**Proposition 3.**


**Proof.** Proof of (i): It is derived by applying Proposition 1.

Proof of (ii): With the validity of the reproductivity property, the statement is proven. Proof of (iii): Since *Hγ*<sup>∗</sup> is a quasihypergroup and considering part (i), we have *Hγ*<sup>∗</sup> as a hypergroup. By Theorem 4, we have *xγ*∗*y* for all *x*, *y* ∈ *Hγ*<sup>∗</sup> if and only if the SBG of *G* is connected.

Proof of (iv): The statement is attained from Equation (3).

**Example 1.** *Consider* (*H*, ◦) *as a semihypergroup that is given in Table 1.*

**Table 1.** Semihypergroup (*H*, ◦)


*It is seen that* 1 ∈ 1 ◦ 2, 1 ∈ 2 ◦ 1, *then* 1*γ*1. *Furthermore, we have* 0*γ*0, 2*γ*2 *and* 0*γ*2. *The corresponding SBG of G is depicted in Figure 2. Moreover, H*/*γ*<sup>∗</sup> = {{0, 2}, 1} *and* |*H*/*γ*∗| = 1.
