delta_{b_{ij}}\right\}\right\rangle\right]\).

Remark 1. Clearly, for each two FFMs A and B, \([(A @ B),(A \$ B),(A \# B),(A\ast B),(A\rightarrow B)]\) are as yet an FFM. Some basic representations are appear as follows:

• For (i), \(0\leq \left(\sqrt[3]{\dfrac{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}}{2}}\right)^3+\left(\sqrt[3]{\dfrac{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}}{2}}\right)^3=\dfrac{\zeta^3_{a_{ij}}+\delta^3_{a_{ij}}}{2}+\dfrac{\zeta^3_{b_{ij}}+\delta^3_{b_{ij}}}{2} \leq \dfrac{1}{2}+\dfrac{1}{2}=1.\)
• For (ii), if \(\delta_{a_{ij}}\geq \zeta_{b_{ij}}\) and \(\zeta_{a_{ij}}\geq \delta_{b_{ij}}\), then \(0\leq \max\left\{ \delta^3_{a_{ij}},\zeta^3_{b_{ij}}\right\}+\min\left\{\zeta^3_{a_{ij}},\delta^3_{b_{ij}}\right\}\leq \delta^3_{a_{ij}}+\zeta^3_{a_{ij}}\leq 1.\)

  If \(\delta_{a_{ij}}\geq \zeta_{b_{ij}}\) and \(\zeta_{a_{ij}}\leq \delta_{b_{ij}}\), then \(0\leq \max\left\{ \delta^3_{a_{ij}},\zeta^3_{b_{ij}}\right\}+\min\left\{\zeta^3_{a_{ij}},\delta^3_{b_{ij}}\right\}\leq \delta^3_{a_{ij}}+\zeta^3_{a_{ij}}\leq 1.\)

  If \(\delta_{a_{ij}}\leq \zeta_{b_{ij}}\) and \(\zeta_{a_{ij}}\geq \delta_{b_{ij}}\), then \(0\leq \max\left\{ \delta^3_{a_{ij}},\zeta^3_{b_{ij}}\right\}+\min\left\{\zeta^3_{a_{ij}},\delta^3_{b_{ij}}\right\}\leq \delta^3_{b_{ij}}+\zeta^3_{b_{ij}}\leq 1.\)

  If \(\delta_{a_{ij}}\leq \zeta_{b_{ij}}\) and \(\zeta_{a_{ij}}\leq \delta_{b_{ij}}\), then \(0\leq \max\left\{ \delta^3_{a_{ij}},\zeta^3_{b_{ij}}\right\}+\min\left\{\zeta^3_{a_{ij}},\delta^3_{b_{ij}}\right\}\leq \delta^3_{b_{ij}}+\zeta^3_{b_{ij}}\leq 1.\)

• For (iii), \(0\leq \left(\sqrt[3]{\zeta_{a_{ij}}\zeta_{b_{ij}}}\right)^3+\left(\sqrt[3]{\delta_{a_{ij}}\delta_{b_{ij}}}\right)^3=\zeta_A(x)\zeta_B(x)+\delta_A(x)\delta_B(x)=\dfrac{\zeta^3_{a_{ij}}+\delta^3_{a_{ij}}}{2}+\dfrac{\zeta^3_{b_{ij}}+\delta^3_{b_{ij}}}{2}\leq \dfrac{1}{2}+\dfrac{1}{2}=1.\)
• For (iv), \(0\leq \left(\dfrac{\sqrt[3]{2}\zeta_{a_{ij}}\zeta_{b_{ij}}}{\sqrt[3]{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}}}\right)^3+\left( \dfrac{\sqrt[3]{2}\delta_{a_{ij}}\delta_{b_{ij}}}{\sqrt[3]{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}}}\right)^3=\dfrac{2\zeta^3_{a_{ij}}\zeta^3_{b_{ij}}}{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}}+ \dfrac{2\delta^3_{a_{ij}}\delta^3_{b_{ij}}}{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}}\leq 1.\)
• For (v), \(0\leq \left(\sqrt[3]{\dfrac{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}}{2(\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}+1)}}\right)^3+\left(\sqrt[3]{\dfrac{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}}{2(\delta^3_{a_{ij}}+\delta^3_{b_{ij}}+1)}}\right)^3={\dfrac{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}}{2(\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}+1)}},{\dfrac{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}}{2(\delta^3_{a_{ij}}+\delta^3_{b_{ij}}+1)}}\leq 1.\)

Lemma 1.[6] For any two numbers \(a,b\in [0,1]\), then \[a.b\leq \min\left\{a,b\right\}\leq \dfrac{2(a.b)}{a+b}\leq \sqrt{a.b}\leq \max\left\{a,b\right\}\leq a+b-a.b,\] and \[a.b\leq \dfrac{a+b}{2(a+b+1)}\leq \dfrac{a+b}{2}.\]

Theorem 1. For \(A,B\in F_{m \times n}\), we have

• (i) \( A @ B=B @ A=(A^C @ B^C)^C\),
• (ii) \( A\$ B=B\$ A=(A^C \$ B^C)^C\),
• (iii) \( A\# B=B\# A=(A^C\# B^C)^C\),
• (iv) \( A\ast B=B\ast A=(A^C\ast B^C)^C\).

Proof. Here, we prove only (i). Others can be proved similarly.

Let A and B be two given FFMs, then

\begin{align*}A @ B &=\left[\left\langle \sqrt[3]{\dfrac{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}}{2}},\sqrt[3]{\dfrac{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}}{2}}\right\rangle\right]=\left[\left\langle \sqrt[3]{\dfrac{\zeta^3_{b_{ij}}+\zeta^3_{a_{ij}}}{2}},\sqrt[3]{\dfrac{\delta^3_{b_{ij}}+\delta^3_{a_{ij}}}{2}}\right\rangle\right]=B@A,\\ A^C @ B^C&=\left[\left\langle \sqrt[3]{\dfrac{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}}{2}},\sqrt[3]{\dfrac{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}}{2}}\right\rangle\right],\\ (A^C @ B^C)^C&=\left[\left\langle \sqrt[3]{\dfrac{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}}{2}},\sqrt[3]{\dfrac{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}}{2}}\right\rangle\right]=A@B.\end{align*} Hence, \( A@B=B@A=(A^C@B^C)^C.\)

The following theorems are obvious:

Theorem 2. For \(A,B,C\in F_{m \times n}\), we have

• (i)  \( (A\wedge B)@ C=(A@C)\wedge(B@C)\),
• (ii)   \((A\vee B)@ C=(A@C)\vee(B@C)\),
• (iii)   \( (A\wedge B)\$ C=(A\$ C)\wedge(B\$ C),\)
• (iv)   \((A\vee B)\$ C=(A\$ C)\vee(B\$ C),\)
• (v)   \( (A\wedge B)\# C=(A\# C)\wedge(B\# C),\)
• (vi)   \( (A\vee B)\# C=(A\# C)\vee(B\# C),\)
• (vii)   \( (A\wedge B)\ast C=(A\ast C)\wedge(B\ast C),\)
• (viii)   \( (A\vee B)\ast C=(A\ast C)\vee(B\ast C).\)

Theorem 3. For \(A,B,C\in F_{m \times n}\), we have

• (i) \( (A\boxplus_{F} B)@ C\leq(A @ C)\boxplus_{F}(B@ C),\)
• (ii)   \( (A\boxtimes_{F} B)@ C\geq(A @ C)\boxtimes_{F}(B@ C),\)
• (iii)  \( (A\boxplus_{F} B)\$ C\leq(A \$ C)\boxplus_{F}(B\$ C),\)
• (vi)   \( (A\boxtimes_{F} B)\$ C\geq(A \$ C)\boxtimes_{F}(B\$ C),\)
• (v)   \( (A\boxplus_{F} B)\ast C\leq(A \ast C)\boxplus_{F}(B\ast C),\)
• (vi)   \( (A\boxtimes_{F} B)\ast C\geq(A \ast C)\boxtimes_{F}(B\ast C)\).

