Mixing
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Need help simplifying this Boolean expression
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
From my textbook, by using distributive law, its able to simplify:
$[(p land lnot q) lor (p land q)] land q$
To:
$[p land (lnot q lor q)] land q $
I don't know how to get to this step, and here is how I've tried by distributing the first expression :
$[(p lor p) land (p lor q) land (lnot q lor p) land (lnot q lor q)] land q$
If I try continue to expand this, it will become very complex. I think I'm doing this in the wrong way and I will appreciate it very much if anyone can explain how the textbook got the simplified answer.
Thanks
logic propositional-calculus
edited Sep 9 at 0:16
amWhy
190k26221433
asked Sep 8 at 23:52
IhavelowIQ
616
add a comment |Â
up vote
1
down vote
favorite
From my textbook, by using distributive law, its able to simplify:
$[(p land lnot q) lor (p land q)] land q$
To:
$[p land (lnot q lor q)] land q $
I don't know how to get to this step, and here is how I've tried by distributing the first expression :
$[(p lor p) land (p lor q) land (lnot q lor p) land (lnot q lor q)] land q$
If I try continue to expand this, it will become very complex. I think I'm doing this in the wrong way and I will appreciate it very much if anyone can explain how the textbook got the simplified answer.
Thanks
logic propositional-calculus
edited Sep 9 at 0:16
amWhy
190k26221433
asked Sep 8 at 23:52
IhavelowIQ
616
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
From my textbook, by using distributive law, its able to simplify:
$[(p land lnot q) lor (p land q)] land q$
To:
$[p land (lnot q lor q)] land q $
I don't know how to get to this step, and here is how I've tried by distributing the first expression :
$[(p lor p) land (p lor q) land (lnot q lor p) land (lnot q lor q)] land q$
If I try continue to expand this, it will become very complex. I think I'm doing this in the wrong way and I will appreciate it very much if anyone can explain how the textbook got the simplified answer.
Thanks
logic propositional-calculus
edited Sep 9 at 0:16
amWhy
190k26221433
asked Sep 8 at 23:52
IhavelowIQ
616
From my textbook, by using distributive law, its able to simplify:
$[(p land lnot q) lor (p land q)] land q$
To:
$[p land (lnot q lor q)] land q $
I don't know how to get to this step, and here is how I've tried by distributing the first expression :
$[(p lor p) land (p lor q) land (lnot q lor p) land (lnot q lor q)] land q$
If I try continue to expand this, it will become very complex. I think I'm doing this in the wrong way and I will appreciate it very much if anyone can explain how the textbook got the simplified answer.
Thanks
logic propositional-calculus
logic propositional-calculus
edited Sep 9 at 0:16
amWhy
190k26221433
asked Sep 8 at 23:52
IhavelowIQ
616
edited Sep 9 at 0:16
amWhy
190k26221433
asked Sep 8 at 23:52
IhavelowIQ
616
edited Sep 9 at 0:16
amWhy
190k26221433
edited Sep 9 at 0:16
amWhy
190k26221433
edited Sep 9 at 0:16
amWhy
190k26221433
190k26221433
asked Sep 8 at 23:52
IhavelowIQ
616
asked Sep 8 at 23:52
IhavelowIQ
616
asked Sep 8 at 23:52
IhavelowIQ
616
616
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
The Distributive law goes both ways, because it is an equivalence.
That is, using the Distributive Law you can go from $p land (neg q lor q)$ to $(p land neg q) lor (p land q)$, but you can use that same Distributive Law to go from the latter to the former, and that is what they did here.
I believe the name is somewhat to blame for the fact that so many students, like you, only think of the Distributive Law going one way, because going from $(p land neg q) lor (p land q)$ to $p land (neg q lor q)$ does not feel like Distribution, but rather as a kind of 'Un-Distribution' or 'Reverse Distribution', or maybe even 'Taking out a common Factor' ... but it is still called Distribution in logic.
answered Sep 8 at 23:56
Bram28
55.5k33982
@Bram28 ahh I see, thank you. I've tried to test the result on truth table, and the two expression are indeed equivalent, however, is there a way to prove their equivalency without using truth table? Is it possible to use the distributive law on [(p∧¬q)∨(p∧q)] and arrive at [p∧(¬q∨q)] ? It seems it just goes on forever with the expansion.
