Question

Simplify the given boolean expression

(~ this symbol represents NOT)

F = A + ~A*B + ~A*~B*C + ~A*~B*~C*D + ~A*~B*~C*~D*E

Answer 1

we have,

F = A + ~A*B + ~A*~B*C + ~A*~B*~C*D + ~A*~B*~C*~D*E

we can write,

F = A + ~A * B + ~A *~B*C + ~A*~B*~C*D + ~A*~B*~C*~D*E we can write, F = A + A^BAR B + A^bsr B^bar C + A^bar B^bar C^bar D + A^bar B^bar C^bar D^bar E F = A +A^bar B + A^bar B^bar C^bar(D + D^bar E) Using distributive for the last term law we can say that, F = A + A^bar B + A^bar B^bar C + A^bar B^bar C^bar((D + D^bar) (D + E)) we know that D + D^' = 1 hence, F = A+ A^bar B + A^bar B^bar C + A^bar B^bar C^bar ((1) (D + E)) F = A + A^bar B + A^bar B^bar C + A^bar B^bar C^bar(D + E) F = A + A^bar B + A^bar B^bar(C + C^bar(D + E)) Using distributive for the last term law we can say that, F = A + A^bar B

$\small F = A + \bar{A}B + \bar{A}\bar{B}C+\bar{A}\bar{B}\bar{C}\left (D+\bar{D}E \right )$

Using ditributive for the last term law we can say that,

$\small F = A + \bar{A}B + \bar{A}\bar{B}C+\bar{A}\bar{B}\bar{C}\left ((D+\bar{D})(D+E) \right )$

we know that D + D' = 1 hence,

$\small F = A + \bar{A}B + \bar{A}\bar{B}C+\bar{A}\bar{B}\bar{C}\left ((1)(D+E) \right )$

$\small F = A + \bar{A}B + \bar{A}\bar{B}C+\bar{A}\bar{B}\bar{C}\left (D+E \right )$

$\small F = A + \bar{A}B + \bar{A}\bar{B}\left (C+\bar{C} (D+E ) \right )$

Using ditributive for the last term law we can say that,

$\small F = A + \bar{A}B + \bar{A}\bar{B}((C+\bar{C})(C+D+E))$

we know that C + C' = 1 hence,

$\small F = A + \bar{A}B + \bar{A}\bar{B}((1)(C+D+E))$

$\small F = A + \bar{A}B + \bar{A}\bar{B}(C+D+E)$

$\small F = A + \bar{A}(B + \bar{B}(C+D+E))$

Using ditributive for the last term law we can say that,

$\small F = A + \bar{A}((B + \bar{B})(B+C+D+E))$

we know that B + B' = 1 hence,

$\small F = A + \bar{A}((1)(B+C+D+E))$

$\small F = A + \bar{A}(B+C+D+E)$

Using ditributive law we can say that,

$\small F = (A + \bar{A})(A+B+C+D+E)$

we know that A + A' = 1 hence,

$\small F =(1)(A+B+C+D+E)$

$\small F =(A+B+C+D+E)$

Hence we can write,

$\small F =A+B+C+D+E$