Theorem 4.For \(A,B,C\in F_{m \times n}\), we have

• (i)   \( (A@ B)\boxplus_{F} C=(A \boxplus_{F} C)@(B\boxplus_{F} C),\)
• (ii)  \( (A@ B)\boxtimes_{F} C=(A \boxtimes_{F} C)@(B\boxtimes_{F} C),\)
• (iii)   \( (A\$ B)\boxplus_{F} C\leq(A \boxplus_{F} C)\$(B\boxplus_{F} C),\)
• (iv)   \( (A\$ B)\boxtimes_{F} C\geq (A \boxtimes_{F} C)\$(B\boxtimes_{F} C),\)
• (v)   \( (A\# B)\boxplus_{F} C\leq(A \boxplus_{F} C)\#(B\boxplus_{F} C),\)
• (vi)   \( (A\# B)\boxtimes_{F} C\geq (A \boxtimes_{F} C)\#(B\boxtimes_{F} C),\)
• (vii)  \( (A\ast B)\boxplus_{F} C\leq(A \boxplus_{F} C)\ast(B\boxplus_{F} C),\)
• (viii)  \( (A\ast B)\boxtimes_{F} C\geq (A \boxtimes_{F} C)\ast(B\boxtimes_{F} C)\).

4. Necessity and possibility operators on Fermatean fuzzy matrices

In this section, we prove the necessity and possibility operators of Fermatean fuzzy matrices. Then we compile some relevent properties of these operators are discussed.

Definition 7. [18] The necessity and possibility operators on a Fermatean fuzzy matrix A is denoted by \(\Box A , \Diamond A\) and is \(\Box A =\left[\left\langle \zeta_{a_{ij}}, \sqrt[3]{1-\zeta^3_{a_{ij}}} \right\rangle\right], \Diamond A =\left[\left\langle \sqrt[3]{1-\delta^3_{a_{ij}}},\delta_{a_{ij}} \right\rangle\right]\).

Theorem 5. For \(A,B \in F_{m \times n}\), we have

• (i)   \( \Box A @ \Box B=\Box(A@B)\leq \Diamond A @ \Diamond B=\Diamond(A@B),\)
• (ii)   \( \Box(A \$ B)\leq \Box A \$ \Box B\leq \Diamond A \$ \Diamond B\leq \Diamond(A\$ B),\)
• (iii)  \( \Box(A \# B)\leq \Box A \# \Box B\leq \Diamond A \# \Diamond B\leq \Diamond(A\# B),\)
• (iv)   \( ~\Box(A \ast B)\leq \Box A \ast \Box B\leq \Diamond A \ast \Diamond B\leq \Diamond(A\ast B).\)

Proof. Here we prove (i) and (iii). (ii) and (iv) can be proved similarly.

• (i)  Since \[\Box A @ \Box B=\left[\left\langle \sqrt[3]{\dfrac{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}}{2}},\sqrt[3]{\dfrac{1-\zeta^3_{a_{ij}}+1-\zeta^3_{b_{ij}}}{2}}\right\rangle\right]=\Box(A@B),\] and \[\Diamond A @ \Diamond B=\left[\left\langle \sqrt[3]{\dfrac{1-\delta^3_{a_{ij}}+1-\delta^3_{b_{ij}}}{2}},\sqrt[3]{\dfrac{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}}{2}}\right\rangle\right]=\Diamond(A@B).\] So, \[\Box A @ \Box B=\Box(A@B)\leq \Diamond A @ \Diamond B=\Diamond(A@B).\]
• (iii)   Since \[ ~\Box A \# \Box B=\left[\left\langle \dfrac{\sqrt[3]{2}\zeta_{a_{ij}}\zeta_{b_{ij}}}{\sqrt[3]{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}}}, \dfrac{\sqrt[3]{2}\sqrt[3]{1-\zeta^3_{a_{ij}}}\sqrt[3]{1-\zeta^3_{b_{ij}}}}{\sqrt[3]{1-\zeta^3_{a_{ij}}+1-\zeta^3_{b_{ij}}}}\right\rangle\right]\leq \Box(A \# B),\] and \[\Diamond A \# \Diamond B=\Bigg[\left\langle \dfrac{\sqrt[3]{2}\sqrt[3]{1-\delta^3_{a_{ij}}}\sqrt[3]{1-\delta^3_{b_{ij}}}}{\sqrt[3]{1-\delta^3_{a_{ij}}+1-\delta^3_{b_{ij}}}},\dfrac{\sqrt[3]{2}\delta_{a_{ij}}\delta_{b_{ij}}}{\sqrt[3]{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}}}\right\rangle\Bigg]\leq \Diamond(A \# B).\] So, \[\Box(A \# B)\leq \Box A \# \Box B\leq \Diamond A \# \Diamond B\leq \Diamond(A\# B).\]

The following theorems are obvious:

Theorem 6. For \(A,B \in F_{m \times n}\), we have

• (i)   \( \Box \left[\left(\Diamond A @ \Diamond B \right)^C \right]=\left[\Diamond (A @ B) \right]^C,\)
• (ii)   \( \Diamond \left[\left(\Box A @ \Box B \right)^C \right]=\left[\Box (A @ B) \right]^C,\)
• (iii)   \( \Box \left[\left(\Diamond A \$ \Diamond B \right)^C \right]=\left[\Diamond (A \$ B) \right]^C,\)
• (vi)   \( \Diamond \left[\left(\Box A \$ \Box B \right)^C \right]=\left[\Box (A \$ B) \right]^C,\)
• (v)   \(\Box \left[\left(\Diamond A \# \Diamond B \right)^C \right]=\left[\Diamond (A \# B) \right]^C,\)
• (vi)   \( \Diamond \left[\left(\Box A \# \Box B \right)^C \right]=\left[\Box (A \# B) \right]^C,\)
• (vii)   \( \Box \left[\left(\Diamond A \ast \Diamond B \right)^C \right]=\left[\Diamond (A \ast B) \right]^C,\)
• (viii)   \( \Diamond \left[\left(\Box A \ast \Box B \right)^C \right]=\left[\Box (A \ast B) \right]^C\).