– IhavelowIQ
Sep 9 at 0:07
@IhavelowIQ Yes. This is what I tried to say in my Answer: going from $[(p land neg q) lor (p land q)]$ to $[p land (neg q lor q)]$ is using the Distributive Law.
– Bram28
Sep 9 at 0:11
@IhavelowIQ All these laws in Boolean algebra go both ways. For example, using DeMorgan you can go from $neg (p lor q)$ to $neg p land neg q$, but going from $neg p land neg q$ to $neg (p lor q)$ is also using the DeMorgan's law, since the DeMorgan's law is an equivalence. So you can go back and forth using the same law. Same for Distribution.
– Bram28
Sep 9 at 0:14
Oh sorry, I missed your edited message. You are right, I was only thinking of going one way (taught in the book), and [(p∧¬q)∨(p∧q)] to [p∧(¬q∨q)] was definitely not apparent for me. But thanks for clarifying this.
– IhavelowIQ
Sep 9 at 0:15
@IhavelowIQ ok, glad I could help! :)
– Bram28
Sep 9 at 0:16
add a comment |Â
up vote
5
down vote
$$[(colorbluep land lnot q) lor (colorbluep land q)] land q equiv [colorbluep land (colorredlnot q lor q)] land qtagdistributivity$$
$$ equiv (colorbluep land colorred top) land q$$
$$equiv colorbluep land q$$
That is, the distributive law declares that $$[pland (lnot q lor q)] equiv [(p land lnot q) lor (p land q)]$$ That means that it is the same law when we write it: $$[(p land lnot q) lor (p land q)] equiv [pland (lnot q lor q)]$$
Addendum
answered Sep 9 at 0:07
amWhy
190k26221433
and if one wants to see why the law is true, truth tables can be really helpful
– The Great Duck
Sep 9 at 4:31
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The Distributive law goes both ways, because it is an equivalence.
That is, using the Distributive Law you can go from $p land (neg q lor q)$ to $(p land neg q) lor (p land q)$, but you can use that same Distributive Law to go from the latter to the former, and that is what they did here.
I believe the name is somewhat to blame for the fact that so many students, like you, only think of the Distributive Law going one way, because going from $(p land neg q) lor (p land q)$ to $p land (neg q lor q)$ does not feel like Distribution, but rather as a kind of 'Un-Distribution' or 'Reverse Distribution', or maybe even 'Taking out a common Factor' ... but it is still called Distribution in logic.
answered Sep 8 at 23:56
Bram28
55.5k33982
@Bram28 ahh I see, thank you. I've tried to test the result on truth table, and the two expression are indeed equivalent, however, is there a way to prove their equivalency without using truth table? Is it possible to use the distributive law on [(p∧¬q)∨(p∧q)] and arrive at [p∧(¬q∨q)] ? It seems it just goes on forever with the expansion.
– IhavelowIQ
Sep 9 at 0:07
@IhavelowIQ Yes. This is what I tried to say in my Answer: going from $[(p land neg q) lor (p land q)]$ to $[p land (neg q lor q)]$ is using the Distributive Law.
– Bram28
Sep 9 at 0:11
@IhavelowIQ All these laws in Boolean algebra go both ways. For example, using DeMorgan you can go from $neg (p lor q)$ to $neg p land neg q$, but going from $neg p land neg q$ to $neg (p lor q)$ is also using the DeMorgan's law, since the DeMorgan's law is an equivalence. So you can go back and forth using the same law. Same for Distribution.
– Bram28
Sep 9 at 0:14
Oh sorry, I missed your edited message. You are right, I was only thinking of going one way (taught in the book), and [(p∧¬q)∨(p∧q)] to [p∧(¬q∨q)] was definitely not apparent for me. But thanks for clarifying this.
– IhavelowIQ
Sep 9 at 0:15
@IhavelowIQ ok, glad I could help! :)
– Bram28
Sep 9 at 0:16
add a comment |Â
up vote
2
down vote
accepted
The Distributive law goes both ways, because it is an equivalence.
