Proving AND distributive law using Boolean algebra
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$$ X(Y+Z) = (XY) + (XZ) $$
I can’t seem to derive the proper steps to prove this equation using Boolean axioms. The hint I’ve been given is using demorgans laws proofs but I still can’t seem to figure it out.
These are the axioms I’ve been given to prove this. I know that this is typically a given axiom but I have to use the others to prove that this is true.
https://encrypted-tbn0.gstatic.com/images?q=tbn:ANd9GcTPdbHp-cPxgLiwXS1zP2npvK92CG-I8IPxvN52574xEs2pd_Q
discrete-mathematics boolean-algebra
asked Sep 10 at 21:19
Jakemathbad
305
 |Â
show 4 more comments
up vote
1
down vote
favorite
$$ X(Y+Z) = (XY) + (XZ) $$
I can’t seem to derive the proper steps to prove this equation using Boolean axioms. The hint I’ve been given is using demorgans laws proofs but I still can’t seem to figure it out.
These are the axioms I’ve been given to prove this. I know that this is typically a given axiom but I have to use the others to prove that this is true.
https://encrypted-tbn0.gstatic.com/images?q=tbn:ANd9GcTPdbHp-cPxgLiwXS1zP2npvK92CG-I8IPxvN52574xEs2pd_Q
discrete-mathematics boolean-algebra
asked Sep 10 at 21:19
Jakemathbad
305
Using the Boolean axioms, means that you should set up a truth table, right ?
– Ahmad Bazzi
Sep 10 at 21:25
No it means using any of these axioms to prove that both sides are equivalent encrypted-tbn0.gstatic.com/…
– Jakemathbad
Sep 10 at 21:27
This is usually one of the axioms for a Boolean algebra. Since this isn't the case for you: What precisely are the axioms of a Boolean algebra that you've been given?
– Stefan Mesken
Sep 10 at 21:28
Then it is the distributive law
– Ahmad Bazzi
Sep 10 at 21:35
@AhmadBazzi yes it is the distributive law, but I have to prove that both sides of the equation are equivalent using only Boolean equations (not the AND distributive law) the point is to derive the equation
– Jakemathbad
Sep 10 at 21:39
 |Â
show 4 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
$$ X(Y+Z) = (XY) + (XZ) $$
I can’t seem to derive the proper steps to prove this equation using Boolean axioms. The hint I’ve been given is using demorgans laws proofs but I still can’t seem to figure it out.
These are the axioms I’ve been given to prove this. I know that this is typically a given axiom but I have to use the others to prove that this is true.
https://encrypted-tbn0.gstatic.com/images?q=tbn:ANd9GcTPdbHp-cPxgLiwXS1zP2npvK92CG-I8IPxvN52574xEs2pd_Q
discrete-mathematics boolean-algebra
asked Sep 10 at 21:19
Jakemathbad
305
$$ X(Y+Z) = (XY) + (XZ) $$
I can’t seem to derive the proper steps to prove this equation using Boolean axioms. The hint I’ve been given is using demorgans laws proofs but I still can’t seem to figure it out.
These are the axioms I’ve been given to prove this. I know that this is typically a given axiom but I have to use the others to prove that this is true.
https://encrypted-tbn0.gstatic.com/images?q=tbn:ANd9GcTPdbHp-cPxgLiwXS1zP2npvK92CG-I8IPxvN52574xEs2pd_Q
discrete-mathematics boolean-algebra
discrete-mathematics boolean-algebra
asked Sep 10 at 21:19
Jakemathbad
305
asked Sep 10 at 21:19
Jakemathbad
305
edited Sep 10 at 21:30
asked Sep 10 at 21:19
Jakemathbad
305
asked Sep 10 at 21:19
Jakemathbad
305
asked Sep 10 at 21:19
Jakemathbad
305
305
Using the Boolean axioms, means that you should set up a truth table, right ?
– Ahmad Bazzi
Sep 10 at 21:25
No it means using any of these axioms to prove that both sides are equivalent encrypted-tbn0.gstatic.com/…
– Jakemathbad
Sep 10 at 21:27
This is usually one of the axioms for a Boolean algebra. Since this isn't the case for you: What precisely are the axioms of a Boolean algebra that you've been given?
– Stefan Mesken
Sep 10 at 21:28
Then it is the distributive law
– Ahmad Bazzi
Sep 10 at 21:35
@AhmadBazzi yes it is the distributive law, but I have to prove that both sides of the equation are equivalent using only Boolean equations (not the AND distributive law) the point is to derive the equation
– Jakemathbad
Sep 10 at 21:39
 |Â
show 4 more comments
Using the Boolean axioms, means that you should set up a truth table, right ?
