sum of products Boolean algebra simplification
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I have a question that states the following:
Use algebraic manipulation to show that for three input variables x1 , x2 , and x3
∑(1,2,3,4,5,6,7) = x1 + x2 + x3
I'm assuming it wants the minimum sum of products to prove its equality to x1 + x2 + x3 (or x + y + z). This is what I've narrowed it down to so far: y'z + x'y + xz' + xyz.
However, I'm unable to understand which property's will reduce this to just x + y + z (or again, x1 + x2 + x3).
boolean-algebra
asked Sep 13 '15 at 22:39
cellsheet
1
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up vote
0
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favorite
I have a question that states the following:
Use algebraic manipulation to show that for three input variables x1 , x2 , and x3
∑(1,2,3,4,5,6,7) = x1 + x2 + x3
I'm assuming it wants the minimum sum of products to prove its equality to x1 + x2 + x3 (or x + y + z). This is what I've narrowed it down to so far: y'z + x'y + xz' + xyz.
However, I'm unable to understand which property's will reduce this to just x + y + z (or again, x1 + x2 + x3).
boolean-algebra
asked Sep 13 '15 at 22:39
cellsheet
1
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have a question that states the following:
Use algebraic manipulation to show that for three input variables x1 , x2 , and x3
∑(1,2,3,4,5,6,7) = x1 + x2 + x3
I'm assuming it wants the minimum sum of products to prove its equality to x1 + x2 + x3 (or x + y + z). This is what I've narrowed it down to so far: y'z + x'y + xz' + xyz.
However, I'm unable to understand which property's will reduce this to just x + y + z (or again, x1 + x2 + x3).
boolean-algebra
asked Sep 13 '15 at 22:39
cellsheet
1
I have a question that states the following:
Use algebraic manipulation to show that for three input variables x1 , x2 , and x3
∑(1,2,3,4,5,6,7) = x1 + x2 + x3
I'm assuming it wants the minimum sum of products to prove its equality to x1 + x2 + x3 (or x + y + z). This is what I've narrowed it down to so far: y'z + x'y + xz' + xyz.
However, I'm unable to understand which property's will reduce this to just x + y + z (or again, x1 + x2 + x3).
boolean-algebra
boolean-algebra
asked Sep 13 '15 at 22:39
cellsheet
1
asked Sep 13 '15 at 22:39
cellsheet
1
asked Sep 13 '15 at 22:39
cellsheet
1
asked Sep 13 '15 at 22:39
cellsheet
1
asked Sep 13 '15 at 22:39
cellsheet
1
1
add a comment |
add a comment |
1 Answer
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In the list of terms 1, 2, 3, 4, 5, 6, 7 only term 0 is missing. Term 0 corresponds to a conjunction of the three inverted variables. Therefore, you can look at the inverse function and invert it to get the desired function:
$F = (F')' = (x' y' z')'$
The first of De Morgan's law states:
The negation of a conjunction is the disjunction of the negations.
This leads to:
$F = x + y + z$
answered Sep 14 '15 at 20:28
Axel Kemper
3,20611318
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1 Answer
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1 Answer
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active
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oldest
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up vote
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In the list of terms 1, 2, 3, 4, 5, 6, 7 only term 0 is missing. Term 0 corresponds to a conjunction of the three inverted variables. Therefore, you can look at the inverse function and invert it to get the desired function:
$F = (F')' = (x' y' z')'$
The first of De Morgan's law states:
The negation of a conjunction is the disjunction of the negations.
This leads to:
$F = x + y + z$
answered Sep 14 '15 at 20:28
Axel Kemper
3,20611318
add a comment |
up vote
1
down vote
In the list of terms 1, 2, 3, 4, 5, 6, 7 only term 0 is missing. Term 0 corresponds to a conjunction of the three inverted variables. Therefore, you can look at the inverse function and invert it to get the desired function:
$F = (F')' = (x' y' z')'$
The first of De Morgan's law states:
The negation of a conjunction is the disjunction of the negations.
This leads to:
$F = x + y + z$
answered Sep 14 '15 at 20:28
Axel Kemper
3,20611318
add a comment |
up vote
1
down vote
up vote
1
down vote
In the list of terms 1, 2, 3, 4, 5, 6, 7 only term 0 is missing. Term 0 corresponds to a conjunction of the three inverted variables. Therefore, you can look at the inverse function and invert it to get the desired function:
$F = (F')' = (x' y' z')'$
The first of De Morgan's law states:
The negation of a conjunction is the disjunction of the negations.
This leads to:
$F = x + y + z$
answered Sep 14 '15 at 20:28
Axel Kemper
3,20611318
In the list of terms 1, 2, 3, 4, 5, 6, 7 only term 0 is missing. Term 0 corresponds to a conjunction of the three inverted variables. Therefore, you can look at the inverse function and invert it to get the desired function:
$F = (F')' = (x' y' z')'$
The first of De Morgan's law states:
The negation of a conjunction is the disjunction of the negations.
This leads to:
$F = x + y + z$
answered Sep 14 '15 at 20:28
Axel Kemper
3,20611318
answered Sep 14 '15 at 20:28
Axel Kemper
3,20611318
answered Sep 14 '15 at 20:28
Axel Kemper
3,20611318
answered Sep 14 '15 at 20:28
Axel Kemper
3,20611318
3,20611318
add a comment |
add a comment |
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