That is, using the Distributive Law you can go from $p land (neg q lor q)$ to $(p land neg q) lor (p land q)$, but you can use that same Distributive Law to go from the latter to the former, and that is what they did here.
I believe the name is somewhat to blame for the fact that so many students, like you, only think of the Distributive Law going one way, because going from $(p land neg q) lor (p land q)$ to $p land (neg q lor q)$ does not feel like Distribution, but rather as a kind of 'Un-Distribution' or 'Reverse Distribution', or maybe even 'Taking out a common Factor' ... but it is still called Distribution in logic.
answered Sep 8 at 23:56
Bram28
55.5k33982
@Bram28 ahh I see, thank you. I've tried to test the result on truth table, and the two expression are indeed equivalent, however, is there a way to prove their equivalency without using truth table? Is it possible to use the distributive law on [(p∧¬q)∨(p∧q)] and arrive at [p∧(¬q∨q)] ? It seems it just goes on forever with the expansion.
– IhavelowIQ
Sep 9 at 0:07
@IhavelowIQ Yes. This is what I tried to say in my Answer: going from $[(p land neg q) lor (p land q)]$ to $[p land (neg q lor q)]$ is using the Distributive Law.
– Bram28
Sep 9 at 0:11
@IhavelowIQ All these laws in Boolean algebra go both ways. For example, using DeMorgan you can go from $neg (p lor q)$ to $neg p land neg q$, but going from $neg p land neg q$ to $neg (p lor q)$ is also using the DeMorgan's law, since the DeMorgan's law is an equivalence. So you can go back and forth using the same law. Same for Distribution.
– Bram28
Sep 9 at 0:14
Oh sorry, I missed your edited message. You are right, I was only thinking of going one way (taught in the book), and [(p∧¬q)∨(p∧q)] to [p∧(¬q∨q)] was definitely not apparent for me. But thanks for clarifying this.
– IhavelowIQ
Sep 9 at 0:15
@IhavelowIQ ok, glad I could help! :)
– Bram28
Sep 9 at 0:16
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The Distributive law goes both ways, because it is an equivalence.
That is, using the Distributive Law you can go from $p land (neg q lor q)$ to $(p land neg q) lor (p land q)$, but you can use that same Distributive Law to go from the latter to the former, and that is what they did here.
I believe the name is somewhat to blame for the fact that so many students, like you, only think of the Distributive Law going one way, because going from $(p land neg q) lor (p land q)$ to $p land (neg q lor q)$ does not feel like Distribution, but rather as a kind of 'Un-Distribution' or 'Reverse Distribution', or maybe even 'Taking out a common Factor' ... but it is still called Distribution in logic.
answered Sep 8 at 23:56
Bram28
55.5k33982
The Distributive law goes both ways, because it is an equivalence.
That is, using the Distributive Law you can go from $p land (neg q lor q)$ to $(p land neg q) lor (p land q)$, but you can use that same Distributive Law to go from the latter to the former, and that is what they did here.
I believe the name is somewhat to blame for the fact that so many students, like you, only think of the Distributive Law going one way, because going from $(p land neg q) lor (p land q)$ to $p land (neg q lor q)$ does not feel like Distribution, but rather as a kind of 'Un-Distribution' or 'Reverse Distribution', or maybe even 'Taking out a common Factor' ... but it is still called Distribution in logic.
answered Sep 8 at 23:56
Bram28
55.5k33982
edited Sep 9 at 0:02
answered Sep 8 at 23:56
Bram28
55.5k33982
answered Sep 8 at 23:56
Bram28
55.5k33982
answered Sep 8 at 23:56
Bram28
55.5k33982
55.5k33982
@Bram28 ahh I see, thank you. I've tried to test the result on truth table, and the two expression are indeed equivalent, however, is there a way to prove their equivalency without using truth table? Is it possible to use the distributive law on [(p∧¬q)∨(p∧q)] and arrive at [p∧(¬q∨q)] ? It seems it just goes on forever with the expansion.
– IhavelowIQ
Sep 9 at 0:07
@IhavelowIQ Yes. This is what I tried to say in my Answer: going from $[(p land neg q) lor (p land q)]$ to $[p land (neg q lor q)]$ is using the Distributive Law.