Theorem 7. For \(A,B \in F_{m \times n}\), we have

• (i)   \(\left[\left(\Box A \boxplus_{F} \Diamond B \right)^C @ ((\Box A)^C \boxtimes_{F} \Diamond B) \right]\vee (\Box A)^C=(\Box A)^C,\)
• (ii)   \(\left[\left(\Box A \boxtimes_{F} \Diamond B \right)^C @ ((\Box A)^C \boxplus_{F} \Diamond B) \right]\wedge (\Box A)^C=(\Box A)^C,\)
• (iii)   \( \left[\left(\Box A \boxplus_{F} \Diamond B \right)^C \$ ((\Box A)^C \boxtimes_{F} \Diamond B) \right]\vee (\Box A)^C=(\Box A)^C,\)
• (iv)   \( \left[\left(\Box A \boxtimes_{F} \Diamond B \right)^C \$ ((\Box A)^C \boxplus_{F} \Diamond B) \right]\wedge (\Box A)^C=(\Box A)^C,\)
• (v)   \( \left[\left(\Box A \boxplus_{F} \Diamond B \right)^C \# ((\Box A)^C \boxtimes_{F} \Diamond B) \right]\vee (\Box A)^C=(\Box A)^C,\)
• (vi)  \( \left[\left(\Box A \boxtimes_{F} \Diamond B \right)^C \# ((\Box A)^C \boxplus_{F} \Diamond B) \right]\wedge (\Box A)^C=(\Box A)^C,\)
• (vii)  \( \left[\left(\Diamond A \boxplus_{F} \Box B \right)^C @ ((\Box A)^C \boxtimes_{F} \Diamond B) \right]\vee (\Diamond A)^C=(\Diamond A)^C,\)
• (viii)  \( \left[\left(\Diamond A \boxplus_{F} \Box B \right)^C \$ ((\Box A)^C \boxtimes_{F} \Diamond B) \right]\vee (\Diamond A)^C=(\Diamond A)^C,\)
• (ix)  \( \left[\left(\Diamond A \boxplus_{F} \Box B \right)^C \# ((\Box A)^C \boxtimes_{F} \Diamond B) \right]\vee (\Diamond A)^C=(\Diamond A)^C,\)
• (x)   \(\left[\left(\Diamond A \boxplus_{F} \Box B \right)^C @ ((\Diamond A)^C \boxtimes_{F} \Box B) \right]\vee (\Diamond A)^C=(\Diamond A),\)
• (xi)   \( \left[\left(\Diamond A \boxplus_{F} \Box B \right)^C \$ ((\Diamond A)^C \boxtimes_{F} \Box B)\right]\vee (\Diamond A)^C=(\Diamond A),\)
• (xii)  \( \left[\left(\Diamond A \boxplus_{F} \Box B \right)^C \# ((\Diamond A)^C \boxtimes_{F} \Box B) \right]\vee (\Diamond A)^C=(\Diamond A).\)

In the next section, we state and prove some new results involving implication operator with other FFM operators

5. Some results of FFMs based on implication operator

In this section, the proofs of the following theorems and corollaries follows from the Definitions 5, 6 and Lemma 1.

Theorem 8. For \(A,B \in F_{m \times n}\), we have

• (i)   \(\left( A^C\rightarrow B\right)@\left(A \rightarrow B^C\right)^C=\left(A @ B\right),\)
• (ii)   \(\left( A^C\rightarrow B\right) \boxplus_{F} \left(A \rightarrow B^C\right)^C=\left(A \boxplus_{F} B\right),\)
• (iii)  \(\left( A^C\rightarrow B\right) \boxtimes_{F} \left(A \rightarrow B^C\right)^C=\left(A \boxtimes_{F} B\right)\),
• (iv)   \(\left( A^C\rightarrow B\right) \$ \left(A \rightarrow B^C\right)^C=\left(A \$ B\right),\)
• (v)   \(\left( A^C\rightarrow B\right) \#\left(A \rightarrow B^C\right)^C=\left(A \# B\right),\)
• (vi)   \(\left( A\rightarrow B\right)^C \boxplus_{F}\left(B \rightarrow A\right)=\left(A \boxplus_{F} B^C\right),\)
• (vii)  \(\left( A\rightarrow B\right)^C @\left(B \rightarrow A\right)=\left(A @ B^C\right),\)
• (viii)   \(\left( A\rightarrow B\right)^C \boxtimes_{F}\left(B \rightarrow A\right)=\left(A \boxtimes_{F} B^C\right),\)
• (ix)   \(\left( A\rightarrow B\right)^C \$ \left(B \rightarrow A\right)=\left(A \$ B^C\right),\)
• (x)   \(\left( A\rightarrow B\right)^C \# \left(B \rightarrow A\right)=\left(A \# B^C\right).\)

Proof. We prove only (i) and (vi) and remaining can be proved analogously.

• (i) \begin{align*} \left( A^C\rightarrow B\right)&@\left(A \rightarrow B^C\right)^C\\ &=\left[ \sqrt[3]{\dfrac{\max\left\{\zeta^3_{a_{ij}},\zeta^3_{b_{ij}}\right\}+\min\left\{\zeta^3_{a_{ij}},\zeta^3_{b_{ij}}\right\}}{2}}, \sqrt[3]{\dfrac{\min\left\{\delta^3_{a_{ij}},\delta^3_{b_{ij}}\right\}+\max\left\{\delta^3_{a_{ij}},\delta^3_{b_{ij}}\right\}}{2}} \right]\\ &=\left[\left\langle \sqrt[3]{\dfrac{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}}{2}},\sqrt[3]{\dfrac{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}}{2}}\right\rangle\right]\\ &=(A @ B).\end{align*}
• (vi) \begin{align*}&\left( A\rightarrow B\right)^C \boxplus_{F}\left(B \rightarrow A\right)\\ &=\left[\sqrt[3]{\min\left\{\zeta^3_{a_{ij}},\delta^3_{b_{ij}}\right\}+\max\left\{\delta^3_{b_{ij}},\zeta^3_{a_{ij}}\right\} - \min\left\{\zeta^3_{a_{ij}},\delta^3_{b_{ij}}\right\}\max\left\{\delta^3_{b_{ij}},\zeta^3_{a_{ij}}\right\}}, \max\left\{\delta_{a_{ij}},\zeta_{b_{ij}}\right\}\min\left\{\zeta_{b_{ij}},\delta_{a_{ij}}\right\}\right]\\ &=\left[\left\langle \sqrt[3]{\zeta^3_{a_{ij}}+\delta^3_{b_{ij}}-\zeta^3_{a_{ij}}\delta^3_{b_{ij}}},\delta_{a_{ij}}\zeta_{b_{ij}}\right\rangle\right]\\ &=(A\boxplus_{F} B^C).\end{align*}

