I need help in simplifying a Boolean expression.
My starting point was (A+D)*(A+B+C)*(~A+C+~D)
And I should end at ~A*B*D +A*~D +C*D (according to online solvers.)
But when I do it by hand on a paper I end up with this:
~A*B*D +A*~D +C*D +A*C
And for the life of me I can't figure out how to simplify and get rid of the extra +A*C
boolean-algebra
edited Dec 9 at 16:58
Rócherz
2,7762721
asked Dec 9 at 16:35
Andres Eelma
33
|
show 1 more comment
My starting point was (A+D)*(A+B+C)*(~A+C+~D)
And I should end at ~A*B*D +A*~D +C*D (according to online solvers.)
But when I do it by hand on a paper I end up with this:
~A*B*D +A*~D +C*D +A*C
And for the life of me I can't figure out how to simplify and get rid of the extra +A*C
boolean-algebra
edited Dec 9 at 16:58
Rócherz
2,7762721
asked Dec 9 at 16:35
Andres Eelma
33
1
Does $p+q$ mean $pland q$? Does $pq$ mean $plor q$?
– J.G.
Dec 9 at 16:37
+ is OR, * is AND, ~is NOT; p+q was p∨q and pq was p∧q if i recall correctly
– Andres Eelma
Dec 9 at 16:45
Thanks for clarifying.
– J.G.
Dec 9 at 16:48
My bad, I didn't realize using asteriks to show multiplication would make the text italics and whatnot.
– Andres Eelma
Dec 9 at 16:54
usepland qfor $pland q$, and useplor qfor $plor q$, and usesimfor $sim$ :)
– Nosrati
Dec 9 at 16:55
|
show 1 more comment
My starting point was (A+D)*(A+B+C)*(~A+C+~D)
And I should end at ~A*B*D +A*~D +C*D (according to online solvers.)
But when I do it by hand on a paper I end up with this:
~A*B*D +A*~D +C*D +A*C
And for the life of me I can't figure out how to simplify and get rid of the extra +A*C
boolean-algebra
edited Dec 9 at 16:58
Rócherz
2,7762721
asked Dec 9 at 16:35
Andres Eelma
33
My starting point was (A+D)*(A+B+C)*(~A+C+~D)
And I should end at ~A*B*D +A*~D +C*D (according to online solvers.)
But when I do it by hand on a paper I end up with this:
~A*B*D +A*~D +C*D +A*C
And for the life of me I can't figure out how to simplify and get rid of the extra +A*C
boolean-algebra
boolean-algebra
edited Dec 9 at 16:58
Rócherz
2,7762721
asked Dec 9 at 16:35
Andres Eelma
33
edited Dec 9 at 16:58
Rócherz
2,7762721
asked Dec 9 at 16:35
Andres Eelma
33
edited Dec 9 at 16:58
Rócherz
2,7762721
edited Dec 9 at 16:58
Rócherz
2,7762721
edited Dec 9 at 16:58
Rócherz
2,7762721
2,7762721
asked Dec 9 at 16:35
Andres Eelma
33
asked Dec 9 at 16:35
Andres Eelma
33
asked Dec 9 at 16:35
Andres Eelma
33
33
1
Does $p+q$ mean $pland q$? Does $pq$ mean $plor q$?
– J.G.
Dec 9 at 16:37
+ is OR, * is AND, ~is NOT; p+q was p∨q and pq was p∧q if i recall correctly
– Andres Eelma
Dec 9 at 16:45
Thanks for clarifying.
– J.G.
Dec 9 at 16:48
My bad, I didn't realize using asteriks to show multiplication would make the text italics and whatnot.
– Andres Eelma
Dec 9 at 16:54
usepland qfor $pland q$, and useplor qfor $plor q$, and usesimfor $sim$ :)
– Nosrati
Dec 9 at 16:55
|
show 1 more comment
1
Does $p+q$ mean $pland q$? Does $pq$ mean $plor q$?
– J.G.