– Bram28
Sep 9 at 0:11
@IhavelowIQ All these laws in Boolean algebra go both ways. For example, using DeMorgan you can go from $neg (p lor q)$ to $neg p land neg q$, but going from $neg p land neg q$ to $neg (p lor q)$ is also using the DeMorgan's law, since the DeMorgan's law is an equivalence. So you can go back and forth using the same law. Same for Distribution.
– Bram28
Sep 9 at 0:14
Oh sorry, I missed your edited message. You are right, I was only thinking of going one way (taught in the book), and [(p∧¬q)∨(p∧q)] to [p∧(¬q∨q)] was definitely not apparent for me. But thanks for clarifying this.
– IhavelowIQ
Sep 9 at 0:15
@IhavelowIQ ok, glad I could help! :)
– Bram28
Sep 9 at 0:16
add a comment |Â
@Bram28 ahh I see, thank you. I've tried to test the result on truth table, and the two expression are indeed equivalent, however, is there a way to prove their equivalency without using truth table? Is it possible to use the distributive law on [(p∧¬q)∨(p∧q)] and arrive at [p∧(¬q∨q)] ? It seems it just goes on forever with the expansion.
– IhavelowIQ
Sep 9 at 0:07
@IhavelowIQ Yes. This is what I tried to say in my Answer: going from $[(p land neg q) lor (p land q)]$ to $[p land (neg q lor q)]$ is using the Distributive Law.
– Bram28
Sep 9 at 0:11
@IhavelowIQ All these laws in Boolean algebra go both ways. For example, using DeMorgan you can go from $neg (p lor q)$ to $neg p land neg q$, but going from $neg p land neg q$ to $neg (p lor q)$ is also using the DeMorgan's law, since the DeMorgan's law is an equivalence. So you can go back and forth using the same law. Same for Distribution.
– Bram28
Sep 9 at 0:14
Oh sorry, I missed your edited message. You are right, I was only thinking of going one way (taught in the book), and [(p∧¬q)∨(p∧q)] to [p∧(¬q∨q)] was definitely not apparent for me. But thanks for clarifying this.
– IhavelowIQ
Sep 9 at 0:15
@IhavelowIQ ok, glad I could help! :)
– Bram28
Sep 9 at 0:16
@Bram28 ahh I see, thank you. I've tried to test the result on truth table, and the two expression are indeed equivalent, however, is there a way to prove their equivalency without using truth table? Is it possible to use the distributive law on [(p∧¬q)∨(p∧q)] and arrive at [p∧(¬q∨q)] ? It seems it just goes on forever with the expansion.
– IhavelowIQ
Sep 9 at 0:07
@Bram28 ahh I see, thank you. I've tried to test the result on truth table, and the two expression are indeed equivalent, however, is there a way to prove their equivalency without using truth table? Is it possible to use the distributive law on [(p∧¬q)∨(p∧q)] and arrive at [p∧(¬q∨q)] ? It seems it just goes on forever with the expansion.
– IhavelowIQ
Sep 9 at 0:07
@IhavelowIQ Yes. This is what I tried to say in my Answer: going from $[(p land neg q) lor (p land q)]$ to $[p land (neg q lor q)]$ is using the Distributive Law.
– Bram28
Sep 9 at 0:11
@IhavelowIQ Yes. This is what I tried to say in my Answer: going from $[(p land neg q) lor (p land q)]$ to $[p land (neg q lor q)]$ is using the Distributive Law.
– Bram28
Sep 9 at 0:11
@IhavelowIQ All these laws in Boolean algebra go both ways. For example, using DeMorgan you can go from $neg (p lor q)$ to $neg p land neg q$, but going from $neg p land neg q$ to $neg (p lor q)$ is also using the DeMorgan's law, since the DeMorgan's law is an equivalence. So you can go back and forth using the same law. Same for Distribution.
– Bram28
Sep 9 at 0:14
@IhavelowIQ All these laws in Boolean algebra go both ways. For example, using DeMorgan you can go from $neg (p lor q)$ to $neg p land neg q$, but going from $neg p land neg q$ to $neg (p lor q)$ is also using the DeMorgan's law, since the DeMorgan's law is an equivalence. So you can go back and forth using the same law. Same for Distribution.
