3.11 super-vertices are suspicions about choosing them

3. Neutrosophic Hypergraphs

Proposition 3.28.5. Assume G is a family of n-SuperHyperGraph. Then Gn \ {Xn} is super-resolving set for al l members of G, simultaneously.

Proof. Suppose G is a family of n-SuperHyperGraph. Thus Gn \ {Xn} is superresolving set for every given n-SuperHyperGraph of G. One super-vertex is out of Gn \ {Xn}. It’s super-resolved from any super-vertex in Gn \ {Xn}. Hence every given two super-vertices are super-resolved from any super-vertex in Gn \ {Xn}. It implies Gn \ {Xn} is super-resolving set for all members of G, simultaneously. Proposition 3.28.6. Assume G is a family of n-SuperHyperGraph. Then Gn \ {Xn} isn’t super-coloring set for al l members of G, simultaneously.

Proof. Suppose G is a family of n-SuperHyperGraph. Thus Gn \ {Xn} isn’t super-coloring set for every given n-SuperHyperGraph of G. One super-vertex is out of Gn \ {Xn}. It isn’t super-colored from any super-vertex in Gn \ {Xn}. Hence every given two super-vertices aren’t super-colored from any super-vertex in Gn \ {Xn}. It implies Gn \ {Xn} isn’t super-coloring set for all members of G, simultaneously. Proposition 3.28.7. Assume G is a family of n-SuperHyperGraph. Then union of super-dominating sets from each member of G is super-dominating set for al l members of G, simultaneously.

Proof. Suppose G is a family of n-SuperHyperGraph. For every chosen nSuperHyperGraph, there’s one super-dominating set in the union of superdominating sets from each member of G. Thus union of super-dominating sets from each member of G is super-dominating set for every given nSuperHyperGraph of G. Even one super-vertex isn’t out of the union. It’s super-dominated from any super-vertex in the union. Hence every given two super-vertices are super-dominated from any super-vertex in union of supercoloring sets. It implies union of super-coloring sets is super-dominating set for all members of G, simultaneously. Proposition 3.28.8. Assume G is a family of n-SuperHyperGraph. Then union of super-resolving sets from each member of G is super-resolving set for al l members of G, simultaneously.

Proof. Suppose G is a family of n-SuperHyperGraph. For every chosen nSuperHyperGraph, there’s one super-resolving set in the union of super-resolving sets from each member of G. Thus union of super-resolving sets from each member of G is super-resolving set for every given n-SuperHyperGraph of G. Even one super-vertex isn’t out of the union. It’s super-resolved from any supervertex in the union. Hence every given two super-vertices are super-resolved from any super-vertex in union of super-coloring sets. It implies union of supercoloring sets is super-resolved set for all members of G, simultaneously. Proposition 3.28.9. Assume G is a family of n-SuperHyperGraph. Then union of super-coloring sets from each member of G is super-coloring set for al l members of G, simultaneously.

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