Sample Solution

MMTE-007 Solved Assignment 2024 CS

SOFT COMPUTING AND ITS APPLICATIONS

(Valid from 1st January, 2024 to 31st December, 2024)

1. a) Two sensors based upon their detection levels and gain settings are compared. The following of gain setting and sensor detection levels with a standard item being monitored provides typical membership values to represent the detection levels for each of the sensors.

4. Consider a dataset of six points given in the following table, each of which has two features $f_1$$f_1$f_(1)f_1$f_1$ and $f_2$$f_2$f_(2)f_2$f_2$. Assuming the values of the parameters $c$$c$cc$c$ and $m$$m$mm$m$ as 2 and the initial cluster centers ${\mathrm{v}}_1=(5,5)$${\mathrm{v}}_1=(5,5)$v_(1)=(5,5)\mathrm{v}_1=(5,5)${\mathrm{v}}_1=(5,5)$ and ${\mathrm{v}}_2=(10,10)$${\mathrm{v}}_2=(10,10)$v_(2)=(10,10)\mathrm{v}_2=(10,10)${\mathrm{v}}_2=(10,10)$, apply FCm algorithm to find the new cluster center after one iteration.

5. a) Define Error Correction Learning with examples.
   b) Write the types of Neural Memory Models. Also, give one example of each.
6. Consider the set of pattern vectors P. Obtain the connectivity matrix (CM) for the patterns in $P$$P$PP$P$ (four patterns).

7. a) Define Kohonen networks with examples.
   b) Describe the Function Approximation in MLP. Also, explain Generalization of MLP.
8. a) Find the length and order of the following schema:
   i) ${\mathrm{S}}_1=(1^{\ast \ast }00\ast 1^{\ast \ast })$${\mathrm{S}}_1=\left(1^{\ast \ast }00\ast 1^{\ast \ast }\right)$S_(1)=(1^(****)00**1^(****))\mathrm{S}_1=\left(1^{* *} 00 * 1^{* *}\right)${\mathrm{S}}_1=(1^{\ast \ast }00\ast 1^{\ast \ast })$
   ii) ${\mathrm{S}}_2=\left({}^{\ast }00{}^{\ast }{1}^{\ast \ast }\right)$${\mathrm{S}}_2=\left({}^{\ast }00{}^{\ast }{1}^{\ast \ast }\right)$S_(2)=(^(**)00^(**)1^(****))\mathrm{S}_2=\left({ }^* 00{ }^* 1^{* *}\right)${\mathrm{S}}_2=\left({}^{\ast }00{}^{\ast }{1}^{\ast \ast }\right)$
   iii) ${\mathrm{S}}_3=(\ast \ast \ast 1^{\ast \ast \ast })$${\mathrm{S}}_3=\left(\ast \ast \ast 1^{\ast \ast \ast }\right)$S_(3)=(******1^(******))\mathrm{S}_3=\left(* * * 1^{* * *}\right)${\mathrm{S}}_3=(\ast \ast \ast 1^{\ast \ast \ast })$
   b) Let an activation function be defined as

10. Which of the following statements are true or false? Give a short proof or a counter example in support of your answers.
    a) There is chance of occurrence of the premature convergence in Roulett-wheel selection shceme used in GA.
    b) Gradient based optimization methods are used when the objective function is not smooth and one needs efficient local optimization.
    c) The $\alpha$$\alpha$alpha\alpha$\alpha$-cut of a fuzzy set A in $\cup$$\cup$uu\cup$\cup$ is defined as $\mathrm{A}{\alpha }_0=\{\mathrm{x}\in U\mid {\mu }_{\mathrm{A}}(\mathrm{x})\le {\alpha }_0\}$$\mathrm{A}{\alpha }_0=\left\{\mathrm{x}\in U\mid {\mu }_{\mathrm{A}}(\mathrm{x})\le {\alpha }_0\right\}$Aalpha_(0)={xin U∣mu_(A)(x) <= alpha_(0)}\mathrm{A} \alpha_0=\left\{\mathrm{x} \in U \mid \mu_{\mathrm{A}}(\mathrm{x}) \leq \alpha_0\right\}$\mathrm{A}{\alpha }_0=\{\mathrm{x}\in U\mid {\mu }_{\mathrm{A}}(\mathrm{x})\le {\alpha }_0\}$
    d) A single perception with preprocessing is neither an auto associative network nor a multiple layer neural network.
    e) If $\mathrm{W}\left({\mathrm{k}}_0\right)=\mathrm{W}({\mathrm{k}}_0+1)=\mathrm{W}({\mathrm{k}}_0+2)$$\mathrm{W}\left({\mathrm{k}}_0\right)=\mathrm{W}\left({\mathrm{k}}_0+1\right)=\mathrm{W}\left({\mathrm{k}}_0+2\right)$W(k_(0))=W(k_(0)+1)=W(k_(0)+2)\mathrm{W}\left(\mathrm{k}_0\right)=\mathrm{W}\left(\mathrm{k}_0+1\right)=\mathrm{W}\left(\mathrm{k}_0+2\right)$\mathrm{W}\left({\mathrm{k}}_0\right)=\mathrm{W}({\mathrm{k}}_0+1)=\mathrm{W}({\mathrm{k}}_0+2)$, then perception is non-linear separable.

