## 2 Preliminaries
### Determination of the Fuzzification Coefficients for Supervised Elements
Therefore, \(f^{\prime}(x)>0\), or \(f(x)\) is an increasing function, for all \(x>1\). \(\Box\)  
Applying Proposition 2 to the following formula used to calculate \(M^{\prime}\) for the supervised elements, \(f(M^{\prime})=\left(\frac{1}{M^{\prime}d_{k}a^{2}}\right)^{\frac{1}{M^{\prime }-1}}\), where \({d_{ik}}^{2}\geq 1\), \(M^{\prime}>1\), we see that this function is an increasing function. Therefore, as \(M^{\prime}\) increases, the membership grade \(U_{ik}\) increases. Setting the supervision for an element corresponds to selecting a larger fuzzification coefficient \(M^{\prime}\) for that element compared to other fuzzification coefficients. Then, Proposition 3 is utilized to guide the determination of parameter \(M^{\prime}\). **Proposition 3. **_Let \(U_{ik}\) be the membership grade of element \(X_{i}\) in its semi-supervise placement into cluster \(k\) according to Equations (17)-(20) with a given parameter \(M\) and a to-be-determined parameter \(M^{\prime}\). Let \(U^{\prime}_{ik}\) be the membership grade of element \(X_{i}\) in its unsupervised placement into cluster \(k\) according to Equation (15). With \(\alpha\in(0,1)\), we have \(U_{ik}\geq\alpha\) if the following equation is satisfied:_  
\[M^{\prime}\alpha^{M^{\prime}-1}\leq M\left(\frac{1-\alpha}{\frac{1}{U^{\prime }_{ik}}-1}\right)^{M-1} \tag{23}\]  
**Proof of Proposition 3. ** From Equation (15), we have  
\[U^{\prime}_{ik}=\frac{1}{\sum_{j=1}^{C}\left(\frac{D_{\alpha}}{D_{ij}}\right)^ {\frac{2}{M-1}}}=\frac{1}{\left(\frac{D_{\alpha}}{D_{ik}}\right)^{\frac{2}{M-1 }}+\sum_{j=1,j\
eq k}^{C}\left(\frac{D_{\alpha}}{D_{ij}}\right)^{\frac{2}{M-1 }}}=\frac{1}{1+\sum_{j=1,j\
eq k}^{C}\left(\frac{D_{\alpha}}{D_{ij}}\right)^{ \frac{2}{M-1}}}\]  
From Equation (17), we have \(\frac{d_{ik}}{d_{ij}}=\frac{D_{\alpha}}{D_{ij}}\), hence, \(U^{\prime}_{ik}=\frac{1}{1+\sum_{j=1,j\
eq k}^{C}\left(\frac{d_{ik}}{d_{ij}} \right)^{\frac{2}{M-1}}}\). Substitute into (23), we have the equivalent  
\[M^{\prime}\alpha^{M^{\prime}-1}\leq\frac{M(1-\alpha)^{M-1}}{\left(\sum_{j=1,j \
eq k}^{C}\left(\frac{d_{ik}}{d_{ij}}\right)^{\frac{2}{M-1}}\right)^{M-1}}= \frac{(1-\alpha)^{M-1}}{{d_{ik}}^{2}\left(\sum_{j=1,j\
eq k}^{C}\left(\frac{1}{ Md_{j}^{2}}\right)^{\frac{1}{M-1}}\right)^{M-1}}\]  
or,  
\[\frac{1}{M^{\prime}{d_{ik}}^{2}}\geq\frac{\alpha^{M^{\prime}-1}}{(1-\alpha)^{M- 1}}\left(\sum_{j=1,j\
eq k}^{C}\left(\frac{1}{M{d_{ij}}^{2}}\right)^{\frac{1}{M -1}}\right)^{M-1}\]  
Substitute in \(\mu_{ij}\) from Equation (18), we have \(\frac{1}{M^{\prime}{d_{ik}}^{2}}\geq\frac{\alpha^{M^{\prime}-1}}{(1-\alpha)^{M- 1}}\left(\sum_{j=1,j\
eq k}^{C}\mu_{ij}\right)^{M-1}\). Substitute \(\frac{1}{M^{\prime}{d_{ik}}^{2}}\) in the above with Equation (19), we have\[\frac{\mu_{ik}^{M^{\prime}-1}}{\left(\mu_{ik}+\sum_{j=1,j\
eq k}^{C}\mu_{ij}\right) ^{M^{\prime}-M}}\geq\frac{\alpha^{M^{\prime}-1}}{\left(1-\alpha\right)^{M-1}} \Bigg{(}\sum_{j=1,j\
eq k}^{C}\mu_{ij}\Bigg{)}^{M-1} \tag{24}\]  
Next, from Equation (20), we have \(U_{ik}=\frac{\mu_{ik}}{\sum_{j=1}^{C}\mu_{ij}}=\frac{\mu_{ik}}{\mu_{ik}+\sum_{ j=1,j\
eq k}^{C}\mu_{ij}}\), hence \(\sum_{j=1,j\
eq k}^{C}\mu_{ij}=\frac{\mu_{ik}}{U_{ik}}-\mu_{ik}\), which is substituted into Equation (24) to have  
\[\frac{\mu_{ik}^{M^{\prime}-1}}{\left(\frac{\mu_{ik}}{U_{ik}}\right)^{M^{\prime }-M}}\geq\frac{\alpha^{M^{\prime}-1}}{\left(1-\alpha\right)^{M-1}}\bigg{(} \frac{\mu_{ik}(1-U_{ik})}{U_{ik}}\bigg{)}^{M-1}\]  
Since \(M^{\prime}>M\), and from the above, we have \(\mu_{ik}^{M-1}U_{ik}^{M^{\prime}-M}\geq\frac{\alpha^{M^{\prime}-1}\mu_{ik}^{ M-1}}{\left(1-\alpha\right)^{M-1}}\Big{(}\frac{1-U_{ik}}{U_{ik}}\Big{)}^{M-1}\).