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Given the following truth tables AND truth table OR truth table NOT table OR tru

ID: 3719155 • Letter: G

Question

Given the following truth tables         AND truth table             OR truth table                 NOT table OR truth table NOT truth table input input output input input output input output X Y XY X Y X+Y X X' 0 0 0 0 0 0 0 1 0 1 0 0 1 1 1 0 1 0 0 1 0 1 1 1 1 1 1 1 Using a truth table, prove that    XZ=(X+Y)(X+Y’)(X’+Z) X Y Z X+Y X+Y' X'+Z (X+Y)(X+Y’)(X’+Z) XZ 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1 Given the following truth tables         AND truth table             OR truth table                 NOT table OR truth table NOT truth table input input output input input output input output X Y XY X Y X+Y X X' 0 0 0 0 0 0 0 1 0 1 0 0 1 1 1 0 1 0 0 1 0 1 1 1 1 1 1 1 Using a truth table, prove that    XZ=(X+Y)(X+Y’)(X’+Z) X Y Z X+Y X+Y' X'+Z (X+Y)(X+Y’)(X’+Z) XZ 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1

Explanation / Answer

X

Y

Z

X+Y

X+Y'

X'+Z

(X+Y)(X+Y’)(X’+Z)

XZ

0

0

0

0

1

1

0

0

0

0

1

0

1

1

0

0

0

1

0

1

0

1

0

0

0

1

1

1

0

1

0

0

1

0

0

1

1

0

0

0

1

0

1

1

1

1

1

1

1

1

0

1

1

0

0

0

1

1

1

1

1

1

1

1

Since the last two columns are equal in values. Therefore it is proved that  XZ=(X+Y)(X+Y’)(X’+Z).

X

Y

Z

X+Y

X+Y'

X'+Z

(X+Y)(X+Y’)(X’+Z)

XZ

0

0

0

0

1

1

0

0

0

0

1

0

1

1

0

0

0

1

0

1

0

1

0

0

0

1

1

1

0

1

0

0

1

0

0

1

1

0

0

0

1

0

1

1

1

1

1

1

1

1

0

1

1

0

0

0

1

1

1

1

1

1

1

1

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