Theorem 9. For \(A,B\in F_{m \times n}\), we have

• (i)  \(\left((A\boxplus_{F} B)\rightarrow(A @ B)^C\right)^C=\left((A @ B)\rightarrow(A \boxplus_{F} B)^C\right)^C=(A @ B),\)
• (ii)  \(\left((A\boxplus_{F} B)^C\rightarrow (A @ B)\right)=\left((A @ B)^C\rightarrow (A \boxplus_{F} B)\right)=(A\boxplus_{F} B),\)
• (iii)  \(\left((A\boxtimes_{F} B)\rightarrow(A @ B)^C\right)^C=\left((A @ B)\rightarrow(A \boxtimes_{F} B)^C\right)^C=(A \boxtimes_{F} B),\)
• (iv)  \(\left((A\boxtimes_{F} B)^C\rightarrow (A @ B)\right)=\left((A @ B)^C\rightarrow (A \boxtimes_{F} B)\right)=(A @ B),\)
• (v)  \(\left((A\boxplus_{F} B)\rightarrow(A \# B)^C\right)^C=\left((A \# B)\rightarrow(A \boxplus_{F} B)^C\right)^C=(A \# B),\)
• (vi)  \(\left((A\boxplus_{F} B)^C\rightarrow (A \# B)\right)=\left((A \# B)^C\rightarrow (A \boxplus_{F} B)\right)=(A\boxplus_{F} B),\)
• (vii)  \(\left((A\boxtimes_{F} B)\rightarrow(A \# B)^C\right)^C=\left((A \# B)\rightarrow(A \boxtimes_{F} B)^C\right)^C=(A \boxtimes_{F} B),\)
• (viii)   \(\left((A\boxtimes_{F} B)^C\rightarrow (A \# B)\right)=\left((A \# B)^C\rightarrow (A \boxtimes_{F} B)\right)=(A \# B),\)
• (ix)  \(\left((A\boxplus_{F} B)\rightarrow(A \$ B)^C\right)^C=\left((A \$ B)\rightarrow(A \boxplus_{F} B)^C\right)^C=(A \$ B),\)
• (x)  \(\left((A\boxplus_{F} B)^C\rightarrow (A \$ B)\right)=\left((A \$ B)^C\rightarrow (A \boxplus_{F} B)\right)=(A\boxplus_{F} B),\)
• (xi)  \(\left((A\boxtimes_{F} B)\rightarrow(A \$ B)^C\right)^C=\left((A \$ B)\rightarrow(A \boxtimes_{F} B)^C\right)^C=(A \boxtimes_{F} B),\)
• (xii)  \(\left((A\boxtimes_{F} B)^C\rightarrow (A \$ B)\right)=\left((A \$ B)^C\rightarrow (A \boxtimes_{F} B)\right)=(A \$ B),\)
• (xiii)  \(\left((A\boxplus_{F} B)\rightarrow(A \boxtimes_{F} B)^C\right)^C=\left((A \boxtimes_{F} B)\rightarrow(A \boxplus_{F} B)^C\right)^C=(A \boxtimes_{F} B),\)
• (xiv)   \(\left((A\boxtimes_{F} B)^C\rightarrow (A \boxplus_{F} B)\right)=\left((A \boxplus_{F} B)^C\rightarrow (A \boxtimes_{F} B)\right)=(A \boxplus_{F} B).\)

Proof. Here, we prove (i), (iii), (v), (vii), (ix) and (xiii). Others can be proved analogously.