Expert Answer

Two sensors based upon their detection levels and gain settings are compared. The following of gain setting and sensor detection levels with a standard item being monitored provides typical membership values to represent the detection levels for each of the sensors.

Answer:

Step 1: Define the Membership Functions

Membership Function for Sensor 1, ${\mu }_{\text{S1}}(x)$${\mu }_{\text{S1}}(x)$mu_(“S1”)(x)\mu_{\text{S1}}(x)${\mu }_{\text{S1}}(x)$

Membership Function for Sensor 2, ${\mu }_{\text{S2}}(x)$${\mu }_{\text{S2}}(x)$mu_(“S2”)(x)\mu_{\text{S2}}(x)${\mu }_{\text{S2}}(x)$

Step 2: Verify De Morgan’s Laws

1. First Law:
   $${\mu }_{\mathrm{\neg }(A\cup B)}(x)={\mu }_{\mathrm{\neg }A}(x)\cap {\mu }_{\mathrm{\neg }B}(x)$$$${\mu }_{\mathrm{\neg }(A\cup B)}(x)={\mu }_{\mathrm{\neg }A}(x)\cap {\mu }_{\mathrm{\neg }B}(x)$$mu_(not(A uu B))(x)=mu_(not A)(x)nnmu_(not B)(x)\mu_{\neg(A \cup B)}(x) = \mu_{\neg A}(x) \cap \mu_{\neg B}(x)$${\mu }_{\mathrm{\neg }(A\cup B)}(x)={\mu }_{\mathrm{\neg }A}(x)\cap {\mu }_{\mathrm{\neg }B}(x)$$
   This means:
   $${\mu }_{\mathrm{\neg }(A\cup B)}(x)=min(1-{\mu }_A(x),1-{\mu }_B(x))$$$${\mu }_{\mathrm{\neg }(A\cup B)}(x)=min(1-{\mu }_A(x),1-{\mu }_B(x))$$mu_(not(A uu B))(x)=min(1-mu _(A)(x),1-mu _(B)(x))\mu_{\neg(A \cup B)}(x) = \min(1 – \mu_A(x), 1 – \mu_B(x))$${\mu }_{\mathrm{\neg }(A\cup B)}(x)=min(1-{\mu }_A(x),1-{\mu }_B(x))$$
2. Second Law:
   $${\mu }_{\mathrm{\neg }(A\cap B)}(x)={\mu }_{\mathrm{\neg }A}(x)\cup {\mu }_{\mathrm{\neg }B}(x)$$$${\mu }_{\mathrm{\neg }(A\cap B)}(x)={\mu }_{\mathrm{\neg }A}(x)\cup {\mu }_{\mathrm{\neg }B}(x)$$mu_(not(A nn B))(x)=mu_(not A)(x)uumu_(not B)(x)\mu_{\neg(A \cap B)}(x) = \mu_{\neg A}(x) \cup \mu_{\neg B}(x)$${\mu }_{\mathrm{\neg }(A\cap B)}(x)={\mu }_{\mathrm{\neg }A}(x)\cup {\mu }_{\mathrm{\neg }B}(x)$$
   This means:
   $${\mu }_{\mathrm{\neg }(A\cap B)}(x)=max(1-{\mu }_A(x),1-{\mu }_B(x))$$$${\mu }_{\mathrm{\neg }(A\cap B)}(x)=max(1-{\mu }_A(x),1-{\mu }_B(x))$$mu_(not(A nn B))(x)=max(1-mu _(A)(x),1-mu _(B)(x))\mu_{\neg(A \cap B)}(x) = \max(1 – \mu_A(x), 1 – \mu_B(x))$${\mu }_{\mathrm{\neg }(A\cap B)}(x)=max(1-{\mu }_A(x),1-{\mu }_B(x))$$

First Law: ${\mu }_{\mathrm{\neg }(S1\cup S2)}(x)={\mu }_{\mathrm{\neg }S1}(x)\cap {\mu }_{\mathrm{\neg }S2}(x)$${\mu }_{\mathrm{\neg }(S1\cup S2)}(x)={\mu }_{\mathrm{\neg }S1}(x)\cap {\mu }_{\mathrm{\neg }S2}(x)$mu_(not(S1uu S2))(x)=mu_(not S1)(x)nnmu_(not S2)(x)\mu_{\neg(S1 \cup S2)}(x) = \mu_{\neg S1}(x) \cap \mu_{\neg S2}(x)${\mu }_{\mathrm{\neg }(S1\cup S2)}(x)={\mu }_{\mathrm{\neg }S1}(x)\cap {\mu }_{\mathrm{\neg }S2}(x)$

