**Proof.**

(1) For two T-SFNs, P and Q, and by definition of PN, we have *PN*-*nP*, *nQ*. = *<sup>t</sup>*+*ntP* + *ntQ* − *<sup>n</sup>tpntQ*. Thus, we have PN(1,1) = 1 and PN(0,0) = 0. Further, since *nP*, *nQ* ∈ [0, 1] and *t* ∈ *Z*, which implies that *ntP* + *ntQ* − *ntPntQ* = 1 − (1 − *ntP*)(<sup>1</sup> − *ntQ*) ≤ 1. Also, *PN*(*nP*, *nQ*) ≥ 0. Therefore, 0 ≤ *PN*(*nP*, *nQ*) ≤ 1.

(2) Since *nP* ≤ *nR* and *nQ* ≤ *nD*. Thus, for any *t* ∈ *Z*, we ge<sup>t</sup> 1 − *ntP* ≥ 1 − *ntR* and 1 − *ntQ* ≥ 1 − *<sup>n</sup>tD*, and hence 1 − (1 − *ntP*)(<sup>1</sup> − *ntQ*) ≤ 1 − (1 − *ntR*)(<sup>1</sup> − *<sup>n</sup>tD*). Thus, *PN*-*nP*, *nQ*. ≤ *PN*(*nR*, *nD*) holds. (3) Holds trivial. -

**Theorem 2.** *Let P* = *mP*, *iP*, *nP*, *Q* = 8*mQ*, *iQ*, *nQ*9, *R* = *mR*, *iR*, *nR and S* = *mS*, *iS*, *nS be four T-SFN. Then:*


**Proof.** Similar to Theorem 1, so we omit here.

**Theorem 3.** *If P and Q are two T-SFNs and λ* > 0 *is a real number, then P* ⊗ *Q and P<sup>λ</sup> are also T-SFNs.*

**Proof.** Follows from the definition easily, so we omit here.

**Theorem 4.** *Let P*1 = *<sup>m</sup>*, *i*, *<sup>n</sup>, P*2 = *m* , *i* , *n be a T-SFNs, λ*, *λ*1, *λ*2 > 0 *be real numbers. Then we have*

(1) *P*1 ⊗ *P*2 = *P*2 ⊗ *P*1

(2) (*<sup>P</sup>*1⊗ *<sup>P</sup>*2)*<sup>λ</sup>* = *Pλ*1⊗ *Pλ*2

(3) *Pλ*1 1⊗ *Pλ*2 1 = *Pλ*1+*λ*<sup>2</sup> 1.

**Proof.** Follows from the definition easily, so we omit here.