– Ahmad Bazzi
Sep 10 at 21:25
No it means using any of these axioms to prove that both sides are equivalent encrypted-tbn0.gstatic.com/…
– Jakemathbad
Sep 10 at 21:27
This is usually one of the axioms for a Boolean algebra. Since this isn't the case for you: What precisely are the axioms of a Boolean algebra that you've been given?
– Stefan Mesken
Sep 10 at 21:28
Then it is the distributive law
– Ahmad Bazzi
Sep 10 at 21:35
@AhmadBazzi yes it is the distributive law, but I have to prove that both sides of the equation are equivalent using only Boolean equations (not the AND distributive law) the point is to derive the equation
– Jakemathbad
Sep 10 at 21:39
Using the Boolean axioms, means that you should set up a truth table, right ?
– Ahmad Bazzi
Sep 10 at 21:25
Using the Boolean axioms, means that you should set up a truth table, right ?
– Ahmad Bazzi
Sep 10 at 21:25
No it means using any of these axioms to prove that both sides are equivalent encrypted-tbn0.gstatic.com/…
– Jakemathbad
Sep 10 at 21:27
No it means using any of these axioms to prove that both sides are equivalent encrypted-tbn0.gstatic.com/…
– Jakemathbad
Sep 10 at 21:27
This is usually one of the axioms for a Boolean algebra. Since this isn't the case for you: What precisely are the axioms of a Boolean algebra that you've been given?
– Stefan Mesken
Sep 10 at 21:28
This is usually one of the axioms for a Boolean algebra. Since this isn't the case for you: What precisely are the axioms of a Boolean algebra that you've been given?
– Stefan Mesken
Sep 10 at 21:28
Then it is the distributive law
– Ahmad Bazzi
Sep 10 at 21:35
Then it is the distributive law
– Ahmad Bazzi
Sep 10 at 21:35
@AhmadBazzi yes it is the distributive law, but I have to prove that both sides of the equation are equivalent using only Boolean equations (not the AND distributive law) the point is to derive the equation
– Jakemathbad
Sep 10 at 21:39
@AhmadBazzi yes it is the distributive law, but I have to prove that both sides of the equation are equivalent using only Boolean equations (not the AND distributive law) the point is to derive the equation
– Jakemathbad
Sep 10 at 21:39
 |Â
show 4 more comments
1 Answer
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up vote
1
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I'm using the notation you use in the question.
It might not be obvious which law is each, since in the image your axioms have a different notation, but I thing with the tags in each step you'll get it.
beginalign
XY + XZ
&= (XY + X)(XY + Z) tagdistributivity\
&= X (XY + Z) tagabsorption\
&= X (Z + XY) tagcommutativity\
&= X ((Z + X)(Z + Y)) tagdistributivity\
&= (X(Z + X))(Z + Y) tagassociativity\
&= X(Z + Y) tagabsorption\
&= X(Y + Z). tagcommutativity
endalign
Notice you still have to pick the right law (there is a pair of each), except for the distributivity, in which case we're using here the one there we're not proving.
answered Sep 11 at 13:39
amrsa
3,3532518
I'm confused on the very first step, how did you use the OR distributive law to make XY + XZ = (XY + X) (XY + Z ), maybe I just can't see it, but it looks as if you used the AND law of distributivity
– Jakemathbad
Sep 11 at 20:17
@Jakemathbad Make $W=XY$, and use $W+XZ=(W+X)(W+Z)$.
– amrsa
Sep 11 at 20:34
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
I'm using the notation you use in the question.
It might not be obvious which law is each, since in the image your axioms have a different notation, but I thing with the tags in each step you'll get it.
beginalign
XY + XZ
&= (XY + X)(XY + Z) tagdistributivity\
&= X (XY + Z) tagabsorption\
&= X (Z + XY) tagcommutativity\
&= X ((Z + X)(Z + Y)) tagdistributivity\
&= (X(Z + X))(Z + Y) tagassociativity\
&= X(Z + Y) tagabsorption\
&= X(Y + Z). tagcommutativity
endalign
Notice you still have to pick the right law (there is a pair of each), except for the distributivity, in which case we're using here the one there we're not proving.
answered Sep 11 at 13:39
amrsa
3,3532518
I'm confused on the very first step, how did you use the OR distributive law to make XY + XZ = (XY + X) (XY + Z ), maybe I just can't see it, but it looks as if you used the AND law of distributivity
– Jakemathbad
Sep 11 at 20:17
@Jakemathbad Make $W=XY$, and use $W+XZ=(W+X)(W+Z)$.
– amrsa
Sep 11 at 20:34
add a comment |Â
up vote
1
down vote
I'm using the notation you use in the question.