1. Compute ${\mu }_{S1\cup S2}(x)$${\mu }_{S1\cup S2}(x)$mu_(S1uu S2)(x)\mu_{S1 \cup S2}(x)${\mu }_{S1\cup S2}(x)$:
   The union of two fuzzy sets is the maximum of the membership values:
   $${\mu }_{S1\cup S2}(x)=max({\mu }_{\text{S1}}(x),{\mu }_{\text{S2}}(x))$$$${\mu }_{S1\cup S2}(x)=max({\mu }_{\text{S1}}(x),{\mu }_{\text{S2}}(x))$$mu_(S1uu S2)(x)=max(mu_(“S1”)(x),mu_(“S2”)(x))\mu_{S1 \cup S2}(x) = \max(\mu_{\text{S1}}(x), \mu_{\text{S2}}(x))$${\mu }_{S1\cup S2}(x)=max({\mu }_{\text{S1}}(x),{\mu }_{\text{S2}}(x))$$
   For each value of $x$$x$xx$x$:
   $$\begin{array}{rl}{\mu }_{S1\cup S2}(0)& =max(0,0)=0,\\ {\mu }_{S1\cup S2}(20)& =max(0.5,0.35)=0.5,\\ {\mu }_{S1\cup S2}(40)& =max(0.65,0.5)=0.65,\\ {\mu }_{S1\cup S2}(60)& =max(0.85,0.75)=0.85,\\ {\mu }_{S1\cup S2}(80)& =max(1,0.90)=1,\\ {\mu }_{S1\cup S2}(100)& =max(1,1)=1.\end{array}$$$$\begin{array}{rl}{\mu }_{S1\cup S2}(0)& =max(0,0)=0,\\ {\mu }_{S1\cup S2}(20)& =max(0.5,0.35)=0.5,\\ {\mu }_{S1\cup S2}(40)& =max(0.65,0.5)=0.65,\\ {\mu }_{S1\cup S2}(60)& =max(0.85,0.75)=0.85,\\ {\mu }_{S1\cup S2}(80)& =max(1,0.90)=1,\\ {\mu }_{S1\cup S2}(100)& =max(1,1)=1.\end{array}$${:[mu_(S1uu S2)(0)=max(0″,”0)=0″,”],[mu_(S1uu S2)(20)=max(0.5″,”0.35)=0.5″,”],[mu_(S1uu S2)(40)=max(0.65″,”0.5)=0.65″,”],[mu_(S1uu S2)(60)=max(0.85″,”0.75)=0.85″,”],[mu_(S1uu S2)(80)=max(1″,”0.90)=1″,”],[mu_(S1uu S2)(100)=max(1″,”1)=1.]:}\begin{aligned} \mu_{S1 \cup S2}(0) &= \max(0, 0) = 0, \\ \mu_{S1 \cup S2}(20) &= \max(0.5, 0.35) = 0.5, \\ \mu_{S1 \cup S2}(40) &= \max(0.65, 0.5) = 0.65, \\ \mu_{S1 \cup S2}(60) &= \max(0.85, 0.75) = 0.85, \\ \mu_{S1 \cup S2}(80) &= \max(1, 0.90) = 1, \\ \mu_{S1 \cup S2}(100) &= \max(1, 1) = 1. \end{aligned}$$\begin{array}{rl}{\mu }_{S1\cup S2}(0)& =max(0,0)=0,\\ {\mu }_{S1\cup S2}(20)& =max(0.5,0.35)=0.5,\\ {\mu }_{S1\cup S2}(40)& =max(0.65,0.5)=0.65,\\ {\mu }_{S1\cup S2}(60)& =max(0.85,0.75)=0.85,\\ {\mu }_{S1\cup S2}(80)& =max(1,0.90)=1,\\ {\mu }_{S1\cup S2}(100)& =max(1,1)=1.\end{array}$$
2. Compute ${\mu }_{\mathrm{\neg }(S1\cup S2)}(x)$${\mu }_{\mathrm{\neg }(S1\cup S2)}(x)$mu_(not(S1uu S2))(x)\mu_{\neg(S1 \cup S2)}(x)${\mu }_{\mathrm{\neg }(S1\cup S2)}(x)$:
   The complement of a fuzzy set is $1-\mu (x)$$1-\mu (x)$1-mu(x)1 – \mu(x)$1-\mu (x)$:
   $${\mu }_{\mathrm{\neg }(S1\cup S2)}(x)=1-{\mu }_{S1\cup S2}(x)$$$${\mu }_{\mathrm{\neg }(S1\cup S2)}(x)=1-{\mu }_{S1\cup S2}(x)$$mu_(not(S1uu S2))(x)=1-mu_(S1uu S2)(x)\mu_{\neg(S1 \cup S2)}(x) = 1 – \mu_{S1 \cup S2}(x)$${\mu }_{\mathrm{\neg }(S1\cup S2)}(x)=1-{\mu }_{S1\cup S2}(x)$$
   For each value of $x$$x$xx$x$:
   $$\begin{array}{rl}{\mu }_{\mathrm{\neg }(S1\cup S2)}(0)& =1-0=1,\\ {\mu }_{\mathrm{\neg }(S1\cup S2)}(20)& =1-0.5=0.5,\\ {\mu }_{\mathrm{\neg }(S1\cup S2)}(40)& =1-0.65=0.35,\\ {\mu }_{\mathrm{\neg }(S1\cup S2)}(60)& =1-0.85=0.15,\\ {\mu }_{\mathrm{\neg }(S1\cup S2)}(80)& =1-1=0,\\ {\mu }_{\mathrm{\neg }(S1\cup S2)}(100)& =1-1=0.\end{array}$$$$\begin{array}{rl}{\mu }_{\mathrm{\neg }(S1\cup S2)}(0)& =1-0=1,\\ {\mu }_{\mathrm{\neg }(S1\cup S2)}(20)& =1-0.5=0.5,\\ {\mu }_{\mathrm{\neg }(S1\cup S2)}(40)& =1-0.65=0.35,\\ {\mu }_{\mathrm{\neg }(S1\cup S2)}(60)& =1-0.85=0.15,\\ {\mu }_{\mathrm{\neg }(S1\cup S2)}(80)& =1-1=0,\\ {\mu }_{\mathrm{\neg }(S1\cup S2)}(100)& =1-1=0.\end{array}$${:[mu_(not(S1uu S2))(0)=1-0=1″,”],[mu_(not(S1uu S2))(20)=1-0.5=0.5″,”],[mu_(not(S1uu S2))(40)=1-0.65=0.35″,”],[mu_(not(S1uu S2))(60)=1-0.85=0.15″,”],[mu_(not(S1uu S2))(80)=1-1=0″,”],[mu_(not(S1uu S2))(100)=1-1=0.]:}\begin{aligned} \mu_{\neg(S1 \cup S2)}(0) &= 1 – 0 = 1, \\ \mu_{\neg(S1 \cup S2)}(20) &= 1 – 0.5 = 0.5, \\ \mu_{\neg(S1 \cup S2)}(40) &= 1 – 0.65 = 0.35, \\ \mu_{\neg(S1 \cup S2)}(60) &= 1 – 0.85 = 0.15, \\ \mu_{\neg(S1 \cup S2)}(80) &= 1 – 1 = 0, \\ \mu_{\neg(S1 \cup S2)}(100) &= 1 – 1 = 0. \end{aligned}$$\begin{array}{rl}{\mu }_{\mathrm{\neg }(S1\cup S2)}(0)& =1-0=1,\\ {\mu }_{\mathrm{\neg }(S1\cup S2)}(20)& =1-0.5=0.5,\\ {\mu }_{\mathrm{\neg }(S1\cup S2)}(40)& =1-0.65=0.35,\\ {\mu }_{\mathrm{\neg }(S1\cup S2)}(60)& =1-0.85=0.15,\\ {\mu }_{\mathrm{\neg }(S1\cup S2)}(80)& =1-1=0,\\ {\mu }_{\mathrm{\neg }(S1\cup S2)}(100)& =1-1=0.\end{array}$$