Dec 9 at 16:37
+ is OR, * is AND, ~is NOT; p+q was p∨q and pq was p∧q if i recall correctly
– Andres Eelma
Dec 9 at 16:45
Thanks for clarifying.
– J.G.
Dec 9 at 16:48
My bad, I didn't realize using asteriks to show multiplication would make the text italics and whatnot.
– Andres Eelma
Dec 9 at 16:54
usepland qfor $pland q$, and useplor qfor $plor q$, and usesimfor $sim$ :)
– Nosrati
Dec 9 at 16:55
1
1
Does $p+q$ mean $pland q$? Does $pq$ mean $plor q$?
– J.G.
Dec 9 at 16:37
Does $p+q$ mean $pland q$? Does $pq$ mean $plor q$?
– J.G.
Dec 9 at 16:37
+ is OR, * is AND, ~is NOT; p+q was p∨q and pq was p∧q if i recall correctly
– Andres Eelma
Dec 9 at 16:45
+ is OR, * is AND, ~is NOT; p+q was p∨q and pq was p∧q if i recall correctly
– Andres Eelma
Dec 9 at 16:45
Thanks for clarifying.
– J.G.
Dec 9 at 16:48
Thanks for clarifying.
– J.G.
Dec 9 at 16:48
My bad, I didn't realize using asteriks to show multiplication would make the text italics and whatnot.
– Andres Eelma
Dec 9 at 16:54
My bad, I didn't realize using asteriks to show multiplication would make the text italics and whatnot.
– Andres Eelma
Dec 9 at 16:54
use
pland q for $pland q$, and use plor q for $plor q$, and use sim for $sim$ :)– Nosrati
Dec 9 at 16:55
use
pland q for $pland q$, and use plor q for $plor q$, and use sim for $sim$ :)– Nosrati
Dec 9 at 16:55
|
show 1 more comment
1 Answer
1
active
oldest
votes
By applying the distributive law you get that $(A lor D) land (A lor B lor C)land ( sim A lor C , lor sim D)$ is equivalent to
$$
(A land A land sim A) lor (A land A land C) lor (A land A land sim D) lor (A land B land sim A) lor (A land B land C) lor (A land B land sim D) lor (A land C land sim A) lor (A land C land C) lor (A land C land sim D) lor (D land A land sim A) lor (D land A land C) lor (D land A land sim D) lor (D land B land sim A) lor (D land B land C) lor (D land B land sim D) lor (D land C land sim A) lor (D land C land C) lor (D land C land sim D)
$$
We now delete all the disjuncts that are equivalent to $0$ and we get rid of all the redundant terms. We get that the original expression is equivalent to
$$
(A land C) lor (A land sim D) lor (A land B land C) lor (A land B land sim D) lor (A land C ) lor (A land C land sim D) lor (D land A land C) lor (D land B land sim A) lor (D land B land C) lor (D land C land sim A) lor (D land C)
$$
We now erase the disjuncts that are smaller than another disjunct.
$$
(A land C) lor (A land sim D) lor (D land B land sim A) lor (D land C)
$$
Which, by commutativity of $land$, is equivalent to the expression you got. Therefore your solution was right.
Note that your solution is equivalent to $ (sim A land B land D) lor (A land sim D) lor (C land D)$. Indeed, by the distributive law and the fact that $D , lor sim D$ is equivalent to $1$, your solution is equivalent to
$$
(sim A land B land D) lor (A land sim D) lor (C land D) lor (A land C land D) lor (A land C land sim D).
$$
Since the two last disjuncts are smaller than the third and second ones, it is equivalent to
$$
(sim A land B land D) lor (A land sim D) lor (C land D) .
