# A Classification Model Based on Interval Rule Inference Network with Interpretability
## 4 Interval Rule Inference Network
### Parameter Determination
#### 4.3.1 The Theoretical Foundations of IRIN  
The traditional gradient descent algorithm is adopted to adjust the parameters in IRIN. In order to update the parameters by partial derivation in different directions for different parameters, it is necessary to prove the differentiability of the approximate reasoning algorithm based on IER in IRIN. **Theorem 1**. : _If \(\sum\limits_{j=1}^{I_{D}}\prod\limits_{k=1}^{L}\beta_{j,k}^{+}\
eq 0\), \(\sum\limits_{j=1}^{I_{D}}\prod\limits_{k=1}^{L}\beta_{j,k}^{-}\
eq 0\) and \(0<w_{k}<1,0\leq\beta_{j,k}^{-}\leq\beta_{j,k}^{+}\leq 1\), the partial derivatives of the overall combined degree \(y_{i}\) in Equations (25)-(26) to \(\beta_{j,k}^{-},\beta_{j,k}^{+}\) and \(w_{k}\) exist._  
**Proof of Theorem 1. ** Let  
\[f_{j}=\left[f_{j}^{-},f_{j}^{+}\right]=\left[\prod\limits_{k=1}^{L}\left(w_{k} \beta_{j,k}^{-}+1-w_{k}\right),\prod\limits_{k=1}^{L}\left(w_{k}\beta_{j,k}^{ +}+1-w_{k}\right)\right] \tag{31}\]  
\[g=\prod\limits_{k=1}^{L}\left(1-w_{k}\right) \tag{32}\]  
It can be seen from the differentiability theorem of functions that the partial differentiable of \(f_{j}\) with respect to \(\beta_{j,k}^{-},\beta_{j,k}^{+}\) and \(w_{k}\) exist, and g is partially differentiable to \(w_{k}\). Let  
\[ \begin{array}{l}\mu_{0}=[u_{0}^{-},u_{0}^{+}]=[\sum\limits_{j=1}^{I_{D}}\prod \limits_{k=1}^{L}\left(w_{k}\beta_{j,k}^{-}+1-w_{k}\right)-\left(J_{D}-1 \right)\prod\limits_{k=1}^{L}\left(1-w_{k}\right),\\ \sum\limits_{j=1}^{I_{D}}\prod\limits_{k=1}^{L}\left(w_{k}\beta_{j,k}^{+}+1-w_ {k}\right)-\left(J_{D}-1\right)\prod\limits_{k=1}^{L}\left(1-w_{k}\right)] \end{array}  \tag{33}\]  
Figure 2: Flowchart of IRIN parameter determination. since,  
\[ \begin{array}{l}\mu_{0}^{-}=\sum\limits_{j=1}^{I_{D}}\prod\limits_{k=1}^{L} \left(w_{k}\beta_{j,k}^{-}+1-w_{k}\right)-\left(J_{D}-1\right)\prod\limits_{k=1 }^{L}\left(1-w_{k}\right)\\ >\mu_{0}^{-}=\sum\limits_{j=1}^{I_{D}}\prod\limits_{k=1}^{L}\left(1-w_{k} \right)-\left(J_{D}-1\right)\prod\limits_{k=1}^{L}\left(1-w_{k}\right)\\ =J_{D}\sum\limits_{j=1}^{I_{D}}\left(1-w_{k}\right)-\left(J_{D}-1\right)\prod \limits_{k=1}^{L}\left(1-w_{k}\right)\\ =\prod\limits_{k=1}^{L}\left(1-w_{k}\right)>0\end{array}  \tag{34}\]  
\[ \begin{array}{l}\mu_{0}^{+}=\sum\limits_{j=1}^{I_{D}}\prod\limits_{k=1}^{L} \left(w_{k}\beta_{j,k}^{+}+1-w_{k}\right)-\left(J_{D}-1\right)\prod\limits_{k =1}^{L}\left(1-w_{k}\right)\\ >\mu_{0}^{+}=\sum\limits_{j=1}^{I_{D}}\prod\limits_{k=1}^{L}\left(1-w_{k} \right)-\left(J_{D}-1\right)\prod\limits_{k=1}^{L}\left(1-w_{k}\right)\\ =J_{D}\sum\limits_{j=1}^{I_{D}}\left(1-w_{k}\right)-\left(J_{D}-1\right)\prod \limits_{k=1}^{L}\left(1-w_{k}\right)\\ =\prod\limits_{k=1}^{L}\left(1-w_{k}\right)>0\end{array}  \tag{35}\]  
Hence, \(\mu_{0}^{-}\) and \(\mu_{0}^{+}\) are partially differentiable with respect to \(\beta_{j,k}^{-}\), \(\beta_{j,k}^{+}\), and \(w_{k}\). And since \(\mu=\frac{1}{\mu_{0}}\), \(\mu\) is partially differentiable with respect to \(\beta_{j,k}^{-}\), \(\beta_{j,k}^{+}\), and \(w_{k}\). Hence, \(y_{j}=\frac{\mu\cdot[\hat{f}_{j}-\hat{g}]}{1-\mu\cdot g}\) is partially differentiable with respect to \(\beta_{j,k}^{-}\), \(\beta_{j,k}^{+}\), and \(w_{k}\). **Theorem 2**. : _If \(\sum_{k=1}^{L}(\theta_{k}\times\prod_{i=1}^{T_{k}}\left(s_{i}^{k}\right)^{\bar {\delta}_{i}})\
eq 0\), the partial derivatives of activation weight \(w_{k}\) in Equation (5) to rule weight \(\theta_{k}\) and attribute weight \(\delta_{i}\) exist._  
Proof of Theorem 2.: Let  
\[z=\prod_{i=1}^{T_{k}}\left(s_{i}^{k}\right)^{\bar{\delta}_{i}} \tag{36}\]  
\(z\) is an exponential function about \(\bar{\delta}_{i}\), so \(z\) is a partial differentiable to. \[w_{k}=\frac{\theta_{k}\cdot z}{\sum_{j=1}^{L}(\theta_{k}\cdot z)} \tag{37}\]  
It is obvious that \(\theta_{k},z\) is differentiable and \(\sum_{k=1}^{L}(\theta_{k}\cdot z)\
eq 0\). Hence, \(w_{k}\) is a partially differentiable to \(\bar{\delta}_{i},z\). From above, \(w_{k}\) is partially differentiable to \(\bar{\delta}_{i},\theta_{k}\).