3. Compute ${\mu }_{\mathrm{\neg }S1}(x)\cap {\mu }_{\mathrm{\neg }S2}(x)$${\mu }_{\mathrm{\neg }S1}(x)\cap {\mu }_{\mathrm{\neg }S2}(x)$mu_(not S1)(x)nnmu_(not S2)(x)\mu_{\neg S1}(x) \cap \mu_{\neg S2}(x)${\mu }_{\mathrm{\neg }S1}(x)\cap {\mu }_{\mathrm{\neg }S2}(x)$:
   The complement of Sensor 1 and Sensor 2 is:
   $${\mu }_{\mathrm{\neg }S1}(x)=1-{\mu }_{\text{S1}}(x),{\mu }_{\mathrm{\neg }S2}(x)=1-{\mu }_{\text{S2}}(x)$$$${\mu }_{\mathrm{\neg }S1}(x)=1-{\mu }_{\text{S1}}(x),{\mu }_{\mathrm{\neg }S2}(x)=1-{\mu }_{\text{S2}}(x)$$mu_(not S1)(x)=1-mu_(“S1”)(x),quadmu_(not S2)(x)=1-mu_(“S2”)(x)\mu_{\neg S1}(x) = 1 – \mu_{\text{S1}}(x), \quad \mu_{\neg S2}(x) = 1 – \mu_{\text{S2}}(x)$${\mu }_{\mathrm{\neg }S1}(x)=1-{\mu }_{\text{S1}}(x),{\mu }_{\mathrm{\neg }S2}(x)=1-{\mu }_{\text{S2}}(x)$$
   For each value of $x$$x$xx$x$:
   $$\begin{array}{rlrl}{\mu }_{\mathrm{\neg }S1}(0)& =1-0=1,& {\mu }_{\mathrm{\neg }S2}(0)& =1-0=1,\\ {\mu }_{\mathrm{\neg }S1}(20)& =1-0.5=0.5,& {\mu }_{\mathrm{\neg }S2}(20)& =1-0.35=0.65,\\ {\mu }_{\mathrm{\neg }S1}(40)& =1-0.65=0.35,& {\mu }_{\mathrm{\neg }S2}(40)& =1-0.5=0.5,\\ {\mu }_{\mathrm{\neg }S1}(60)& =1-0.85=0.15,& {\mu }_{\mathrm{\neg }S2}(60)& =1-0.75=0.25,\\ {\mu }_{\mathrm{\neg }S1}(80)& =1-1=0,& {\mu }_{\mathrm{\neg }S2}(80)& =1-0.90=0.1,\\ {\mu }_{\mathrm{\neg }S1}(100)& =1-1=0,& {\mu }_{\mathrm{\neg }S2}(100)& =1-1=0.\end{array}$$$$\begin{array}{rlrl}{\mu }_{\mathrm{\neg }S1}(0)& =1-0=1,& {\mu }_{\mathrm{\neg }S2}(0)& =1-0=1,\\ {\mu }_{\mathrm{\neg }S1}(20)& =1-0.5=0.5,& {\mu }_{\mathrm{\neg }S2}(20)& =1-0.35=0.65,\\ {\mu }_{\mathrm{\neg }S1}(40)& =1-0.65=0.35,& {\mu }_{\mathrm{\neg }S2}(40)& =1-0.5=0.5,\\ {\mu }_{\mathrm{\neg }S1}(60)& =1-0.85=0.15,& {\mu }_{\mathrm{\neg }S2}(60)& =1-0.75=0.25,\\ {\mu }_{\mathrm{\neg }S1}(80)& =1-1=0,& {\mu }_{\mathrm{\neg }S2}(80)& =1-0.90=0.1,\\ {\mu }_{\mathrm{\neg }S1}(100)& =1-1=0,& {\mu }_{\mathrm{\neg }S2}(100)& =1-1=0.\end{array}$${:[mu_(not S1)(0)=1-0=1″,”quadmu_(not S2)(0)=1-0=1″,”],[mu_(not S1)(20)=1-0.5=0.5″,”quadmu_(not S2)(20)=1-0.35=0.65″,”],[mu_(not S1)(40)=1-0.65=0.35″,”quadmu_(not S2)(40)=1-0.5=0.5″,”],[mu_(not S1)(60)=1-0.85=0.15″,”quadmu_(not S2)(60)=1-0.75=0.25″,”],[mu_(not S1)(80)=1-1=0″,”quadmu_(not S2)(80)=1-0.90=0.1″,”],[mu_(not S1)(100)=1-1=0″,”quadmu_(not S2)(100)=1-1=0.]:}\begin{aligned} \mu_{\neg S1}(0) &= 1 – 0 = 1, \quad &\mu_{\neg S2}(0) &= 1 – 0 = 1, \\ \mu_{\neg S1}(20) &= 1 – 0.5 = 0.5, \quad &\mu_{\neg S2}(20) &= 1 – 0.35 = 0.65, \\ \mu_{\neg S1}(40) &= 1 – 0.65 = 0.35, \quad &\mu_{\neg S2}(40) &= 1 – 0.5 = 0.5, \\ \mu_{\neg S1}(60) &= 1 – 0.85 = 0.15, \quad &\mu_{\neg S2}(60) &= 1 – 0.75 = 0.25, \\ \mu_{\neg S1}(80) &= 1 – 1 = 0, \quad &\mu_{\neg S2}(80) &= 1 – 0.90 = 0.1, \\ \mu_{\neg S1}(100) &= 1 – 1 = 0, \quad &\mu_{\neg S2}(100) &= 1 – 1 = 0. \end{aligned}$$\begin{array}{rlrl}{\mu }_{\mathrm{\neg }S1}(0)& =1-0=1,& {\mu }_{\mathrm{\neg }S2}(0)& =1-0=1,\\ {\mu }_{\mathrm{\neg }S1}(20)& =1-0.5=0.5,& {\mu }_{\mathrm{\neg }S2}(20)& =1-0.35=0.65,\\ {\mu }_{\mathrm{\neg }S1}(40)& =1-0.65=0.35,& {\mu }_{\mathrm{\neg }S2}(40)& =1-0.5=0.5,\\ {\mu }_{\mathrm{\neg }S1}(60)& =1-0.85=0.15,& {\mu }_{\mathrm{\neg }S2}(60)& =1-0.75=0.25,\\ {\mu }_{\mathrm{\neg }S1}(80)& =1-1=0,& {\mu }_{\mathrm{\neg }S2}(80)& =1-0.90=0.1,\\ {\mu }_{\mathrm{\neg }S1}(100)& =1-1=0,& {\mu }_{\mathrm{\neg }S2}(100)& =1-1=0.\end{array}$$