$$
answered Dec 10 at 8:14
Luca Carai
19618
add a comment |
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1 Answer
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1 Answer
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By applying the distributive law you get that $(A lor D) land (A lor B lor C)land ( sim A lor C , lor sim D)$ is equivalent to
$$
(A land A land sim A) lor (A land A land C) lor (A land A land sim D) lor (A land B land sim A) lor (A land B land C) lor (A land B land sim D) lor (A land C land sim A) lor (A land C land C) lor (A land C land sim D) lor (D land A land sim A) lor (D land A land C) lor (D land A land sim D) lor (D land B land sim A) lor (D land B land C) lor (D land B land sim D) lor (D land C land sim A) lor (D land C land C) lor (D land C land sim D)
$$
We now delete all the disjuncts that are equivalent to $0$ and we get rid of all the redundant terms. We get that the original expression is equivalent to
$$
(A land C) lor (A land sim D) lor (A land B land C) lor (A land B land sim D) lor (A land C ) lor (A land C land sim D) lor (D land A land C) lor (D land B land sim A) lor (D land B land C) lor (D land C land sim A) lor (D land C)
$$
We now erase the disjuncts that are smaller than another disjunct.
$$
(A land C) lor (A land sim D) lor (D land B land sim A) lor (D land C)
$$
Which, by commutativity of $land$, is equivalent to the expression you got. Therefore your solution was right.
Note that your solution is equivalent to $ (sim A land B land D) lor (A land sim D) lor (C land D)$. Indeed, by the distributive law and the fact that $D , lor sim D$ is equivalent to $1$, your solution is equivalent to
$$
(sim A land B land D) lor (A land sim D) lor (C land D) lor (A land C land D) lor (A land C land sim D).
$$
Since the two last disjuncts are smaller than the third and second ones, it is equivalent to
$$
(sim A land B land D) lor (A land sim D) lor (C land D) .
$$
answered Dec 10 at 8:14
Luca Carai
19618
add a comment |
By applying the distributive law you get that $(A lor D) land (A lor B lor C)land ( sim A lor C , lor sim D)$ is equivalent to
$$
(A land A land sim A) lor (A land A land C) lor (A land A land sim D) lor (A land B land sim A) lor (A land B land C) lor (A land B land sim D) lor (A land C land sim A) lor (A land C land C) lor (A land C land sim D) lor (D land A land sim A) lor (D land A land C) lor (D land A land sim D) lor (D land B land sim A) lor (D land B land C) lor (D land B land sim D) lor (D land C land sim A) lor (D land C land C) lor (D land C land sim D)
$$
We now delete all the disjuncts that are equivalent to $0$ and we get rid of all the redundant terms. We get that the original expression is equivalent to
$$
(A land C) lor (A land sim D) lor (A land B land C) lor (A land B land sim D) lor (A land C ) lor (A land C land sim D) lor (D land A land C) lor (D land B land sim A) lor (D land B land C) lor (D land C land sim A) lor (D land C)
$$
We now erase the disjuncts that are smaller than another disjunct.
$$
(A land C) lor (A land sim D) lor (D land B land sim A) lor (D land C)
$$
Which, by commutativity of $land$, is equivalent to the expression you got. Therefore your solution was right.
Note that your solution is equivalent to $ (sim A land B land D) lor (A land sim D) lor (C land D)$. Indeed, by the distributive law and the fact that $D , lor sim D$ is equivalent to $1$, your solution is equivalent to
$$
(sim A land B land D) lor (A land sim D) lor (C land D) lor (A land C land D) lor (A land C land sim D).
$$
Since the two last disjuncts are smaller than the third and second ones, it is equivalent to
$$
(sim A land B land D) lor (A land sim D) lor (C land D) .
$$
answered Dec 10 at 8:14
Luca Carai
19618
add a comment |
By applying the distributive law you get that $(A lor D) land (A lor B lor C)land ( sim A lor C , lor sim D)$ is equivalent to
$$
(A land A land sim A) lor (A land A land C) lor (A land A land sim D) lor (A land B land sim A) lor (A land B land C) lor (A land B land sim D) lor (A land C land sim A) lor (A land C land C) lor (A land C land sim D) lor (D land A land sim A) lor (D land A land C) lor (D land A land sim D) lor (D land B land sim A) lor (D land B land C) lor (D land B land sim D) lor (D land C land sim A) lor (D land C land C) lor (D land C land sim D)
$$
We now delete all the disjuncts that are equivalent to $0$ and we get rid of all the redundant terms. We get that the original expression is equivalent to
$$
(A land C) lor (A land sim D) lor (A land B land C) lor (A land B land sim D) lor (A land C ) lor (A land C land sim D) lor (D land A land C) lor (D land B land sim A) lor (D land B land C) lor (D land C land sim A) lor (D land C)
$$
We now erase the disjuncts that are smaller than another disjunct.