• (i)   Since \begin{align*} \left((A\boxplus_{F} B)\rightarrow(A @ B)^C\right)^C &=\left[ \min\left\{\sqrt[3]{\zeta^3_{a_{ij}}+\zeta^3_{a_{ij}}-\zeta^3_{a_{ij}}\zeta^3_{a_{ij}}},\sqrt[3]{\dfrac{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}}{2}}\right\}, \max\left\{\delta_{a_{ij}}\delta_{b_{ij}},\sqrt[3]{\dfrac{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}}{2}}\right\}\right]\\ &=\left[\left\langle \sqrt[3]{\dfrac{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}}{2}},\sqrt[3]{\dfrac{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}}{2}}\right\rangle\right]\\ &=A @ B,\end{align*} and \begin{align*} \left((A @ B)\rightarrow(A \boxplus_{F} B)^C\right)^C &=\left[ \min\left\{\sqrt[3]{\dfrac{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}}{2}},\sqrt[3]{\zeta^3_{a_{ij}}+\zeta^3_{a_{ij}}-\zeta^3_{a_{ij}}\zeta^3_{a_{ij}}}\right\}, \max\left\{\sqrt[3]{\dfrac{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}}{2}} ,\delta_{a_{ij}}\delta_{b_{ij}}\right\}\right]\\ &=\left[\left\langle \sqrt[3]{\dfrac{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}}{2}},\sqrt[3]{\dfrac{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}}{2}}\right\rangle\right]\\ &=A @ B.\end{align*} So, \[\left((A\boxplus_{F} B)\rightarrow(A @ B)^C\right)^C=\left((A @ B)\rightarrow(A \boxplus_{F} B)^C\right)^C=(A @ B).\]
• (iii)   Since \begin{align*} \left((A\boxtimes_{F} B)\rightarrow (A @ B)^C\right)^C &=\left[ \min\left\{\zeta_{a_{ij}}\zeta_{b_{ij}},\sqrt[3]{\dfrac{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}}{2}}\right\}, \max \left\{\sqrt[3]{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}-\delta^3_{a_{ij}}\delta^3_{b_{ij}}},\sqrt[3]{\dfrac{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}}{2}}\right\}\right]\\ &=\left[\left\langle \zeta_{a_{ij}}\zeta_{b_{ij}},\sqrt[3]{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}-\delta^3_{a_{ij}}\delta^3_{b_{ij}}}\right\rangle\right]\\ &=A\boxtimes_{F} B,\end{align*} and \begin{align*} \left((A @ B)\rightarrow(A \boxtimes_{F} B)^C\right)^C &=\left[ \min\left\{\sqrt[3]{\dfrac{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}}{2}},\zeta_{a_{ij}}\zeta_{b_{ij}}\right\}, ~\max \left\{\sqrt[3]{\dfrac{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}}{2}},\sqrt[3]{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}-\delta^3_{a_{ij}}\delta^3_{b_{ij}}}\right\}\right]\\ &=\left[\left\langle \zeta_{a_{ij}}\zeta_{b_{ij}},\sqrt[3]{\delta^3_{a_{ij}}+\delta^3_{a_{ij}}-\delta^3_{a_{ij}}\delta^3_{a_{ij}}}\right\rangle\right]\\ &=A\boxtimes_{F} B.\end{align*} So, \[\left((A\boxtimes_{F} B)\rightarrow(A @ B)^C\right)^C=\left((A @ B)\rightarrow(A \boxtimes_{F} B)^C\right)^C=(A \boxtimes_{F} B).\]
• (v) Since \begin{align*} \left((A\boxplus_{F} B)\rightarrow(A \# B)^C\right)^C &=\left[ \min\left\{\sqrt[3]{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}-\zeta^3_{a_{ij}}\zeta^3_{b_{ij}}},\dfrac{\sqrt[3]{2}\zeta_{a_{ij}}\zeta_{b_{ij}}}{\sqrt[3]{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}}}\right\}, \max\left\{\delta_{a_{ij}}\delta_{b_{ij}},\dfrac{\sqrt[3]{2}\delta_{a_{ij}}\delta_{b_{ij}}}{\sqrt[3]{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}}}\right\}\right]\\ &=\left[\left\langle \dfrac{\sqrt[3]{2}\zeta_{a_{ij}}\zeta_{b_{ij}}}{\sqrt[3]{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}}}, \dfrac{\sqrt[3]{2}\delta_{a_{ij}}\delta_{b_{ij}}}{\sqrt[3]{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}}}\right\rangle\right]\\ &=A\# B,\end{align*} and \begin{align*} \left((A \# B)\rightarrow(A \boxplus_{F} B)^C\right)^C &=\left[ \min\left\{\dfrac{\sqrt[3]{2}\zeta_{a_{ij}}\zeta_{b_{ij}}}{\sqrt[3]{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}}},\sqrt[3]{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}-\zeta^3_{a_{ij}}\zeta^3_{b_{ij}}}\right\}, \max\left\{\dfrac{\sqrt[3]{2}\delta_{a_{ij}}\delta_{b_{ij}}}{\sqrt[3]{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}}},\delta_{a_{ij}}\delta_{b_{ij}}\right\}\right]\\ &=\left[\left\langle \dfrac{\sqrt[3]{2}\zeta_{a_{ij}}\zeta_{b_{ij}}}{\sqrt[3]{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}}}, \dfrac{\sqrt[3]{2}\delta_{a_{ij}}\delta_{b_{ij}}}{\sqrt[3]{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}}}\right\rangle\right]\\ &=A\# B.\end{align*} So, \[\left((A\boxplus_{F} B)\rightarrow(A \# B)^C\right)^C=\left((A \# B)\rightarrow(A \boxplus_{F} B)^C\right)^C=(A \# B).\]
• (vii)   Since \begin{align*}\left((A\boxtimes_{F} B)\rightarrow(A \# B)^C\right)^C& =\left[ x,\min\left\{\zeta_{a_{ij}}\zeta_{b_{ij}},\dfrac{\sqrt[3]{2}\zeta_{a_{ij}}\zeta_{b_{ij}}}{\sqrt[3]{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}}}\right\}, \max \left\{\sqrt[3]{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}-\delta^3_{a_{ij}}\delta^3_{b_{ij}}},\dfrac{\sqrt[3]{2}\delta_{a_{ij}}\delta_{b_{ij}}}{\sqrt[3]{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}}}\right\}\right]\\ &=\left[\left\langle \zeta_{a_{ij}}\zeta_{b_{ij}},\sqrt[3]{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}-\delta^3_{a_{ij}}\delta^3_{b_{ij}}}\right\rangle\right]\\ &=A\boxtimes_{F} B,\end{align*} and \begin{align*} \left((A \# B)\rightarrow(A \boxtimes_{F} B)^C\right)^C &=\left[ x,\min\left\{\dfrac{\sqrt[3]{2}\zeta_{a_{ij}}\zeta_{b_{ij}}}{\sqrt[3]{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}}},\zeta_{a_{ij}}\zeta_{b_{ij}}\right\}, \max \left\{\dfrac{\sqrt[3]{2}\delta_{a_{ij}}\delta_{b_{ij}}}{\sqrt[3]{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}}},\sqrt[3]{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}-\delta^3_{a_{ij}}\delta^3_{b_{ij}}}\right\}\right]\\ &=\left[\left\langle \zeta_{a_{ij}}\zeta_{b_{ij}},\sqrt[3]{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}-\delta^3_{a_{ij}}\delta^3_{b_{ij}}}\right\rangle\right]\\ &=A\boxtimes_{F} B.\end{align*} So, \[\left((A\boxtimes_{F} B)\rightarrow(A \# B)^C\right)^C=\left((A \# B)\rightarrow(A \boxtimes_{F} B)^C\right)^C=(A \boxtimes_{F} B).\]
• (ix) Since \begin{align*}\left((A\boxplus_{F} B)\rightarrow(A \$ B)^C\right)^C &=\left[ \min\left\{\sqrt[3]{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}-\zeta^3_{a_{ij}}\zeta^3_{b_{ij}}},\sqrt[3]{\zeta_{a_{ij}}\zeta_{b_{ij}}}\right\}, \max\left\{\delta_{a_{ij}}\delta_{b_{ij}},\sqrt[3]{\delta_{a_{ij}}\delta_{b_{ij}}}\right\}\right]\\ &=\left[\left\langle \sqrt[3]{\zeta_{a_{ij}}\zeta_{b_{ij}}},\sqrt[3]{\delta_{a_{ij}}\delta_{b_{ij}}}\right\rangle\right]\\ &=A\$ B,\end{align*} and \begin{align*} \left((A \$ B)\rightarrow(A \boxplus_{F} B)^C\right)^C &=\left[x, \min\left\{\sqrt[3]{\zeta_{a_{ij}}\zeta_{b_{ij}}},\sqrt[3]{\zeta^3_{a_{ij}}+\zeta^3_{a_{ij}}-\zeta^3_{a_{ij}}\zeta^3_{a_{ij}}}\right\}, \max\left\{\sqrt[3]{\delta_{a_{ij}}\delta_{b_{ij}}},\delta_{a_{ij}}\delta_{b_{ij}}\right\}\right]\\ &=\left[\left\langle \sqrt[3]{\zeta_{a_{ij}}\zeta_{b_{ij}}},\sqrt[3]{\delta_{a_{ij}}\delta_{b_{ij}}}\right\rangle\right]\\ &=A\$ B.\end{align*} So, \[\left((A\boxplus_{F} B)\rightarrow(A \$ B)^C\right)^C=\left((A \$ B)\rightarrow(A \boxplus_{F} B)^C\right)^C=(A \$ B).\]
• (xi) Since \begin{align*} \left((A\boxtimes_{F} B)\rightarrow(A \$ B)^C\right)^C &=\left[\min\left\{\zeta_{a_{ij}}\zeta_{b_{ij}},\sqrt[3]{\zeta_{a_{ij}}\zeta_{b_{ij}}}\right\}, \max\left\{\sqrt[3]{\delta^3_{a_{ij}}+\delta^3_{a_{ij}}-\delta^3_{a_{ij}}\delta^3_{a_{ij}}},\sqrt[3]{\delta_{a_{ij}}\delta_{b_{ij}}}\right\}~\right]\\ &=\left[\left\langle \zeta_{a_{ij}}\zeta_{b_{ij}},\sqrt[3]{\delta^3_{a_{ij}}+\delta^3_{a_{ij}}-\delta^3_{a_{ij}}\delta^3_{a_{ij}}}\right\rangle\right]\\ &=A\boxtimes_{F} B,\end{align*} and \begin{align*} \left((A \$ B)\rightarrow(A \boxtimes_{F} B)^C\right)^C &=\left[\min\left\{\sqrt[3]{\zeta_{a_{ij}}\zeta_{b_{ij}}},\zeta_{a_{ij}}\zeta_{b_{ij}}\right\}, \max\left\{\sqrt[3]{\delta_{a_{ij}}\delta_{b_{ij}}},\sqrt[3]{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}-\delta^3_{a_{ij}}\delta^3_{b_{ij}}}\right\}\right]\\ &=\left[\left\langle \zeta_{a_{ij}}\zeta_{b_{ij}},\sqrt[3]{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}-\delta^3_{a_{ij}}\delta^3_{b_{ij}}}\right\rangle\right]\\ &=A\boxtimes_{F} B.\end{align*} So, \[\left((A\boxtimes_{F} B)\rightarrow(A \$ B)^C\right)^C=\left((A \$ B)\rightarrow(A \boxtimes_{F} B)^C\right)^C=(A \boxtimes_{F} B).\]
• (xiii) Since \begin{align*} \left((A\boxplus_{F} B)\rightarrow(A \boxtimes_{F} B)^C\right)^C &=\left[ \min\left\{\sqrt[3]{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}-\zeta^3_{a_{ij}}\zeta^3_{b_{ij}}},\zeta_{a_{ij}}\zeta_{b_{ij}}\right\}, \max\left\{\delta_{a_{ij}}\delta_{b_{ij}},\sqrt[3]{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}-\delta^3_{a_{ij}}\delta^3_{b_{ij}}}\right\}~\right]\\ &=\left[\left\langle \zeta_{a_{ij}}\zeta_{b_{ij}},\sqrt[3]{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}-\delta^3_{a_{ij}}\delta^3_{b_{ij}}}\right\rangle\right]\\ &=A\boxtimes_{F} B,\end{align*} and \begin{align*} \left((A \boxtimes_{F} B)\rightarrow(A \boxplus_{F} B)^C\right)^C &=\left[ \min\left\{\zeta_{a_{ij}}\zeta_{b_{ij}},\sqrt[3]{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}-\zeta^3_{a_{ij}}\zeta^3_{b_{ij}}}\right\}, \max\left\{\sqrt[3]{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}-\delta^3_{a_{ij}}\delta^3_{b_{ij}}},\delta_{a_{ij}}\delta_{b_{ij}}\right\}~\right]\\ &=\left[\left\langle \zeta_{a_{ij}}\zeta_{b_{ij}},\sqrt[3]{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}-\delta^3_{a_{ij}}\delta^3_{b_{ij}}}\right\rangle\right]\\ &=A\boxtimes_{F} B.\end{align*} So, \[\left((A\boxplus_{F} B)\rightarrow(A \boxtimes_{F} B)^C\right)^C=\left((A \boxtimes_{F} B)\rightarrow(A \boxplus_{F} B)^C\right)^C=(A \boxtimes_{F} B).\]

