Can any one help me solving this expression using boolean Algebra F=x'(y.z'+y'.z)+x.y'(y+x.z') ...
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Can any one help me solving this expression using boolean Algebra F=x'(y.z'+y'.z)+x.y'(y+x.z')
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$begingroup$
I have solved some expression but can't proceed further
F=x'(y.z'+y'.z)+x.y'(y+x.z')
=x'y.z'+x'.y'.z+x.y'.y+x.x.y'.z'(by distributive law a(b+c)=ab+ac)
=x'y.z'+x'.y'.z+x.x.y'.z'(by using rule y'.y=0)
=x'y.z'+x'.y'.z+x.y'.z'(by using rule x.x=x)
boolean-algebra
asked Mar 24 at 12:40
Alex FernandezAlex Fernandez
11
$endgroup$
add a comment |
$begingroup$
I have solved some expression but can't proceed further
F=x'(y.z'+y'.z)+x.y'(y+x.z')
=x'y.z'+x'.y'.z+x.y'.y+x.x.y'.z'(by distributive law a(b+c)=ab+ac)
=x'y.z'+x'.y'.z+x.x.y'.z'(by using rule y'.y=0)
=x'y.z'+x'.y'.z+x.y'.z'(by using rule x.x=x)
boolean-algebra
asked Mar 24 at 12:40
Alex FernandezAlex Fernandez
11
$endgroup$
add a comment |
$begingroup$
I have solved some expression but can't proceed further
F=x'(y.z'+y'.z)+x.y'(y+x.z')
=x'y.z'+x'.y'.z+x.y'.y+x.x.y'.z'(by distributive law a(b+c)=ab+ac)
=x'y.z'+x'.y'.z+x.x.y'.z'(by using rule y'.y=0)
=x'y.z'+x'.y'.z+x.y'.z'(by using rule x.x=x)
boolean-algebra
asked Mar 24 at 12:40
Alex FernandezAlex Fernandez
11
$endgroup$
I have solved some expression but can't proceed further
F=x'(y.z'+y'.z)+x.y'(y+x.z')
=x'y.z'+x'.y'.z+x.y'.y+x.x.y'.z'(by distributive law a(b+c)=ab+ac)
=x'y.z'+x'.y'.z+x.x.y'.z'(by using rule y'.y=0)
=x'y.z'+x'.y'.z+x.y'.z'(by using rule x.x=x)
boolean-algebra
boolean-algebra
asked Mar 24 at 12:40
Alex FernandezAlex Fernandez
11
asked Mar 24 at 12:40
Alex FernandezAlex Fernandez
11
edited Mar 24 at 17:51
Alex Fernandez
asked Mar 24 at 12:40
Alex FernandezAlex Fernandez
11
asked Mar 24 at 12:40
Alex FernandezAlex Fernandez
11
asked Mar 24 at 12:40
Alex FernandezAlex Fernandez
11
11
add a comment |
add a comment |
1 Answer
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$begingroup$
You can factor out $z$ and $z'$ in the first two terms:
$$ =x'yz' + x'yz + xy'z'$$
$$ = x'y (z'+z) + xyz' $$
$$ = x'y + xyz' $$
answered Mar 24 at 17:25
NeilNeil
1014
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
You can factor out $z$ and $z'$ in the first two terms:
$$ =x'yz' + x'yz + xy'z'$$
$$ = x'y (z'+z) + xyz' $$
$$ = x'y + xyz' $$
answered Mar 24 at 17:25
NeilNeil
1014
$endgroup$
add a comment |
$begingroup$
You can factor out $z$ and $z'$ in the first two terms:
$$ =x'yz' + x'yz + xy'z'$$
$$ = x'y (z'+z) + xyz' $$
$$ = x'y + xyz' $$
answered Mar 24 at 17:25
NeilNeil
1014
$endgroup$
add a comment |
$begingroup$
You can factor out $z$ and $z'$ in the first two terms:
$$ =x'yz' + x'yz + xy'z'$$
$$ = x'y (z'+z) + xyz' $$
$$ = x'y + xyz' $$
answered Mar 24 at 17:25
NeilNeil
1014
$endgroup$
You can factor out $z$ and $z'$ in the first two terms:
$$ =x'yz' + x'yz + xy'z'$$
$$ = x'y (z'+z) + xyz' $$
$$ = x'y + xyz' $$
answered Mar 24 at 17:25
NeilNeil
1014
answered Mar 24 at 17:25
NeilNeil
1014
answered Mar 24 at 17:25
NeilNeil
1014
answered Mar 24 at 17:25
NeilNeil
1014
1014
add a comment |
add a comment |
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