Simplify Product of Sums
$begingroup$
Similar question to:
Boolean Algebra - Product of Sums
I was given a truth table and asked to give the sums-of-products and the product-of-sums expressions.
I reduced the sums-of-products expression to this, which I believe is correct:
F(x,y,z) = xy + yz + xz
I have so far reduced my product-of-sums expression to this:
F(x,y,z) = (x+y+z)(x+y+z’)(x+y’+z)(x’+y+z)
but I can't figure out how to further reduce my product-of-sums, whilst still retaining the "product-of-sums" form and not converting back to "sums-of-products" form. If someone could show me how to further reduce the product-of-sums, I would be appreciative if you could explain which Boolean identity is being used at each stage of of the simplification (i.e. Distributive Law, DeMorgan's Law etc). Thanks!
logic boolean-algebra
edited Apr 13 '17 at 12:19
Community♦
1
asked Apr 15 '14 at 5:42
jdubbingjdubbing
112
$endgroup$
add a comment |
$begingroup$
Similar question to:
Boolean Algebra - Product of Sums
I was given a truth table and asked to give the sums-of-products and the product-of-sums expressions.
I reduced the sums-of-products expression to this, which I believe is correct:
F(x,y,z) = xy + yz + xz
I have so far reduced my product-of-sums expression to this:
F(x,y,z) = (x+y+z)(x+y+z’)(x+y’+z)(x’+y+z)
but I can't figure out how to further reduce my product-of-sums, whilst still retaining the "product-of-sums" form and not converting back to "sums-of-products" form. If someone could show me how to further reduce the product-of-sums, I would be appreciative if you could explain which Boolean identity is being used at each stage of of the simplification (i.e. Distributive Law, DeMorgan's Law etc). Thanks!
logic boolean-algebra
edited Apr 13 '17 at 12:19
Community♦
1
asked Apr 15 '14 at 5:42
jdubbingjdubbing
112
$endgroup$
$begingroup$
How about en.wikipedia.org/wiki/Karnaugh_map
$endgroup$
– evil999man
Apr 15 '14 at 5:46
$begingroup$
I preferably want to do this question without using K maps, as we haven't been taught them yet. I am expected to be able to achieve this without a K map I believe.
$endgroup$
– jdubbing
Apr 15 '14 at 6:41
$begingroup$
What I can think of is applying negation of negation and then de morgan in sop.
$endgroup$
– evil999man
Apr 15 '14 at 6:46
$begingroup$
Ok, so I did that. Can you confirm that, for the same truth table, if the sum-of-products is: F(x,y,z) = xy + yz + xz, then the product-of-sums would be: F(x,y,z) = (x+y)(y+z)(x+z) ? Thanks so far
$endgroup$
– jdubbing
Apr 15 '14 at 8:08
$begingroup$
Seems like it. You can always verify by a truth table. or a program.
$endgroup$
– evil999man
Apr 15 '14 at 9:05
add a comment |
$begingroup$
Similar question to:
Boolean Algebra - Product of Sums
I was given a truth table and asked to give the sums-of-products and the product-of-sums expressions.
I reduced the sums-of-products expression to this, which I believe is correct:
F(x,y,z) = xy + yz + xz
I have so far reduced my product-of-sums expression to this:
F(x,y,z) = (x+y+z)(x+y+z’)(x+y’+z)(x’+y+z)
but I can't figure out how to further reduce my product-of-sums, whilst still retaining the "product-of-sums" form and not converting back to "sums-of-products" form. If someone could show me how to further reduce the product-of-sums, I would be appreciative if you could explain which Boolean identity is being used at each stage of of the simplification (i.e. Distributive Law, DeMorgan's Law etc). Thanks!
logic boolean-algebra
edited Apr 13 '17 at 12:19
Community♦
1
asked Apr 15 '14 at 5:42
jdubbingjdubbing
112
$endgroup$
Similar question to:
Boolean Algebra - Product of Sums
I was given a truth table and asked to give the sums-of-products and the product-of-sums expressions.