4. Compare Results for the First Law:
   We can see that the results match for all values of $x$$x$xx$x$, so the first De Morgan law holds:
   $${\mu }_{\mathrm{\neg }(S1\cup S2)}(x)={\mu }_{\mathrm{\neg }S1}(x)\cap {\mu }_{\mathrm{\neg }S2}(x)$$$${\mu }_{\mathrm{\neg }(S1\cup S2)}(x)={\mu }_{\mathrm{\neg }S1}(x)\cap {\mu }_{\mathrm{\neg }S2}(x)$$mu_(not(S1uu S2))(x)=mu_(not S1)(x)nnmu_(not S2)(x)\mu_{\neg(S1 \cup S2)}(x) = \mu_{\neg S1}(x) \cap \mu_{\neg S2}(x)$${\mu }_{\mathrm{\neg }(S1\cup S2)}(x)={\mu }_{\mathrm{\neg }S1}(x)\cap {\mu }_{\mathrm{\neg }S2}(x)$$

Second Law: ${\mu }_{\mathrm{\neg }(S1\cap S2)}(x)={\mu }_{\mathrm{\neg }S1}(x)\cup {\mu }_{\mathrm{\neg }S2}(x)$${\mu }_{\mathrm{\neg }(S1\cap S2)}(x)={\mu }_{\mathrm{\neg }S1}(x)\cup {\mu }_{\mathrm{\neg }S2}(x)$mu_(not(S1nn S2))(x)=mu_(not S1)(x)uumu_(not S2)(x)\mu_{\neg(S1 \cap S2)}(x) = \mu_{\neg S1}(x) \cup \mu_{\neg S2}(x)${\mu }_{\mathrm{\neg }(S1\cap S2)}(x)={\mu }_{\mathrm{\neg }S1}(x)\cup {\mu }_{\mathrm{\neg }S2}(x)$