$$
(A land C) lor (A land sim D) lor (D land B land sim A) lor (D land C)
$$
Which, by commutativity of $land$, is equivalent to the expression you got. Therefore your solution was right.
Note that your solution is equivalent to $ (sim A land B land D) lor (A land sim D) lor (C land D)$. Indeed, by the distributive law and the fact that $D , lor sim D$ is equivalent to $1$, your solution is equivalent to
$$
(sim A land B land D) lor (A land sim D) lor (C land D) lor (A land C land D) lor (A land C land sim D).
$$
Since the two last disjuncts are smaller than the third and second ones, it is equivalent to
$$
(sim A land B land D) lor (A land sim D) lor (C land D) .
$$
answered Dec 10 at 8:14
Luca Carai
19618
By applying the distributive law you get that $(A lor D) land (A lor B lor C)land ( sim A lor C , lor sim D)$ is equivalent to
$$
(A land A land sim A) lor (A land A land C) lor (A land A land sim D) lor (A land B land sim A) lor (A land B land C) lor (A land B land sim D) lor (A land C land sim A) lor (A land C land C) lor (A land C land sim D) lor (D land A land sim A) lor (D land A land C) lor (D land A land sim D) lor (D land B land sim A) lor (D land B land C) lor (D land B land sim D) lor (D land C land sim A) lor (D land C land C) lor (D land C land sim D)
$$
We now delete all the disjuncts that are equivalent to $0$ and we get rid of all the redundant terms. We get that the original expression is equivalent to
$$
(A land C) lor (A land sim D) lor (A land B land C) lor (A land B land sim D) lor (A land C ) lor (A land C land sim D) lor (D land A land C) lor (D land B land sim A) lor (D land B land C) lor (D land C land sim A) lor (D land C)
$$
We now erase the disjuncts that are smaller than another disjunct.
$$
(A land C) lor (A land sim D) lor (D land B land sim A) lor (D land C)
$$
Which, by commutativity of $land$, is equivalent to the expression you got. Therefore your solution was right.
Note that your solution is equivalent to $ (sim A land B land D) lor (A land sim D) lor (C land D)$. Indeed, by the distributive law and the fact that $D , lor sim D$ is equivalent to $1$, your solution is equivalent to
$$
(sim A land B land D) lor (A land sim D) lor (C land D) lor (A land C land D) lor (A land C land sim D).
$$
Since the two last disjuncts are smaller than the third and second ones, it is equivalent to
$$
(sim A land B land D) lor (A land sim D) lor (C land D) .
$$
answered Dec 10 at 8:14
Luca Carai
19618
answered Dec 10 at 8:14
Luca Carai
19618
answered Dec 10 at 8:14
Luca Carai
19618
answered Dec 10 at 8:14
Luca Carai
19618
19618
add a comment |
add a comment |
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1
Does $p+q$ mean $pland q$? Does $pq$ mean $plor q$?
– J.G.
Dec 9 at 16:37
+ is OR, * is AND, ~is NOT; p+q was p∨q and pq was p∧q if i recall correctly
– Andres Eelma
Dec 9 at 16:45
Thanks for clarifying.
– J.G.
Dec 9 at 16:48
My bad, I didn't realize using asteriks to show multiplication would make the text italics and whatnot.
– Andres Eelma
Dec 9 at 16:54
use
pland qfor $pland q$, and useplor qfor $plor q$, and usesimfor $sim$ :)– Nosrati
Dec 9 at 16:55