3.12 Vertices are suspicions about choosing them

3.28. Optimal Sets and Numbers For Family of n-SuperHyperGraph

Proof. Suppose G is a family of n-SuperHyperGraph. For every chosen nSuperHyperGraph, there’s one super-coloring set in the union of super-coloring sets from each member of G. Thus union of super-coloring sets from each member of G is super-coloring set for every given n-SuperHyperGraph of G. Even one super-vertex isn’t out of the union. It’s super-colored from any super-vertex in the union. Hence every given two super-vertices are super-colored from any super-vertex in union of super-coloring sets. It implies union of super-coloring sets is super-colored set for all members of G, simultaneously. Proposition 3.28.10. Assume G is a family of n-SuperHyperGraph. For every given super-vertex, there’s one n-SuperHyperGraph such that the super-vertex has another super-vertex which are incident to a super-edge. If for given supervertex, al l super-vertices have a common super-edge in this way, then Gn \ {Xn} is optimal-super-dominating set for al l members of G, simultaneously.

Proof. Suppose G is a family of n-SuperHyperGraph. For all nSuperHyperGraph, there’s no super-dominating set from any of member of G. Thus Gn \ {Xn} is super-dominating set for every given n-SuperHyperGraph of G. For every given super-vertex, there’s one n-SuperHyperGraph such that the super-vertex has another super-vertex which are incident to a super-edge. Only one super-vertex is out of V \ {x}. It’s super-dominated from any super-vertex in the V \ {x}. Hence every given two super-vertices are super-dominated from any super-vertex in Gn \ {Xn} It implies Gn \ {Xn} is super-dominating set for all members of G, simultaneously. If for given super-vertex, all super-vertices have a common super-edge in this way, then Gn \ {Xn} is optimal-super-dominating set for all members of G, simultaneously. Proposition 3.28.11. Assume G is a family of n-SuperHyperGraph. For every given super-vertex, there’s one n-SuperHyperGraph such that the super-vertex has another super-vertex which are incident to a super-edge. If for given supervertex, al l super-vertices have a common super-edge in this way, then Gn \ {Xn} is optimal-super-resolving set for al l members of G, simultaneously.

Proof. Suppose G is a family of n-SuperHyperGraph. For all nSuperHyperGraph, there’s no super-resolving set from any of member of G. Thus Gn \ {Xn} is super-resolving set for every given n-SuperHyperGraph of G. For every given super-vertex, there’s one n-SuperHyperGraph such that the super-vertex has another super-vertex which are incident to a super-edge. Only one super-vertex is out of Gn \ {Xn}. It’s super-resolved from any super-vertex in the Gn \ {Xn}. Hence every given two super-vertices are super-resolving from any super-vertex in Gn \ {Xn}. It implies Gn \ {Xn} is super-resolved set for all members of G, simultaneously. If for given super-vertex, all super-vertices have a common super-edge in this way, then Gn \ {Xn} is optimal-super-resolving set for all members of G, simultaneously. Proposition 3.28.12. Assume G is a family of n-SuperHyperGraph. For every given super-vertex, there’s one n-SuperHyperGraph such that the super-vertex has another super-vertex which are incident to a super-edge. If for given supervertex, al l super-vertices have a common super-edge in this way, then Gn is optimal-super-coloring set for al l members of G, simultaneously.

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