The rules of precedence for Boolean operators are: Boolean complement; Boolean product and Boolean sum. Find the values of the following: (i) (1+0) · 1+(1 ·0) (ii) (1 v 0) ^1 v 0 v 1
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- F= (~A xor ~B) or ~CF= ~A or ~B or ~CF= ((~A or B) and (~A or ~B)) or (C+~A) Use the functions above to create digital circuits.2- Use Boolean Algebra to simplify the following functions.F= (~A+B) (A+B)(C+A)F= ABC+ A(~B+~C)+C(A+B)F=(A+B)(A+C)(B+C)Simplify the following expressions by applying Boolean rules. L(LM + M)* :The correct representation to the wave function in the Figure attached is uo(t-to) - { t to U₂(t+to) Vout open terminals خیار 2 R t = T Vout خیار 3
- True Or False Variable, complement, and literal are all terms used in Boolean algebra. Addition in Boolean algebra is equivalent to the NOR function. Multiplication in Boolean algebra is equivalent to the AND function. The commutative law, associative law, and distributive law are all laws in Boolean algebra. The complement of 0 is 0 itself. When a Boolean variable is multiplied by its complement, the result is the variable.Write the distributive law:Convert this regular expression using the Closure Properties of FA Regular Expression: (a+b)* aabb (a+b)*
- Question Number Four: Answer each one of the following questions in the space assigned: 1- Simplify each of the following expressions using boolean algebra identities: [4 Marks] a- AB + A( CD + CD') b- ( BC' + A'D) ( AB' + CD') 2- Optimize (simplify) the following boolean function F together with the don't care conditions d in Product-of-sums using k-map: F(A,B.C.D) = IIM (1,3,4,6,9,11)+Ed (0,2,5,8,10,12,14)Simplify B (x, y, z) = X’Z+X’Y+YZactoring is a powerful simplification technique in Boolean algebra, just as it is in real-number algebra. Show how you can use factoring to help simplify the following Boolean expressions: A. F = Y + YZB. F = XY’Z + XY’Z’