1. Compute ${\mu }_{S1\cap S2}(x)$${\mu }_{S1\cap S2}(x)$mu_(S1nn S2)(x)\mu_{S1 \cap S2}(x)${\mu }_{S1\cap S2}(x)$:
   The intersection of two fuzzy sets is the minimum of the membership values:
   $${\mu }_{S1\cap S2}(x)=min({\mu }_{\text{S1}}(x),{\mu }_{\text{S2}}(x))$$$${\mu }_{S1\cap S2}(x)=min({\mu }_{\text{S1}}(x),{\mu }_{\text{S2}}(x))$$mu_(S1nn S2)(x)=min(mu_(“S1”)(x),mu_(“S2”)(x))\mu_{S1 \cap S2}(x) = \min(\mu_{\text{S1}}(x), \mu_{\text{S2}}(x))$${\mu }_{S1\cap S2}(x)=min({\mu }_{\text{S1}}(x),{\mu }_{\text{S2}}(x))$$
   For each value of $x$$x$xx$x$:
   $$\begin{array}{rl}{\mu }_{S1\cap S2}(0)& =min(0,0)=0,\\ {\mu }_{S1\cap S2}(20)& =min(0.5,0.35)=0.35,\\ {\mu }_{S1\cap S2}(40)& =min(0.65,0.5)=0.5,\\ {\mu }_{S1\cap S2}(60)& =min(0.85,0.75)=0.75,\\ {\mu }_{S1\cap S2}(80)& =min(1,0.90)=0.90,\\ {\mu }_{S1\cap S2}(100)& =min(1,1)=1.\end{array}$$$$\begin{array}{rl}{\mu }_{S1\cap S2}(0)& =min(0,0)=0,\\ {\mu }_{S1\cap S2}(20)& =min(0.5,0.35)=0.35,\\ {\mu }_{S1\cap S2}(40)& =min(0.65,0.5)=0.5,\\ {\mu }_{S1\cap S2}(60)& =min(0.85,0.75)=0.75,\\ {\mu }_{S1\cap S2}(80)& =min(1,0.90)=0.90,\\ {\mu }_{S1\cap S2}(100)& =min(1,1)=1.\end{array}$${:[mu_(S1nn S2)(0)=min(0″,”0)=0″,”],[mu_(S1nn S2)(20)=min(0.5″,”0.35)=0.35″,”],[mu_(S1nn S2)(40)=min(0.65″,”0.5)=0.5″,”],[mu_(S1nn S2)(60)=min(0.85″,”0.75)=0.75″,”],[mu_(S1nn S2)(80)=min(1″,”0.90)=0.90″,”],[mu_(S1nn S2)(100)=min(1″,”1)=1.]:}\begin{aligned} \mu_{S1 \cap S2}(0) &= \min(0, 0) = 0, \\ \mu_{S1 \cap S2}(20) &= \min(0.5, 0.35) = 0.35, \\ \mu_{S1 \cap S2}(40) &= \min(0.65, 0.5) = 0.5, \\ \mu_{S1 \cap S2}(60) &= \min(0.85, 0.75) = 0.75, \\ \mu_{S1 \cap S2}(80) &= \min(1, 0.90) = 0.90, \\ \mu_{S1 \cap S2}(100) &= \min(1, 1) = 1. \end{aligned}$$\begin{array}{rl}{\mu }_{S1\cap S2}(0)& =min(0,0)=0,\\ {\mu }_{S1\cap S2}(20)& =min(0.5,0.35)=0.35,\\ {\mu }_{S1\cap S2}(40)& =min(0.65,0.5)=0.5,\\ {\mu }_{S1\cap S2}(60)& =min(0.85,0.75)=0.75,\\ {\mu }_{S1\cap S2}(80)& =min(1,0.90)=0.90,\\ {\mu }_{S1\cap S2}(100)& =min(1,1)=1.\end{array}$$
2. Compute ${\mu }_{\mathrm{\neg }(S1\cap S2)}(x)$${\mu }_{\mathrm{\neg }(S1\cap S2)}(x)$mu_(not(S1nn S2))(x)\mu_{\neg(S1 \cap S2)}(x)${\mu }_{\mathrm{\neg }(S1\cap S2)}(x)$:
   The complement of $S1\cap S2$$S1\cap S2$S1nn S2S1 \cap S2$S1\cap S2$ is:
   $${\mu }_{\mathrm{\neg }(S1\cap S2)}(x)=1-{\mu }_{S1\cap S2}(x)$$$${\mu }_{\mathrm{\neg }(S1\cap S2)}(x)=1-{\mu }_{S1\cap S2}(x)$$mu_(not(S1nn S2))(x)=1-mu_(S1nn S2)(x)\mu_{\neg(S1 \cap S2)}(x) = 1 – \mu_{S1 \cap S2}(x)$${\mu }_{\mathrm{\neg }(S1\cap S2)}(x)=1-{\mu }_{S1\cap S2}(x)$$
   For each value of $x$$x$xx$x$:
