** **

but we still have x and x' in expression.

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Fill $f(x,y,z)=xy+x'z$ in K-map

Hazard occurred because a variable and its complement passed through different number of logic operations.

In other words, if a variable and its complement are present in different product terms, may cause Hazards.

Here it is variable $x$, that may cause Hazard.

So make an extra pair in K-map that doesn't contain variable $x$.

so $f(x,y,z)=xy+x'z +yz$ is Hazard Free.

+1

C) Using Redundant theorem

AB + A'C + BC = AB + A'C + BC(A+A')

= AB + A'C + ABC + A'BC

= AB(1 + C) + A'C(1 + B)

=AB.1 + A'C.1 [A +1 = 1 ]

=AB + A'C

Similarly

F = XY + ZX' + YZ.1

F = XY + ZX' + YZ(X+X') [ A +A' = 1 ]

F= XY + ZX' + XYZ + X'YZ

F = XY(1 + Z) + ZX'(1 + Y)

F = XY.1 + ZX'.1 [A +1 = 1 ]

F = XY + ZX'

AB + A'C + BC = AB + A'C + BC(A+A')

= AB + A'C + ABC + A'BC

= AB(1 + C) + A'C(1 + B)

=AB.1 + A'C.1 [A +1 = 1 ]

=AB + A'C

Similarly

F = XY + ZX' + YZ.1

F = XY + ZX' + YZ(X+X') [ A +A' = 1 ]

F= XY + ZX' + XYZ + X'YZ

F = XY(1 + Z) + ZX'(1 + Y)

F = XY.1 + ZX'.1 [A +1 = 1 ]

F = XY + ZX'

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