Boolean Algebra - Binary logic


Keep in mind, this is not the only order in which these can be simplified.

Part 2

A.  A*B*C+A*B+C = A*B+C

 ABC+AB+C = ABC+AB1+C  Identity
 ABC+AB1+C = AB(C+1)+C  Distribution
 AB(C+1)+C = AB(1)+C  Null 
 AB(1)+C = AB+C  Identity

B. A*(B+A+C) = A

  A(B+A+C) = AB+AA+AC  Distribution
  AB+AA+AC = AB+A+AC   Itempotent
  AB+A+AC = AB+1A+AC   Identity
  AB+1A+AC = A(B+1+C)  Distribution
  A(B+1+C) = A(1)      Null
  A(1) = A             Identity


C. A*C+C*(!A+A*B) = C

  AC+C(!A+AB) = AC+!AC+ABC Distribution
  AC+!AC+ABC = C(A+!A)+ABC Distribution
  C(A+!A)+ABC = C(1)+ABC Inverse
  C(1)+ABC = C(1+AB) Distribution
  C(1+AB) = C(1) Null
  C(1) = C Identity
 
D. (A+B*C*D+E)*E*D = D*E

 (A+BCD+E)ED = ADE+BCDEDE+DEE  Distribution
 ADE+BCDEDE+DEE = ADE+BCDE+DE  Itempotent
 ADE+BCDE+DE = ADE+BCDE+1DE   Identity
 ADE+BCDE+1DE = (A+BC+1)DE    Distribution
 (A+BC+1)DE = (1)DE           Null
 1DE = DE                     Identity

E. F=(A+B)(A+C)(!A+C) = (A+B)*C

 (A+B)(A+C)(!A+C) = (A+B)(A!A+C!A+AC+CC)   Distribution
 (A+B)(A!A+C!A+AC+CC) = (A+B)(0+C!A+AC+CC) Inverse
 (A+B)(0+C!A+AC+CC) = (A+B)(C!A+AC+CC)     Identity
 (A+B)(C!A+AC+CC) = (A+B)(C!A+AC+C)        Itempotent
 (A+B)(C!A+AC+C) = (A+B)(C!A+AC+1C)        Identity
 (A+B)(C!A+AC+1C) = (A+B)(!A+A+1)C         Distribution
 (A+B)(!A+A+1)C  = (A+B)(1)C               Null
 (A+B)(1)C = (A+B)c                        Identity



Part 3 

A. X+!X+Y

  x y  x+!x+y
  0 0  0+1+0  1
  0 1  0+1+1  1
  1 0  1+0+0  1
  1 1  1+0+1  1

x+!x+y = 1+y  Inverse
1+y = 1  Null


B. X*(!X+Y)

  x y  x*(!x+y)
  0 0  0*(1+0) 0
  0 1  0*(1+1) 0
  1 0  1*(0+0) 0
  1 1  1*(0+1) 1

x*(!x+y) = x*!x+x*y  Distribution
x*!x+x*y = 0+x*y     Inverse
0+x*y = x*y          Identity
 
  
C. X*(!X*Y)

  x y x*(!x*y)
  0 0 0*(1*0) 0
  0 1 0*(1*1) 0
  1 0 1*(0*0) 0
  1 1 1*(0*1) 0

x*(!x*y) = (x*!x)*y  Associative
(x*!x)*y = 0*y       Inverse
0*y = 0              Null

D. !X*(X+!Y)

x y !x*(x+!y) 0 0 1*(0+1) 1 0 1 1*(0+0) 0 1 0 0*(1+1) 0 1 1 0*(1+0) 0 !x*(x+!y) = !x*x+!x*!y Distribution !x*x+!x*!y = 0+!x*!y Inverse 0+!x*!y = !x*!y Identity !x*!y = !(x+y) DeMorgan E. !X+X*Y Use substitute A for !X, B for X, C for Y and use distribution. Then substitute back and simplify. Substitution is not required but it makes the opportunity to use distribution obvious. !x+x*y = a+b*c Substitution a+b*c = (a+b)*(a+c) Distribution (a+b)*(a+c) = (!x+x)*(!x+y) Substitute back (!x+x)*(!x+y) = (1)*(!x+y) Inverse 1*(!x+y) = !x+y Identity x y !x+x*y !x+y 0 0 1+0*0 1 1+0 1 0 1 1+0*1 1 1+1 1 1 0 0+1*0 0 0+0 0 1 1 0+1+1 1 0+1 1 f. !(!X*!X+Y) !(!x+!x+y) = !(!x+y) Itempotent !(!x+y) = !!x*!y DeMorgan !!x*!y = x*!y Cancellation x y !(!x*!x+y) x*!y 0 0 !(1*1+0) 0 0*1 0 0 1 !(1*1+1) 0 0*0 0 1 0 !(0*0+0) 1 1*1 1 1 1 !(0*0+1) 0 1*0 0 Part 4 - More DeMorgan 5 Points each A. C*(A+B+!(!C+!D))*D C*(A+B+!(!C+!D))*D = C*D(A+B+!!C*!!D) DeMorgan C*D(A+B+!!C*!!D) = C*D(A+B+C*D) Cancellation C*D(A+B+C*D) = C*D*A+C*D*B+C*D*C*D Distribution C*D*A+C*D*B+C*D*C*D = C*D*A+C*D*B+C*D Itempotent C*D*A+C*D*B+C*D = C*D*A+C*D*B+1*C*D Identity C*D*A+C*D*B+1*C*D = C*D(A+B+1) Distribution C*D(A+B+1) = C*D(1) Null C*D*1 = C*D Identity B. !((A*B+B*!C)*A) !((A*B+B*!C)*A) = !(A*A*B+A*B*!C) Distribution !(A*A*B+A*B*!C) = !(A*B+A*B*!C) Itempotent !(A*B+A*B*!C) = !(1*A*B+A*B*!C) Identity !(1*A*B+A*B*!C) = !(A*B(1+!C)) Distribution !(A*B(1+!C)) = !(A*B(1)) Null !(A*B(1)) = !(A*B) Identity