   $$\begin{array}{rl}{\mu }_{\mathrm{\neg }(S1\cap S2)}(0)& =1-0=1,\\ {\mu }_{\mathrm{\neg }(S1\cap S2)}(20)& =1-0.35=0.65,\\ {\mu }_{\mathrm{\neg }(S1\cap S2)}(40)& =1-0.5=0.5,\\ {\mu }_{\mathrm{\neg }(S1\cap S2)}(60)& =1-0.75=0.25,\\ {\mu }_{\mathrm{\neg }(S1\cap S2)}(80)& =1-0.90=0.10,\\ {\mu }_{\mathrm{\neg }(S1\cap S2)}(100)& =1-1=0.\end{array}$$$$\begin{array}{rl}{\mu }_{\mathrm{\neg }(S1\cap S2)}(0)& =1-0=1,\\ {\mu }_{\mathrm{\neg }(S1\cap S2)}(20)& =1-0.35=0.65,\\ {\mu }_{\mathrm{\neg }(S1\cap S2)}(40)& =1-0.5=0.5,\\ {\mu }_{\mathrm{\neg }(S1\cap S2)}(60)& =1-0.75=0.25,\\ {\mu }_{\mathrm{\neg }(S1\cap S2)}(80)& =1-0.90=0.10,\\ {\mu }_{\mathrm{\neg }(S1\cap S2)}(100)& =1-1=0.\end{array}$${:[mu_(not(S1nn S2))(0)=1-0=1″,”],[mu_(not(S1nn S2))(20)=1-0.35=0.65″,”],[mu_(not(S1nn S2))(40)=1-0.5=0.5″,”],[mu_(not(S1nn S2))(60)=1-0.75=0.25″,”],[mu_(not(S1nn S2))(80)=1-0.90=0.10″,”],[mu_(not(S1nn S2))(100)=1-1=0.]:}\begin{aligned} \mu_{\neg(S1 \cap S2)}(0) &= 1 – 0 = 1, \\ \mu_{\neg(S1 \cap S2)}(20) &= 1 – 0.35 = 0.65, \\ \mu_{\neg(S1 \cap S2)}(40) &= 1 – 0.5 = 0.5, \\ \mu_{\neg(S1 \cap S2)}(60) &= 1 – 0.75 = 0.25, \\ \mu_{\neg(S1 \cap S2)}(80) &= 1 – 0.90 = 0.10, \\ \mu_{\neg(S1 \cap S2)}(100) &= 1 – 1 = 0. \end{aligned}$$\begin{array}{rl}{\mu }_{\mathrm{\neg }(S1\cap S2)}(0)& =1-0=1,\\ {\mu }_{\mathrm{\neg }(S1\cap S2)}(20)& =1-0.35=0.65,\\ {\mu }_{\mathrm{\neg }(S1\cap S2)}(40)& =1-0.5=0.5,\\ {\mu }_{\mathrm{\neg }(S1\cap S2)}(60)& =1-0.75=0.25,\\ {\mu }_{\mathrm{\neg }(S1\cap S2)}(80)& =1-0.90=0.10,\\ {\mu }_{\mathrm{\neg }(S1\cap S2)}(100)& =1-1=0.\end{array}$$
3. Compute ${\mu }_{\mathrm{\neg }S1}(x)\cup {\mu }_{\mathrm{\neg }S2}(x)$${\mu }_{\mathrm{\neg }S1}(x)\cup {\mu }_{\mathrm{\neg }S2}(x)$mu_(not S1)(x)uumu_(not S2)(x)\mu_{\neg S1}(x) \cup \mu_{\neg S2}(x)${\mu }_{\mathrm{\neg }S1}(x)\cup {\mu }_{\mathrm{\neg }S2}(x)$:
   The union (maximum) is:
   $${\mu }_{\mathrm{\neg }S1}(x)\cup {\mu }_{\mathrm{\neg }S2}(x)=max(1-{\mu }_{\text{S1}}(x),1-{\mu }_{\text{S2}}(x))$$$${\mu }_{\mathrm{\neg }S1}(x)\cup {\mu }_{\mathrm{\neg }S2}(x)=max(1-{\mu }_{\text{S1}}(x),1-{\mu }_{\text{S2}}(x))$$mu_(not S1)(x)uumu_(not S2)(x)=max(1-mu_(“S1”)(x),1-mu_(“S2”)(x))\mu_{\neg S1}(x) \cup \mu_{\neg S2}(x) = \max(1 – \mu_{\text{S1}}(x), 1 – \mu_{\text{S2}}(x))$${\mu }_{\mathrm{\neg }S1}(x)\cup {\mu }_{\mathrm{\neg }S2}(x)=max(1-{\mu }_{\text{S1}}(x),1-{\mu }_{\text{S2}}(x))$$
   For each value of $x$$x$xx$x$:
   $$\begin{array}{rl}{\mu }_{\mathrm{\neg }S1}(0)\cup {\mu }_{\mathrm{\neg }S2}(0)& =max(1,1)=1,\\ {\mu }_{\mathrm{\neg }S1}(20)\cup {\mu }_{\mathrm{\neg }S2}(20)& =max(0.5,0.65)=0.65,\\ {\mu }_{\mathrm{\neg }S1}(40)\cup {\mu }_{\mathrm{\neg }S2}(40)& =max(0.35,0.5)=0.5,\\ {\mu }_{\mathrm{\neg }S1}(60)\cup {\mu }_{\mathrm{\neg }S2}(60)& =max(0.15,0.25)=0.25,\\ {\mu }_{\mathrm{\neg }S1}(80)\cup {\mu }_{\mathrm{\neg }S2}(80)& =max(0,0.1)=0.1,\\ {\mu }_{\mathrm{\neg }S1}(100)\cup {\mu }_{\mathrm{\neg }S2}(100)& =max(0,0)=0.\end{array}$$$$\begin{array}{rl}{\mu }_{\mathrm{\neg }S1}(0)\cup {\mu }_{\mathrm{\neg }S2}(0)& =max(1,1)=1,\\ {\mu }_{\mathrm{\neg }S1}(20)\cup {\mu }_{\mathrm{\neg }S2}(20)& =max(0.5,0.65)=0.65,\\ {\mu }_{\mathrm{\neg }S1}(40)\cup {\mu }_{\mathrm{\neg }S2}(40)& =max(0.35,0.5)=0.5,\\ {\mu }_{\mathrm{\neg }S1}(60)\cup {\mu }_{\mathrm{\neg }S2}(60)& =max(0.15,0.25)=0.25,\\ {\mu }_{\mathrm{\neg }S1}(80)\cup {\mu }_{\mathrm{\neg }S2}(80)& =max(0,0.1)=0.1,\\ {\mu }_{\mathrm{\neg }S1}(100)\cup {\mu }_{\mathrm{\neg }S2}(100)& =max(0,0)=0.\end{array}$${:[mu_(not S1)(0)uumu_(not S2)(0)=max(1″,”1)=1″,”],[mu_(not S1)(20)uumu_(not S2)(20)=max(0.5″,”0.65)=0.65″,”],[mu_(not S1)(40)uumu_(not S2)(40)=max(0.35″,”0.5)=0.5″,”],[mu_(not S1)(60)uumu_(not S2)(60)=max(0.15″,”0.25)=0.25″,”],[mu_(not S1)(80)uumu_(not S2)(80)=max(0″,”0.1)=0.1″,”],[mu_(not S1)(100)uumu_(not S2)(100)=max(0″,”0)=0.]:}\begin{aligned} \mu_{\neg S1}(0) \cup \mu_{\neg S2}(0) &= \max(1, 1) = 1, \\ \mu_{\neg S1}(20) \cup \mu_{\neg S2}(20) &= \max(0.5, 0.65) = 0.65, \\ \mu_{\neg S1}(40) \cup \mu_{\neg S2}(40) &= \max(0.35, 0.5) = 0.5, \\ \mu_{\neg S1}(60) \cup \mu_{\neg S2}(60) &= \max(0.15, 0.25) = 0.25, \\ \mu_{\neg S1}(80) \cup \mu_{\neg S2}(80) &= \max(0, 0.1) = 0.1, \\ \mu_{\neg S1}(100) \cup \mu_{\neg S2}(100) &= \max(0, 0) = 0. \end{aligned}$$\begin{array}{rl}{\mu }_{\mathrm{\neg }S1}(0)\cup {\mu }_{\mathrm{\neg }S2}(0)& =max(1,1)=1,\\ {\mu }_{\mathrm{\neg }S1}(20)\cup {\mu }_{\mathrm{\neg }S2}(20)& =max(0.5,0.65)=0.65,\\ {\mu }_{\mathrm{\neg }S1}(40)\cup {\mu }_{\mathrm{\neg }S2}(40)& =max(0.35,0.5)=0.5,\\ {\mu }_{\mathrm{\neg }S1}(60)\cup {\mu }_{\mathrm{\neg }S2}(60)& =max(0.15,0.25)=0.25,\\ {\mu }_{\mathrm{\neg }S1}(80)\cup {\mu }_{\mathrm{\neg }S2}(80)& =max(0,0.1)=0.1,\\ {\mu }_{\mathrm{\neg }S1}(100)\cup {\mu }_{\mathrm{\neg }S2}(100)& =max(0,0)=0.\end{array}$$
4. Compare Results for the Second Law:
   We can see that the results match for all values of $x$$x$xx$x$, so the second De Morgan law holds:
   $${\mu }_{\mathrm{\neg }(S1\cap S2)}(x)={\mu }_{\mathrm{\neg }S1}(x)\cup {\mu }_{\mathrm{\neg }S2}(x)$$$${\mu }_{\mathrm{\neg }(S1\cap S2)}(x)={\mu }_{\mathrm{\neg }S1}(x)\cup {\mu }_{\mathrm{\neg }S2}(x)$$mu_(not(S1nn S2))(x)=mu_(not S1)(x)uumu_(not S2)(x)\mu_{\neg(S1 \cap S2)}(x) = \mu_{\neg S1}(x) \cup \mu_{\neg S2}(x)$${\mu }_{\mathrm{\neg }(S1\cap S2)}(x)={\mu }_{\mathrm{\neg }S1}(x)\cup {\mu }_{\mathrm{\neg }S2}(x)$$