It might not be obvious which law is each, since in the image your axioms have a different notation, but I thing with the tags in each step you'll get it.
beginalign
XY + XZ
&= (XY + X)(XY + Z) tagdistributivity\
&= X (XY + Z) tagabsorption\
&= X (Z + XY) tagcommutativity\
&= X ((Z + X)(Z + Y)) tagdistributivity\
&= (X(Z + X))(Z + Y) tagassociativity\
&= X(Z + Y) tagabsorption\
&= X(Y + Z). tagcommutativity
endalign
Notice you still have to pick the right law (there is a pair of each), except for the distributivity, in which case we're using here the one there we're not proving.
answered Sep 11 at 13:39
amrsa
3,3532518
I'm confused on the very first step, how did you use the OR distributive law to make XY + XZ = (XY + X) (XY + Z ), maybe I just can't see it, but it looks as if you used the AND law of distributivity
– Jakemathbad
Sep 11 at 20:17
@Jakemathbad Make $W=XY$, and use $W+XZ=(W+X)(W+Z)$.
– amrsa
Sep 11 at 20:34
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I'm using the notation you use in the question.
It might not be obvious which law is each, since in the image your axioms have a different notation, but I thing with the tags in each step you'll get it.
beginalign
XY + XZ
&= (XY + X)(XY + Z) tagdistributivity\
&= X (XY + Z) tagabsorption\
&= X (Z + XY) tagcommutativity\
&= X ((Z + X)(Z + Y)) tagdistributivity\
&= (X(Z + X))(Z + Y) tagassociativity\
&= X(Z + Y) tagabsorption\
&= X(Y + Z). tagcommutativity
endalign
Notice you still have to pick the right law (there is a pair of each), except for the distributivity, in which case we're using here the one there we're not proving.
answered Sep 11 at 13:39
amrsa
3,3532518
I'm using the notation you use in the question.
It might not be obvious which law is each, since in the image your axioms have a different notation, but I thing with the tags in each step you'll get it.
beginalign
XY + XZ
&= (XY + X)(XY + Z) tagdistributivity\
&= X (XY + Z) tagabsorption\
&= X (Z + XY) tagcommutativity\
&= X ((Z + X)(Z + Y)) tagdistributivity\
&= (X(Z + X))(Z + Y) tagassociativity\
&= X(Z + Y) tagabsorption\
&= X(Y + Z). tagcommutativity
endalign
Notice you still have to pick the right law (there is a pair of each), except for the distributivity, in which case we're using here the one there we're not proving.
answered Sep 11 at 13:39
amrsa
3,3532518
answered Sep 11 at 13:39
amrsa
3,3532518
answered Sep 11 at 13:39
amrsa
3,3532518
answered Sep 11 at 13:39
amrsa
3,3532518
3,3532518
I'm confused on the very first step, how did you use the OR distributive law to make XY + XZ = (XY + X) (XY + Z ), maybe I just can't see it, but it looks as if you used the AND law of distributivity
– Jakemathbad
Sep 11 at 20:17
@Jakemathbad Make $W=XY$, and use $W+XZ=(W+X)(W+Z)$.
– amrsa
Sep 11 at 20:34
add a comment |Â
I'm confused on the very first step, how did you use the OR distributive law to make XY + XZ = (XY + X) (XY + Z ), maybe I just can't see it, but it looks as if you used the AND law of distributivity
– Jakemathbad
Sep 11 at 20:17
@Jakemathbad Make $W=XY$, and use $W+XZ=(W+X)(W+Z)$.
– amrsa
Sep 11 at 20:34
I'm confused on the very first step, how did you use the OR distributive law to make XY + XZ = (XY + X) (XY + Z ), maybe I just can't see it, but it looks as if you used the AND law of distributivity
– Jakemathbad
Sep 11 at 20:17
I'm confused on the very first step, how did you use the OR distributive law to make XY + XZ = (XY + X) (XY + Z ), maybe I just can't see it, but it looks as if you used the AND law of distributivity
– Jakemathbad
Sep 11 at 20:17
@Jakemathbad Make $W=XY$, and use $W+XZ=(W+X)(W+Z)$.
– amrsa
Sep 11 at 20:34
@Jakemathbad Make $W=XY$, and use $W+XZ=(W+X)(W+Z)$.
– amrsa
Sep 11 at 20:34
add a comment |Â
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Using the Boolean axioms, means that you should set up a truth table, right ?
– Ahmad Bazzi
Sep 10 at 21:25
No it means using any of these axioms to prove that both sides are equivalent encrypted-tbn0.gstatic.com/…
– Jakemathbad
Sep 10 at 21:27
This is usually one of the axioms for a Boolean algebra. Since this isn't the case for you: What precisely are the axioms of a Boolean algebra that you've been given?
– Stefan Mesken
Sep 10 at 21:28
Then it is the distributive law
– Ahmad Bazzi
Sep 10 at 21:35
@AhmadBazzi yes it is the distributive law, but I have to prove that both sides of the equation are equivalent using only Boolean equations (not the AND distributive law) the point is to derive the equation
– Jakemathbad
Sep 10 at 21:39