# Linguistic Interval-Valued Spherical Fuzzy Soft Set and Its Application in Decision Making
## 3 Linguistic Interval-Valued Spherical Fuzzy Soft Set
### Operations on \(Livesfsss\)
: _Let \((\tilde{F},A)\) and \((\tilde{G},B)\) be two linguistic interval-valued spherical fuzzy soft sets over \(U\), then we have following properties._\(\square\square(\tilde{F},A)=\square(\tilde{F},A)\);  
\((2)\)\(\square((\tilde{F},A)\odot(\tilde{G},B))=\square(\tilde{F},A)\odot\square(\tilde{G},B)\);  
\((3)\)\(\square((\tilde{F},A)\cap(\tilde{G},B))=\square(\tilde{F},A)\odot\square(\tilde{G},B)\);  
\((4)\)\(\square((\tilde{F},A)\wedge(\tilde{G},B))=\square(\tilde{F},A)\wedge\square( \tilde{G},B)\);  
\((5)\)\(\square((\tilde{F},A)\vee(\tilde{G},B))=\square(\tilde{F},A)\vee\square( \tilde{G},B)\). Proof. : We can obtain (1) easily according to Definition 16. And we prove the other properties as follows. Suppose that, for \((\tilde{F},A)\), \(\forall\alpha\in A\), \(\tilde{F}(\alpha)=([s_{a1},s_{b1}],[s_{c1},s_{d1}],[s_{e1},s_{e1}])\) and for \((\tilde{G},B)\), \(\forall\beta\in B\), \(\tilde{G}(\beta)=([s_{a2},s_{b2}],[s_{c2},s_{d2}],[s_{e2},s_{e2}])\). (2). Suppose that \((\tilde{F},A)\odot(\tilde{G},B)=(\tilde{H},C)\), where \(C=A\cup B\) and \(\forall\epsilon\in C\),  
\[\tilde{H}(\epsilon)= \begin{cases}\tilde{F}(\epsilon)=([s_{a1},s_{b1}],[s_{c1},s_{d1}],[s_{e1},s_{e1}]),\epsilon\in A-B;\\ \tilde{G}(\epsilon)=([s_{a2},s_{b2}],[s_{c2},s_{d2}],[s_{e2},s_{e2}]),\epsilon \in B-A;\\ \tilde{F}(\epsilon)\cup G(\epsilon),\ \epsilon\in A\cap B;\end{cases} \]  
where \(\tilde{F}(\epsilon)\cup\tilde{G}(\epsilon)=([max\{s_{a1},s_{a2}\},max\{s_{b1},s_{b2}\}],[min\{s_{c1},s_{c2}\},min\{s_{d1},s_{d2}\}],\)\([min\{s_{e1},s_{e2}\},min\{s_{01},s_{e2}\}])\). Then we have \(\square((\tilde{F},A)\odot(\tilde{G},B))=\square(\tilde{H},C)\). According to Definition 16, we have  
\[\square\tilde{H}(\epsilon)= \begin{cases}\square\tilde{F}(\epsilon)=\Big{(}[s_ {a1},s_{b1}],[s_{0},s_{0}],\Big{[}s_{\sqrt{(h)^{2}-(b1)^{2}}},s_{\sqrt{(h)^{2 }-(a1)^{2}}}\Big{]}\Big{)},\epsilon\in A-B;\\ \square\tilde{G}(\epsilon)=\Big{(}[s_{a2},s_{b2}],[s_{0},s_{0}],\Big{[}s_{ \sqrt{(h)^{2}-(b2)^{2}}},s_{\sqrt{(h)^{2}-(a2)^{2}}}\Big{]}\Big{)},\epsilon \in B-A;\\ \square(\tilde{F}(\epsilon)\cup\tilde{G}(\epsilon)),\ \epsilon\in A\cap B;\end{cases} \]  
where \(\square(\tilde{F}(\epsilon)\cup\tilde{G}(\epsilon))=([max\{s_{a1},s_{a2}\}, max\{s_{b1},s_{b2}\}],[s_{0},s_{0}],[min\{s_{\sqrt{(h)^{2}-(b1)^{2}}},s_{\sqrt{(h)^{2}-(b2)^{2}}} \},\)\(min\{s_{\sqrt{(h)^{2}-(a1)^{2}}},s_{\sqrt{(h)^{2}-(a2)^{2}}}\}])\). According to Definitions 12 and 16, \(\square(F,A)\odot\square(\tilde{G},B)=(\tilde{O},C)\), \(C=A\cup B\) and \(\forall\epsilon\in C\),  
\[\tilde{O}(\epsilon)= \begin{cases}\square\tilde{F}(\epsilon)=\Big{(}[s_{a1},s_{ b1}],[s_{0},s_{0}],\Big{[}s_{\sqrt{(h)^{2}-(b1)^{2}}},s_{\sqrt{(h)^{2}-(a1)^{2}}} \Big{]}\Big{)},\epsilon\in A-B;\\ \square\tilde{G}(\epsilon)=\Big{(}[s_{a2},s_{b2}],[s_{0},s_{0}],\Big{[}s_{ \sqrt{(h)^{2}-(b2)^{2}}},s_{\sqrt{(h)^{2}-(a2)^{2}}}\Big{]}\Big{)},\epsilon \in B-A;\\ \square\tilde{F}(\epsilon)\cup\square\tilde{G}(\epsilon),\ \epsilon\in A\cap B;\end{cases} \]  
where \(\square\tilde{F}(\epsilon)\cup\square\tilde{G}(\epsilon)=([max\{s_{a1},s_{a2}\}, max\{s_{b1},s_{b2}\}],[s_{0},s_{0}],[min\{s_{\sqrt{(h)^{2}-(b1)^{2}}},s_{\sqrt{(h)^{2}-(b2)^{2}}} \}],\)\(min\{s_{\sqrt{(h)^{2}-(a1)^{2}}},s_{\sqrt{(h)^{2}-(a2)^{2}}}\}])\). Consequently, \(\square(\tilde{H},C)=(\tilde{O},C)\). Thus \(\square((\tilde{F},A)\odot(\tilde{G},B))=\square(\tilde{F},A)\odot\square( \tilde{G},B)\). (3). The proof is similar to that of (2). (4). Suppose that \((\tilde{F},A)\wedge(\tilde{G},B)=(\tilde{H},A\times B)\), where \(\forall(\alpha,\beta)\in A\times B\), \(\tilde{H}(\alpha,\beta)=\tilde{F}(\alpha)\cap\tilde{G}(\beta)=([min\{s_{a1},s_{ a2}\},min\{s_{b1},s_{b2}\}],[min\{s_{c1},s_{c2}\},min\{s_{d1},s_{a2}\}],[max\{s_{c1},s_{c2}\},max\{s_{c1},s_{c2}\},max\{s_{c1},s_{c2}\}])\). Then we have \((\tilde{F},A)\wedge(\tilde{G},B)=(\tilde{H},A\times B)\).