I reduced the sums-of-products expression to this, which I believe is correct:
F(x,y,z) = xy + yz + xz
I have so far reduced my product-of-sums expression to this:
F(x,y,z) = (x+y+z)(x+y+z’)(x+y’+z)(x’+y+z)
but I can't figure out how to further reduce my product-of-sums, whilst still retaining the "product-of-sums" form and not converting back to "sums-of-products" form. If someone could show me how to further reduce the product-of-sums, I would be appreciative if you could explain which Boolean identity is being used at each stage of of the simplification (i.e. Distributive Law, DeMorgan's Law etc). Thanks!
logic boolean-algebra
logic boolean-algebra
edited Apr 13 '17 at 12:19
Community♦
1
asked Apr 15 '14 at 5:42
jdubbingjdubbing
112
edited Apr 13 '17 at 12:19
Community♦
1
asked Apr 15 '14 at 5:42
jdubbingjdubbing
112
edited Apr 13 '17 at 12:19
Community♦
1
edited Apr 13 '17 at 12:19
Community♦
1
edited Apr 13 '17 at 12:19
Community♦
1
1
asked Apr 15 '14 at 5:42
jdubbingjdubbing
112
asked Apr 15 '14 at 5:42
jdubbingjdubbing
112
asked Apr 15 '14 at 5:42
jdubbingjdubbing
112
112
$begingroup$
How about en.wikipedia.org/wiki/Karnaugh_map
$endgroup$
– evil999man
Apr 15 '14 at 5:46
$begingroup$
I preferably want to do this question without using K maps, as we haven't been taught them yet. I am expected to be able to achieve this without a K map I believe.
$endgroup$
– jdubbing
Apr 15 '14 at 6:41
$begingroup$
What I can think of is applying negation of negation and then de morgan in sop.
$endgroup$
– evil999man
Apr 15 '14 at 6:46
$begingroup$
Ok, so I did that. Can you confirm that, for the same truth table, if the sum-of-products is: F(x,y,z) = xy + yz + xz, then the product-of-sums would be: F(x,y,z) = (x+y)(y+z)(x+z) ? Thanks so far
$endgroup$
– jdubbing
Apr 15 '14 at 8:08
$begingroup$
Seems like it. You can always verify by a truth table. or a program.
$endgroup$
– evil999man
Apr 15 '14 at 9:05
add a comment |
$begingroup$
How about en.wikipedia.org/wiki/Karnaugh_map
$endgroup$
– evil999man
Apr 15 '14 at 5:46
$begingroup$
I preferably want to do this question without using K maps, as we haven't been taught them yet. I am expected to be able to achieve this without a K map I believe.
$endgroup$
– jdubbing
Apr 15 '14 at 6:41
$begingroup$
What I can think of is applying negation of negation and then de morgan in sop.
$endgroup$
– evil999man
Apr 15 '14 at 6:46
$begingroup$
Ok, so I did that. Can you confirm that, for the same truth table, if the sum-of-products is: F(x,y,z) = xy + yz + xz, then the product-of-sums would be: F(x,y,z) = (x+y)(y+z)(x+z) ? Thanks so far
$endgroup$
– jdubbing
Apr 15 '14 at 8:08
$begingroup$
Seems like it. You can always verify by a truth table. or a program.
$endgroup$
– evil999man
Apr 15 '14 at 9:05
$begingroup$
How about en.wikipedia.org/wiki/Karnaugh_map
$endgroup$
– evil999man
Apr 15 '14 at 5:46
$begingroup$
How about en.wikipedia.org/wiki/Karnaugh_map
$endgroup$
– evil999man
Apr 15 '14 at 5:46
$begingroup$
I preferably want to do this question without using K maps, as we haven't been taught them yet. I am expected to be able to achieve this without a K map I believe.
$endgroup$
– jdubbing
Apr 15 '14 at 6:41
$begingroup$
I preferably want to do this question without using K maps, as we haven't been taught them yet. I am expected to be able to achieve this without a K map I believe.
$endgroup$
– jdubbing
Apr 15 '14 at 6:41
$begingroup$
What I can think of is applying negation of negation and then de morgan in sop.
$endgroup$
– evil999man
Apr 15 '14 at 6:46
$begingroup$
What I can think of is applying negation of negation and then de morgan in sop.
$endgroup$
– evil999man
Apr 15 '14 at 6:46
$begingroup$
Ok, so I did that. Can you confirm that, for the same truth table, if the sum-of-products is: F(x,y,z) = xy + yz + xz, then the product-of-sums would be: F(x,y,z) = (x+y)(y+z)(x+z) ? Thanks so far
$endgroup$
– jdubbing
Apr 15 '14 at 8:08
$begingroup$
Ok, so I did that. Can you confirm that, for the same truth table, if the sum-of-products is: F(x,y,z) = xy + yz + xz, then the product-of-sums would be: F(x,y,z) = (x+y)(y+z)(x+z) ? Thanks so far
$endgroup$
– jdubbing
Apr 15 '14 at 8:08
$begingroup$
Seems like it. You can always verify by a truth table. or a program.
$endgroup$
– evil999man
Apr 15 '14 at 9:05
$begingroup$
Seems like it. You can always verify by a truth table. or a program.
