Слайд 2
![((A*B)+C)*A](/_ipx/f_webp&q_80&fit_contain&s_1440x1080/imagesDir/jpg/94427/slide-1.jpg)
Слайд 3
![](/_ipx/f_webp&q_80&fit_contain&s_1440x1080/imagesDir/jpg/94427/slide-2.jpg)
Слайд 4
![(A*B) + (C’*B) + (A*B’)](/_ipx/f_webp&q_80&fit_contain&s_1440x1080/imagesDir/jpg/94427/slide-3.jpg)
Слайд 5
![LOGIC DIAGRAM 1 1 0](/_ipx/f_webp&q_80&fit_contain&s_1440x1080/imagesDir/jpg/94427/slide-4.jpg)
Слайд 6
![](/_ipx/f_webp&q_80&fit_contain&s_1440x1080/imagesDir/jpg/94427/slide-5.jpg)
Слайд 7
![PROBLEM 1 A system used 3 switches A,B and C;](/_ipx/f_webp&q_80&fit_contain&s_1440x1080/imagesDir/jpg/94427/slide-6.jpg)
PROBLEM 1
A system used 3 switches A,B and C; a combination
of switches determines whether an alarm, X, sounds:
If switch A or Switch B are in the ON position and if switch C is in the OFF position then a signal to sound an alarm, X is produced.
Convert this problem into a logic statement.
Слайд 8
![1) solve logistics problems at an elementary level 2) SIMPLIFY A LOGIC CIRCUIT/EXPRESSION USING BOOLEAN ALGEBRA](/_ipx/f_webp&q_80&fit_contain&s_1440x1080/imagesDir/jpg/94427/slide-7.jpg)
1) solve logistics problems at an elementary level
2) SIMPLIFY A
LOGIC CIRCUIT/EXPRESSION USING BOOLEAN ALGEBRA
Слайд 9
![](/_ipx/f_webp&q_80&fit_contain&s_1440x1080/imagesDir/jpg/94427/slide-8.jpg)
Слайд 10
![COMMUTATIVE LAWS A+B = B+A](/_ipx/f_webp&q_80&fit_contain&s_1440x1080/imagesDir/jpg/94427/slide-9.jpg)
COMMUTATIVE LAWS
A+B = B+A
Слайд 11
![ASSOCIATIVE LAWS : A + (B + C) = (A + B) + C](/_ipx/f_webp&q_80&fit_contain&s_1440x1080/imagesDir/jpg/94427/slide-10.jpg)
ASSOCIATIVE LAWS :
A + (B + C) = (A + B)
+ C
Слайд 12
![DISTRIBUTIVE LAW: A(B + C) = AB + AC](/_ipx/f_webp&q_80&fit_contain&s_1440x1080/imagesDir/jpg/94427/slide-11.jpg)
DISTRIBUTIVE LAW:
A(B + C) = AB + AC
Слайд 13
![DEMORGAN'S THEOREMS X + Y = X Y](/_ipx/f_webp&q_80&fit_contain&s_1440x1080/imagesDir/jpg/94427/slide-12.jpg)
DEMORGAN'S THEOREMS
X + Y = X Y
Слайд 14
![EXAMPLE](/_ipx/f_webp&q_80&fit_contain&s_1440x1080/imagesDir/jpg/94427/slide-13.jpg)
Слайд 15
![Using Boolean algebra techniques, simplify this expression: AB + A(B + C) + B(B + C)](/_ipx/f_webp&q_80&fit_contain&s_1440x1080/imagesDir/jpg/94427/slide-14.jpg)
Using Boolean algebra techniques, simplify this expression:
AB + A(B +
C) + B(B + C)
Слайд 16
![X = A.B.C + A’.C Y = (Q + R)(Q’](/_ipx/f_webp&q_80&fit_contain&s_1440x1080/imagesDir/jpg/94427/slide-15.jpg)
X = A.B.C + A’.C
Y = (Q + R)(Q’ + R’)
W
= A.B.C + A.B’.C + A’
Слайд 17
![X = A.B.C + A’.C Y = (Q + R)(Q’](/_ipx/f_webp&q_80&fit_contain&s_1440x1080/imagesDir/jpg/94427/slide-16.jpg)
X = A.B.C + A’.C
Y = (Q + R)(Q’ + R’)
W
= A.B.C + A.B’.C + A’
Слайд 18
![Y = (Q + R)(Q’ + R’) Answer : =](/_ipx/f_webp&q_80&fit_contain&s_1440x1080/imagesDir/jpg/94427/slide-17.jpg)
Y = (Q + R)(Q’ + R’)
Answer :
= Q.Q’ +
Q.R’ +Q’.R +R.R’
= Q.R’ + Q’.R