The proof of the following Corollaries follows from Theorem 9.

Corollary 10. For \(A, B \in F_{m \times n},\) we have \(\left((A\boxtimes_{F} B)\rightarrow(A @ B)^C\right)^C=\left((A @ B)\rightarrow(A \boxtimes_{F} B)^C\right)^C= \left((A\boxtimes_{F} B)\rightarrow(A \# B)^C\right)^C=\) \(\left((A \# B)\rightarrow(A \boxtimes_{F} B)^C\right)^C=\left((A\boxtimes_{F} B)\rightarrow(A \$ B)^C\right)^C=\left((A \$ B)\rightarrow(A \boxtimes_{F} B)^C\right)^C =\left((A\boxplus_{F} B)\rightarrow(A \boxtimes_{F} B)^C\right)^C=\left((A \boxtimes_{F} B)\rightarrow(A \boxplus_{F} B)^C\right)^C =(A \boxtimes_{F} B)\).

Corollary 11. For \(A, B \in F_{m \times n},\) we have \(\left((A\boxplus_{F} B)^C\rightarrow (A @ B)\right)=\left((A @ B)^C\rightarrow (A \boxplus_{F} B)\right) =\left((A\boxplus_{F} B)^C\rightarrow (A \# B)\right)=\)\(\left((A \# B)^C\rightarrow (A \boxplus_{F} B)\right) =\left((A\boxplus_{F} B)^C\rightarrow (A \$ B)\right)=\left((A \$ B)^C\rightarrow (A \boxplus_{F} B)\right) =\left((A\boxtimes_{F} B)^C\rightarrow (A \boxplus_{F} B)\right)=\left((A \boxplus_{F} B)^C\rightarrow (A \boxtimes_{F} B)\right) =(A \boxplus_{F} B)\).

Theorem 12. For \(A, B \in F_{m \times n},\) we have \(\left[\left(A^C\rightarrow B\right)\boxplus_{F}\left(A\rightarrow B^C\right)^C\right]@ \left[\left(A^C\rightarrow B\right)\boxtimes_{F}\left(A\rightarrow B^C\right)^C\right]=\left(A @ B\right)\).

Proof. Since \[\left[\left(A^C\rightarrow B\right)\boxplus_{F}\left(A\rightarrow B^C\right)^C\right]=\left[\left\langle \sqrt[3]{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}-\zeta^3_{a_{ij}}\zeta^3_{b_{ij}}},\delta_{a_{ij}}\delta_{b_{ij}}\right\rangle\right],\] and \[\left[\left(A^C\rightarrow B\right)\boxtimes_{F}\left(A\rightarrow B^C\right)^C\right]=\left[\left\langle \zeta_{a_{ij}}\zeta_{b_{ij}},\sqrt[3]{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}-\delta^3_{a_{ij}}\delta^3_{b_{ij}}}\right\rangle\right].\] So, \begin{align*} \left[\left(A^C\rightarrow B\right)\right.&\left.\boxplus_{F}\left(A\rightarrow B^C\right)^C\right]@ \left[\left(A^C\rightarrow B\right)\boxtimes_{F}\left(A\rightarrow B^C\right)^C\right]\\ &=\left[ \sqrt[3]{\dfrac{\left(\sqrt[3]{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}-\zeta^3_{a_{ij}}\zeta^3_{b_{ij}}}\right)^3+\zeta^3_{a_{ij}}\zeta^3_{b_{ij}}}{2}}, \sqrt[3]{\dfrac{\delta^3_{a_{ij}}\delta^3_{b_{ij}}+\left(\sqrt[3]{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}-\delta^3_{a_{ij}}\delta^3_{b_{ij}}}\right)^3}{2}}~\right]\\ &=\left[\left\langle \sqrt[3]{\dfrac{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}}{2}},\sqrt[3]{\dfrac{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}}{2}}\right\rangle\right]\\ &=(A @ B).\end{align*}

Theorem 13. For \(A, B \in F_{m \times n},\) we have\begin{align*} \left[\left(\left(A^C\rightarrow B\right)\right.\right.&\left.\left.\boxplus_{F}\left(A\rightarrow B^C\right)^C\right)\bigcap \left(\left(A^C\rightarrow B\right)\boxtimes_{F}\left(A\rightarrow B^C\right)^C\right)\right]\\ &@ \left[\left(\left(A^C\rightarrow B\right)\boxplus_{F}\left(A\rightarrow B^C\right)^C\right)\bigcup \left(\left(A^C\rightarrow B\right)\boxtimes_{F}\left(A\rightarrow B^C\right)^C\right)\right]=(A @ B).\end{align*}