$endgroup$
– evil999man
Apr 15 '14 at 9:05
add a comment |
1 Answer
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$begingroup$
Note that $(x+y+z)(x+y+z')=x+y$ etc.
Enter twice the dummy term $(x+y+z)$ into $F(x,y,z)=(x+y+z)(x+y+z')(x+y'+z)(x'+y+z).$ Then $$F(x,y,z)=underbrace{(x+y+z)(x+y+z')}underbrace{color{blue}{(x+y+z)}(x+y'+z)}underbrace{color{blue}{(x+y+z)}(x'+y+z)}$$ so $$F(x,y,z)=(x+y)(x+z)(y+z).$$
answered Jan 18 at 22:37
user376343user376343
3,9834829
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add a comment |
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$begingroup$
Note that $(x+y+z)(x+y+z')=x+y$ etc.
Enter twice the dummy term $(x+y+z)$ into $F(x,y,z)=(x+y+z)(x+y+z')(x+y'+z)(x'+y+z).$ Then $$F(x,y,z)=underbrace{(x+y+z)(x+y+z')}underbrace{color{blue}{(x+y+z)}(x+y'+z)}underbrace{color{blue}{(x+y+z)}(x'+y+z)}$$ so $$F(x,y,z)=(x+y)(x+z)(y+z).$$
answered Jan 18 at 22:37
user376343user376343
3,9834829
$endgroup$
add a comment |
$begingroup$
Note that $(x+y+z)(x+y+z')=x+y$ etc.
Enter twice the dummy term $(x+y+z)$ into $F(x,y,z)=(x+y+z)(x+y+z')(x+y'+z)(x'+y+z).$ Then $$F(x,y,z)=underbrace{(x+y+z)(x+y+z')}underbrace{color{blue}{(x+y+z)}(x+y'+z)}underbrace{color{blue}{(x+y+z)}(x'+y+z)}$$ so $$F(x,y,z)=(x+y)(x+z)(y+z).$$
answered Jan 18 at 22:37
user376343user376343
3,9834829
$endgroup$
add a comment |
$begingroup$
Note that $(x+y+z)(x+y+z')=x+y$ etc.
Enter twice the dummy term $(x+y+z)$ into $F(x,y,z)=(x+y+z)(x+y+z')(x+y'+z)(x'+y+z).$ Then $$F(x,y,z)=underbrace{(x+y+z)(x+y+z')}underbrace{color{blue}{(x+y+z)}(x+y'+z)}underbrace{color{blue}{(x+y+z)}(x'+y+z)}$$ so $$F(x,y,z)=(x+y)(x+z)(y+z).$$
answered Jan 18 at 22:37
user376343user376343
3,9834829
$endgroup$
Note that $(x+y+z)(x+y+z')=x+y$ etc.
Enter twice the dummy term $(x+y+z)$ into $F(x,y,z)=(x+y+z)(x+y+z')(x+y'+z)(x'+y+z).$ Then $$F(x,y,z)=underbrace{(x+y+z)(x+y+z')}underbrace{color{blue}{(x+y+z)}(x+y'+z)}underbrace{color{blue}{(x+y+z)}(x'+y+z)}$$ so $$F(x,y,z)=(x+y)(x+z)(y+z).$$
answered Jan 18 at 22:37
user376343user376343
3,9834829
answered Jan 18 at 22:37
user376343user376343
3,9834829
answered Jan 18 at 22:37
user376343user376343
3,9834829
answered Jan 18 at 22:37
user376343user376343
3,9834829
3,9834829
add a comment |
add a comment |
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$begingroup$
How about en.wikipedia.org/wiki/Karnaugh_map
$endgroup$
– evil999man
Apr 15 '14 at 5:46
$begingroup$
I preferably want to do this question without using K maps, as we haven't been taught them yet. I am expected to be able to achieve this without a K map I believe.
$endgroup$
– jdubbing
Apr 15 '14 at 6:41
$begingroup$
What I can think of is applying negation of negation and then de morgan in sop.
$endgroup$
– evil999man
Apr 15 '14 at 6:46
$begingroup$
Ok, so I did that. Can you confirm that, for the same truth table, if the sum-of-products is: F(x,y,z) = xy + yz + xz, then the product-of-sums would be: F(x,y,z) = (x+y)(y+z)(x+z) ? Thanks so far
$endgroup$
– jdubbing
Apr 15 '14 at 8:08
$begingroup$
Seems like it. You can always verify by a truth table. or a program.
$endgroup$
– evil999man
Apr 15 '14 at 9:05