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Result of MUX (first one), is, say $f_1 = x\bar z+ \bar yz$

Result of MUX(second one), $f= f_1\bar y +xy$

$\qquad =(x\bar z+\bar yz)\bar y+xy$

$\qquad = x\bar y\bar z +\bar yz +xy$

$\qquad =x(\bar y\bar z +y) +\bar yz$

$\qquad = x(\bar y+y)(\bar z+y) +\bar yz$

$\qquad =xz'+xy+y'z .$

Option A.

Note:

- $f =I_0\bar S +I_1S,$ for $2:1$ MUX, where $I_0$ and $I_1$ are inputs, $S$ is the select line
- Distributive property, $A+BC = (A+B)(A+C)$
- $A+\bar A =1$

+1

See applying these properties may increase your adrenaline in exam if you could not convert your SOP to desired answer. Better minimize the SOP obtained from circuit and then minimize each min term from options. It will be accurate. :)

0

$f(x,y,z) = xy'z' + y'z+xy$

If we take $y'$ is common,then we get

$f(x,y,z) = y'(xz' + z)+xy$

Apply the distributive property $x+yz = (x+y)\cdot(x+z)$

$f(x,y,z) = y'\big[(x + z)\cdot(z'+z)\big]+xy$

$f(x,y,z) = y'(x + z)+xy$

$\boxed{f(x,y,z) = xy' + y'z + xy}$

$f(x,y,z) = x(y'+y) + y'z$

$\boxed{f(x,y,z) = x + y'z}$

This is minimal SOP.

$$(OR)$$

$f(x,y,z) = xy'z' + y'z+xy$

If we take $x$ is common,then we get

$f(x,y,z) = x(y'z' + y) + y'z$

Apply the distributive property $x+yz = (x+y)\cdot(x+z)$

$f(x,y,z) = x\big[(y' + y)\cdot(z'+y)\big] + y'z$

$f(x,y,z) = x(z'+y) + y'z$

$\boxed{f(x,y,z) = xz' + xy + y'z}$

$f(x,y,z)= xz' + (xy+y')\cdot(xy + z)$

$f(x,y,z)= xz' + [(x+y')\cdot(y+y')]\cdot(xy + z)$

$f(x,y,z)= xz' + (x+y')\cdot(xy + z)$

$f(x,y,z)= xz' + xyz + xz + y'z$

$f(x,y,z)= xz' + xz(y + 1) + y'z$

$f(x,y,z)= xz' + xz + y'z$

$f(x,y,z)= x(z' + z) + y'z$

$\boxed{f(x,y,z)= x+ y'z}$

This is minimal SOP.

So, both are same.

$$(OR)$$

$f(x,y,z) = xy'z' + y'z+xy$

We can expand this expression and get the canonical SOP.

$f(x,y,z) = \underbrace{xy'z'}_{100} + \underbrace{x'y'z}_{001} + \underbrace{xy'z}_{101}+\underbrace{xyz'}_{110}+ \underbrace{xyz}_{111}$

$f(x,y,z) = \sum(1,4,5,6,7)$

$\boxed{f(x,y,z)= x+ y'z}$

see this for reference: minimization

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