Proof. Since \begin{align*}\left[\left(\left(A^C\rightarrow B\right)\right.\right.&\left.\left.\boxplus_{F}\left(A\rightarrow B^C\right)^C\right)\bigcap \left(\left(A^C\rightarrow B\right)\boxtimes_{F}\left(A\rightarrow B^C\right)^C\right)\right]\\ &=\left[ \min\left\{\sqrt[3]{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}-\zeta^3_{a_{ij}}\zeta^3_{b_{ij}}},\zeta_{a_{ij}}\zeta_{b_{ij}}\right\}, \max\left\{\delta_{a_{ij}}\delta_{a_{ij}},\sqrt[3]{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}-\delta^3_{a_{ij}}\delta^3_{b_{ij}}}\right\}~\right]\\ &=\left[\left\langle \zeta_{a_{ij}}\zeta_{b_{ij}},\sqrt[3]{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}-\delta^3_{a_{ij}}\delta^3_{b_{ij}}}\right\rangle\right],\end{align*} and \begin{align*} \left[\left(\left(A^C\rightarrow B\right)\right.\right.&\left.\left.\boxplus_{F}\left(A\rightarrow B^C\right)^C\right)\bigcup \left(\left(A^C\rightarrow B\right)\boxtimes_{F}\left(A\rightarrow B^C\right)^C\right)\right]\\ &=\left[ \max\left\{\sqrt[3]{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}-\zeta^3_{a_{ij}}\zeta^3_{b_{ij}}},\zeta_{a_{ij}}\zeta_{b_{ij}}\right\}, \min\left\{\delta_{a_{ij}}\delta_{b_{ij}},\sqrt[3]{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}-\delta^3_{a_{ij}}\delta^3_{b_{ij}}}\right\}~\right]\\ &=\left[\left\langle \sqrt[3]{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}-\zeta^3_{a_{ij}}\zeta^3_{b_{ij}}},\delta_{a_{ij}}\delta_{b_{ij}}\right\rangle\right].\end{align*} Hence \begin{align*} \left[\left(\left(A^C\rightarrow B\right)\right.\right.&\left.\left.\boxplus_{F}\left(A\rightarrow B^C\right)^C\right)\bigcap \left(\left(A^C\rightarrow B\right)\boxtimes_{F}\left(A\rightarrow B^C\right)^C\right)\right]\\ &@ \left[\left(\left(A^C\rightarrow B\right)\boxplus_{F}\left(A\rightarrow B^C\right)^C\right)\bigcup \left(\left(A^C\rightarrow B\right)\boxtimes_{F}\left(A\rightarrow B^C\right)^C\right)\right]\\ =&\left[ \sqrt[3]{\dfrac{\left(\sqrt[3]{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}-\zeta^3_{a_{ij}}\zeta^3_{b_{ij}}}\right)^3+\zeta^3_{a_{ij}}\zeta^3_{b_{ij}}}{2}}, \sqrt[3]{\dfrac{\delta^3_{a_{ij}}\delta^3_{b_{ij}}+\left(\sqrt[3]{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}-\delta^3_{a_{ij}}\delta^3_{b_{ij}}}\right)^3}{2}}\right]\\ =&\left[\left\langle \sqrt[3]{\dfrac{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}}{2}},\sqrt[3]{\dfrac{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}}{2}}\right\rangle\right]\\ =&A @ B.\end{align*}

Theorem 14. For \(A, B \in F_{m \times n},\) we have \(\left[\left((A\boxplus_{F} B)\rightarrow(A @ B)^C\right)^C \bigcup \left((A\boxtimes_{F} B)\rightarrow(A @ B)^C\right)^C\right] \bigcup \left[\left((A\boxplus_{F} B)\rightarrow(A @ B)^C\right)^C \bigcap \left((A\boxtimes_{F} B)\rightarrow(A @ B)^C\right)^C\right] =A @ B.\)

Proof. From Theorem 9, we have \[\left((A\boxplus_{F} B)\rightarrow(A @ B)^C\right)^C=\left[\left\langle \sqrt[3]{\dfrac{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}}{2}},\sqrt[3]{\dfrac{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}}{2}}\right\rangle\right],\] and \[\left((A\boxtimes_{F} B)\rightarrow(A @ B)^C\right)^C=\left[\left\langle \zeta_{a_{ij}}\zeta_{b_{ij}},\sqrt[3]{\delta^3_{a_{ij}}+\delta^3_{a_{ij}}-\delta^3_{a_{ij}}\delta^3_{a_{ij}}}\right\rangle\right].\] So, \begin{align*} \left[\left((A\boxplus_{F} B)\right.\right.&\left.\left.\rightarrow(A @ B)^C\right)^C \bigcup \left((A\boxtimes_{F} B)\rightarrow(A @ B)^C\right)^C\right]\\ &=\left[ \max\left\{\sqrt[3]{\dfrac{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}}{2}},(\zeta_{a_{ij}}\zeta_{b_{ij}})\right\}, \min\left\{\sqrt[3]{\dfrac{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}}{2}},\sqrt[3]{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}-\delta^3_{a_{ij}}\delta^3_{b_{ij}}}\right\}\right]\\ &=\left[\left\langle \sqrt[3]{\dfrac{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}}{2}},\sqrt[3]{\dfrac{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}}{2}}\right\rangle\right],\end{align*} and \begin{align*} \left[\left((A\boxplus_{F} B)\right.\right.&\left.\left.rightarrow(A @ B)^C\right)^C \bigcap \left((A\boxtimes_{F} B)\rightarrow(A @ B)^C\right)^C\right]\\ &=\left[ \min\left\{\sqrt[3]{\dfrac{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}}{2}},(\zeta_{a_{ij}}\zeta_{b_{ij}})\right\}, \max\left\{\sqrt[3]{\dfrac{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}}{2}},\sqrt[3]{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}-\delta^3_{a_{ij}}\delta^3_{b_{ij}}}\right\}\right]\\ &=\left[\left\langle \zeta_{a_{ij}}\zeta_{b_{ij}},\sqrt[3]{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}-\delta^3_{a_{ij}}\delta^3_{b_{ij}}}\right\rangle\right].\end{align*} Hence \begin{align*}&\left[\left((A\boxplus_{F} B)\rightarrow(A @ B)^C\right)^C \bigcup \left((A\boxtimes_{F} B)\rightarrow(A @ B)^C\right)^C\right] \left[\left((A\boxplus_{F} B)\rightarrow(A @ B)^C\right)^C \bigcap \left((A\boxtimes_{F} B)\rightarrow(A @ B)^C\right)^C\right]\\ &=\left[ \max\left\{\sqrt[3]{\dfrac{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}}{2}},(\zeta_{a_{ij}}\zeta_{b_{ij}})\right\}, \min\left\{\sqrt[3]{\dfrac{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}}{2}},\sqrt[3]{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}-\delta^3_{a_{ij}}\delta^3_{b_{ij}}}\right\}\right]\\ &=\left[\left\langle \sqrt[3]{\dfrac{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}}{2}},\sqrt[3]{\dfrac{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}}{2}}\right\rangle\right]=A @ B.\end{align*}

Theorem 15. For \(A, B \in F_{m \times n},\) we have \(\left[\left((A\boxplus_{F} B)\rightarrow(A @ B)^C\right)^C \bigcup \left((A\boxtimes_{F} B)\rightarrow(A @ B)^C\right)^C\right] \bigcap \left[\left((A\boxplus_{F} B)\rightarrow(A @ B)^C\right)^C \bigcap \left((A\boxtimes_{F} B)\rightarrow(A @ B)^C\right)^C\right]=A \boxtimes_{F} B.\)

Proof. The proof is similar to that of Theorem 14.

Theorem 16. For \(A, B \in F_{m \times n},\) we have \(\left((A \boxplus_{F} B)^C\rightarrow (A @ B)\right)@\left((A \boxtimes_{F} B)\rightarrow (A @ B)^C\right)^C=(A @ B)\).