– Bram28
Sep 9 at 0:14
Oh sorry, I missed your edited message. You are right, I was only thinking of going one way (taught in the book), and [(p∧¬q)∨(p∧q)] to [p∧(¬q∨q)] was definitely not apparent for me. But thanks for clarifying this.
– IhavelowIQ
Sep 9 at 0:15
Oh sorry, I missed your edited message. You are right, I was only thinking of going one way (taught in the book), and [(p∧¬q)∨(p∧q)] to [p∧(¬q∨q)] was definitely not apparent for me. But thanks for clarifying this.
– IhavelowIQ
Sep 9 at 0:15
@IhavelowIQ ok, glad I could help! :)
– Bram28
Sep 9 at 0:16
@IhavelowIQ ok, glad I could help! :)
– Bram28
Sep 9 at 0:16
add a comment |Â
up vote
5
down vote
$$[(colorbluep land lnot q) lor (colorbluep land q)] land q equiv [colorbluep land (colorredlnot q lor q)] land qtagdistributivity$$
$$ equiv (colorbluep land colorred top) land q$$
$$equiv colorbluep land q$$
That is, the distributive law declares that $$[pland (lnot q lor q)] equiv [(p land lnot q) lor (p land q)]$$ That means that it is the same law when we write it: $$[(p land lnot q) lor (p land q)] equiv [pland (lnot q lor q)]$$
Addendum
answered Sep 9 at 0:07
amWhy
190k26221433
and if one wants to see why the law is true, truth tables can be really helpful
– The Great Duck
Sep 9 at 4:31
add a comment |Â
up vote
5
down vote
$$[(colorbluep land lnot q) lor (colorbluep land q)] land q equiv [colorbluep land (colorredlnot q lor q)] land qtagdistributivity$$
$$ equiv (colorbluep land colorred top) land q$$
$$equiv colorbluep land q$$
That is, the distributive law declares that $$[pland (lnot q lor q)] equiv [(p land lnot q) lor (p land q)]$$ That means that it is the same law when we write it: $$[(p land lnot q) lor (p land q)] equiv [pland (lnot q lor q)]$$
Addendum
answered Sep 9 at 0:07
amWhy
190k26221433
and if one wants to see why the law is true, truth tables can be really helpful
– The Great Duck
Sep 9 at 4:31
add a comment |Â
up vote
5
down vote
up vote
5
down vote
$$[(colorbluep land lnot q) lor (colorbluep land q)] land q equiv [colorbluep land (colorredlnot q lor q)] land qtagdistributivity$$
$$ equiv (colorbluep land colorred top) land q$$
$$equiv colorbluep land q$$
That is, the distributive law declares that $$[pland (lnot q lor q)] equiv [(p land lnot q) lor (p land q)]$$ That means that it is the same law when we write it: $$[(p land lnot q) lor (p land q)] equiv [pland (lnot q lor q)]$$
Addendum
answered Sep 9 at 0:07
amWhy
190k26221433
$$[(colorbluep land lnot q) lor (colorbluep land q)] land q equiv [colorbluep land (colorredlnot q lor q)] land qtagdistributivity$$
$$ equiv (colorbluep land colorred top) land q$$
$$equiv colorbluep land q$$
That is, the distributive law declares that $$[pland (lnot q lor q)] equiv [(p land lnot q) lor (p land q)]$$ That means that it is the same law when we write it: $$[(p land lnot q) lor (p land q)] equiv [pland (lnot q lor q)]$$
Addendum
answered Sep 9 at 0:07
amWhy
190k26221433
edited 2 days ago
answered Sep 9 at 0:07
amWhy
190k26221433
answered Sep 9 at 0:07
amWhy
190k26221433
answered Sep 9 at 0:07
amWhy
190k26221433
190k26221433
and if one wants to see why the law is true, truth tables can be really helpful
– The Great Duck
Sep 9 at 4:31
add a comment |Â
and if one wants to see why the law is true, truth tables can be really helpful
– The Great Duck
Sep 9 at 4:31
and if one wants to see why the law is true, truth tables can be really helpful
– The Great Duck
Sep 9 at 4:31
and if one wants to see why the law is true, truth tables can be really helpful
– The Great Duck
Sep 9 at 4:31
add a comment |Â
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