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+17 votes

Let $f(w, x, y, z) = \sum {\left(0,4,5,7,8,9,13,15\right)}$. Which of the following expressions are NOT equivalent to $f$?

**P:** $x'y'z' + w'xy' + wy'z + xz$

**Q:** $w'y'z' + wx'y' + xz$

**R:** $w'y'z' + wx'y' + xyz+xy'z$

**S:** $x'y'z' + wx'y'+ w'y$

- P only
- Q and S
- R and S
- S only

+16 votes

Best answer

w'x' | w'x | wx | wx' | |
---|---|---|---|---|

y'z' | 1 | 1 | 1 | |

y'z | 1 | 1 | 1 | |

yz | 1 | 1 | ||

yz' |

So, minimized expression will be

$xz + w'y'z' + wx'y'$ which is Q. From the K-map, we can also get P and R. So, only S is NOT equivalent to $f$.

http://www.eecs.berkeley.edu/~newton/Classes/CS150sp98/lectures/week4_2/sld011.htm

Go with Minterm representation of each option, (Note that order of W,X,Y,Z is should be preserved. )

P : X' Y' Z' + W' X' Y' + W Y' Z + X Z

= () X' Y' Z' + W' X Y' () + W () Y' Z + () X () Z

= () 000 + 010() + 1 () 01 + () 1 () 1

= 0000 + 1000 + 0100 + 0101 + 1001 + 1101 + 0101+0111+1101+1111

= 0 + 8 + 4 + 5 + 9+13 + 5+7+13+15

= ∑ m(0,4,5,7,8,9,13,15)

Q : W' Y' Z' + W X' Y' + XZ

= W' () Y' Z' + W X' Y' () + () X () Z

= 0 () 0 0 + 100 () + () 1 () 1

= 0000+0100 + 1000+1001 + 0101 + 0111+1101+1111

= 0+4 + 8+9 + 5+7+13+15

= ∑ m(0,4,5,7,8,9,13,15)

R : W' Y' Z' + W X' Y' + XYZ + X Y' Z

= W' () Y' Z' + W X' Y' () + () X Y Z + () X Y' Z

= 0 () 0 0 + 100 () + () 111 + () 101

= 0000+ 0100 + 1000+1001 + 0111+1111 + 0101 + 1101

= 0+4 + 8+9 + 7+15 + 5+13

= ∑ m(0,4,5,7,8,9,13,15)

S : X' Y' Z' + W X' Y' + W' Y

= () X' Y' Z' + W X' Y' () + W' () Y ()

= () 000 + 100 () + 0 () 1 ()

= 0000+1000 + 1000+1001 + 0010+0011+0110+0111

= 0+8 + 8+9 + 2+3+6+7

= ∑ m(0,2,3,6,7,8,9)

+21 votes

Let me show u a very simple method

Let w =1 ,x =1 ,y=1 ,z=1 then the value of f is 1

consider each statement

x'y'z' + w' x y' +w y' z +x z = 0.0.0 + 0.1.0 + 1.0.1 + 1.1 =1

w' y' z' +w x' z' y' +x z = 0.0.0 +1.0.0 +1.1 =1

w' y' z' +w x' y' +x y z +x y' z =0.0.0 +1.0.0 +1.1.1 +1.0.1 =1

x' y' z' +w x' y' + w' y = 0.0.0 + 1.0.0 + 0.1 =0

So statement (d) is false because w=1 x=1 y=1 z=1 the value of f is 0 .

(d) does not contain the essential Minterms .

x′y′z′+w′xy′+wy′z+xz

x′y′z′+w′xy′+wy′z+xz

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