Proof. Since \[\left((A \boxplus_{F} B)^C\rightarrow (A @ B)\right)=\left[\left\langle \sqrt[3]{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}-\zeta^3_{a_{ij}}\zeta^3_{b_{ij}}},\delta_{a_{ij}}\delta_{b_{ij}}\right\rangle\right],\] and \[\left((A \boxtimes_{F} B)\rightarrow (A @ B)^C\right)^C=\left[\left\langle \zeta_{a_{ij}}\zeta_{b_{ij}},\sqrt[3]{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}-\delta^3_{a_{ij}}\delta^3_{b_{ij}}}\right\rangle\right].\] So, \begin{align*} \left((A \boxplus_{F} B)^C\right.&\left.\rightarrow (A @ B)\right)@\left((A \boxtimes_{F} B)\rightarrow (A @ B)^C\right)^C\\ &=\left[ \sqrt[3]{\dfrac{\left(\sqrt[3]{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}-\zeta^3_{a_{ij}}\zeta^3_{b_{ij}}}\right)^3+(\zeta_{a_{ij}}\zeta_{b_{ij}})^3}{2}}, \sqrt[3]{\dfrac{(\delta_{a_{ij}}\delta_{b_{ij}})^3+\left(\sqrt[3]{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}-\delta^3_{a_{ij}}\delta^3_{b_{ij}}}\right)^3}{2}}~\right]\end{align*} \begin{align*} &=\left[\left\langle \sqrt[3]{\dfrac{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}}{2}},\sqrt[3]{\dfrac{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}}{2}}\right\rangle\right]=A @ B.\end{align*}

Theorem 17. For \(A, B \in F_{m \times n},\) we have \(\left((A \boxplus_{F} B)^C\rightarrow (A \# B)\right)@\left((A \boxtimes_{F} B)\rightarrow (A \# B)^C\right)^C=(A @ B)\).

Proof. Since \[\left((A \boxplus_{F} B)^C\rightarrow (A \# B)\right)=\left[\left\langle \sqrt[3]{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}-\zeta^3_{a_{ij}}\zeta^3_{b_{ij}}},\delta_{a_{ij}}\delta_{b_{ij}}\right\rangle\right],\] and \[\left((A \boxtimes_{F} B)\rightarrow (A \# B)^C\right)^C=\left[\left\langle \zeta_{a_{ij}}\zeta_{b_{ij}},\sqrt[3]{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}-\delta^3_{a_{ij}}\delta^3_{b_{ij}}}\right\rangle\right].\] So, \begin{align*} \left((A \boxplus_{F} B)^C\right.&\left.\rightarrow (A \# B)\right)@\left((A \boxtimes_{F} B)\rightarrow (A \# B)^C\right)^C\\ &=\left[ \sqrt[3]{\dfrac{\left(\sqrt[3]{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}-\zeta^3_{a_{ij}}\zeta^3_{b_{ij}}}\right)^3+(\zeta_{a_{ij}}\zeta_{b_{ij}})^3}{2}}, \sqrt[3]{\dfrac{(\delta_{a_{ij}}\delta_{b_{ij}})^3+\left(\sqrt[3]{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}-\delta^3_{a_{ij}}\delta^3_{b_{ij}}}\right)^3}{2}}~\right]\\ &=\left[\left\langle \sqrt[3]{\dfrac{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}}{2}},\sqrt[3]{\dfrac{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}}{2}}\right\rangle\right]\\ &=A @ B.\end{align*}

Theorem 18. For \(A, B \in F_{m \times n},\) we have \(\left((A \boxplus_{F} B)^C\rightarrow (A \$ B)\right)@\left((A \boxtimes_{F} B)\rightarrow (A \$ B)^C\right)^C=(A @ B)\).

Proof. Since \[\left((A \boxplus_{F} B)^C\rightarrow (A \$ B)\right)=\left[\left\langle \sqrt[3]{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}-\zeta^3_{a_{ij}}\zeta^3_{b_{ij}}},\delta_{a_{ij}}\delta_{b_{ij}}\right\rangle\right],\] and \[\left((A \boxtimes_{F} B)\rightarrow (A \$ B)^C\right)^C=\left[\left\langle \zeta_{a_{ij}}\zeta_{b_{ij}},\sqrt[3]{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}-\delta^3_{a_{ij}}\delta^3_{b_{ij}}}\right\rangle\right].\] So, \begin{align*}\left((A \boxplus_{F} B)^C\right.&\left.\rightarrow (A \$ B)\right)@\left((A \boxtimes_{F} B)\rightarrow (A \$ B)^C\right)^C\\ &=\left[ \sqrt[3]{\dfrac{\left(\sqrt[3]{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}-\zeta^3_{a_{ij}}\zeta^3_{b_{ij}}}\right)^3+(\zeta_{a_{ij}}\zeta_{b_{ij}})^3}{2}}, \sqrt[3]{\dfrac{(\delta_{a_{ij}}\delta_{b_{ij}})^3+\left(\sqrt[3]{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}-\delta^3_{a_{ij}}\delta^3_{b_{ij}}}\right)^3}{2}}~\right]\\ &=\left[\left\langle \sqrt[3]{\dfrac{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}}{2}},\sqrt[3]{\dfrac{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}}{2}}\right\rangle\right]\\ &=A @ B.\end{align*}

Theorem 19. For \(A, B \in F_{m \times n},\) we have \(\left((A \boxtimes_{F} B)^C\rightarrow (A \boxplus_{F} B)\right)@\left((A \boxplus_{F} B)\rightarrow (A \boxtimes_{F} B)^C\right)^C=(A @ B)\).

Proof. Since \[ \left((A \boxtimes_{F} B)^C\rightarrow (A \boxplus_{F} B)\right)=\left[\left\langle \sqrt[3]{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}-\zeta^3_{a_{ij}}\zeta^3_{b_{ij}}},\delta_{a_{ij}}\delta_{b_{ij}}\right\rangle\right],\] and \[ \left((A \boxplus_{F} B)\rightarrow (A \boxtimes_{F} B)^C\right)^C=\left[\left\langle \zeta_{a_{ij}}\zeta_{b_{ij}},\sqrt[3]{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}-\delta^3_{a_{ij}}\delta^3_{b_{ij}}}\right\rangle\right].\] So, \begin{align*} \left((A \boxtimes_{F} B)^C\right.&\left.\rightarrow (A \boxplus_{F} B)\right)\left((A \boxplus_{F} B)\rightarrow (A \boxtimes_{F} B)^C\right)^C\\ &=\left[ \sqrt[3]{\dfrac{\left(\sqrt[3]{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}-\zeta^3_{a_{ij}}\zeta^3_{b_{ij}}}\right)^3+(\zeta_{a_{ij}}\zeta_{b_{ij}})^3}{2}}, \sqrt[3]{\dfrac{(\delta_{a_{ij}}\delta_{b_{ij}})^3+\left(\sqrt[3]{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}-\delta^3_{a_{ij}}\delta^3_{b_{ij}}}\right)^3}{2}}~\right]\\ &=\left[\left\langle \sqrt[3]{\dfrac{\zeta^3_{a_{ij}}+\zeta^3_{b_{ij}}}{2}},\sqrt[3]{\dfrac{\delta^3_{a_{ij}}+\delta^3_{b_{ij}}}{2}}\right\rangle\right]\\ &=A @ B.\end{align*}

6. Conclusion and future scope

In this paper, some new operators \([(A \$ B),(A \# B),(A\ast B),(A\rightarrow B)]\) of Fermatean fuzzy matrices are defined and investigated their several algebraic properties. Further, the necessity and possibility operators of Fermatean fuzzy matrices are proved. Finally, we have identified and proved several of these properties, particularly those involving the operator \(A\rightarrow B\) defined as Fermatean fuzzy implication with other operators. In further research, we may apply these operators in the field of different areas, for example, dynamic decision and consensus , business and marketing management, design, engineering and manufacturing, information technology and networking applications, human resources management, military applications, energy management, geographic information system applications etc.

Conflicts of Interest

The author declares no conflict of interest.