finding the conjunctive normal form of a statement

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I am currently trying to find the conjunctive normal form of this statement



$(A land B) lor ((lnot C lor A) land lnot B) lor B$.



I think I have found the answer but I am unsure if it is correct.



Am I correct in saying the answer is:



$A lor B lor lnot C$




logic propositional-calculus conjunctive-normal-form




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  • Seems right to me
    – angryavian
    Aug 2 at 5:36














up vote
2
down vote

favorite












I am currently trying to find the conjunctive normal form of this statement



$(A land B) lor ((lnot C lor A) land lnot B) lor B$.



I think I have found the answer but I am unsure if it is correct.



Am I correct in saying the answer is:



$A lor B lor lnot C$




logic propositional-calculus conjunctive-normal-form




share|cite|improve this question





















  • Seems right to me
    – angryavian
    Aug 2 at 5:36












up vote
2
down vote

favorite









up vote
2
down vote

favorite











I am currently trying to find the conjunctive normal form of this statement



$(A land B) lor ((lnot C lor A) land lnot B) lor B$.



I think I have found the answer but I am unsure if it is correct.



Am I correct in saying the answer is:



$A lor B lor lnot C$




logic propositional-calculus conjunctive-normal-form




share|cite|improve this question













I am currently trying to find the conjunctive normal form of this statement



$(A land B) lor ((lnot C lor A) land lnot B) lor B$.



I think I have found the answer but I am unsure if it is correct.



Am I correct in saying the answer is:



$A lor B lor lnot C$





logic propositional-calculus conjunctive-normal-form





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Aug 2 at 6:08









Taroccoesbrocco

3,31941330




3,31941330









asked Aug 2 at 5:26









user10168997

434




434











  • Seems right to me
    – angryavian
    Aug 2 at 5:36
















  • Seems right to me
    – angryavian
    Aug 2 at 5:36















Seems right to me
– angryavian
Aug 2 at 5:36




Seems right to me
– angryavian
Aug 2 at 5:36










1 Answer
1






active

oldest

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up vote
3
down vote













Yes, $A lor B lor lnot C$ is a conjunctive normal form (CNF) of $(A land B) lor ((lnot C lor A) land lnot B) lor B$.



A way to prove it is just to observe that $A lor B lor lnot C$ is a CNF and it is logically equivalent (you can check it by means of truth tables) to $(A land B) lor ((lnot C lor A) land lnot B) lor B$.



An alternative proof is to use logical equivalences as rewriting rules leading from $(A land B) lor ((lnot C lor A) land lnot B) lor B$ to $A lor B lor lnot C$. More precisely, observe first that:



beginalign
((lnot C lor A) land lnot B) lor B &equiv (lnot C lor A lor B) land (lnot B lor B) &&textdistributivity law \
&equiv lnot C lor A lor B &&textidentity law (as $lnot B lor B$ is a tautology).
endalign



Then (the use of the associativity law below is always implicit),
beginalign
(A land B) lor ((lnot C lor A) land lnot B) lor B &equiv (A land B) lor (lnot C lor A lor B) &&textsubstitution \
& equiv (A land B) lor (A lor B) lor lnot C &&textcommutativity law \
&equiv ((A lor A lor B) land (B lor A lor B)) lor lnot C && textdistributivity law \
&equiv ((A lor B) land (A lor B)) lor lnot C && textidempotent law \
&equiv A lor B lor lnot C &&textidempotent law.
endalign






share|cite|improve this answer





















  • would the DNF be the same?
    – user10168997
    Aug 2 at 14:44










  • @user10168997 - Yes, of course. In this case, your starting formula $(A land B) lor ((lnot C lor A) land lnot B) lor B$ has a CNF and a DNF that coincide.
    – Taroccoesbrocco
    Aug 2 at 16:05











  • @user10168997 - Pay attention that a CNF of a given formula is not unique, and similarly for DNF. For instance, $A land (A lor B lor lnot C)$ is a CNF of $(A land B) lor ((lnot C lor A) land lnot B) lor B$ as well. So, properly speaking it is not clear what you refer to when you talk of the CNF of a given formula.
    – Taroccoesbrocco
    Aug 2 at 16:08











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1 Answer
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active

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes








up vote
3
down vote













Yes, $A lor B lor lnot C$ is a conjunctive normal form (CNF) of $(A land B) lor ((lnot C lor A) land lnot B) lor B$.



A way to prove it is just to observe that $A lor B lor lnot C$ is a CNF and it is logically equivalent (you can check it by means of truth tables) to $(A land B) lor ((lnot C lor A) land lnot B) lor B$.



An alternative proof is to use logical equivalences as rewriting rules leading from $(A land B) lor ((lnot C lor A) land lnot B) lor B$ to $A lor B lor lnot C$. More precisely, observe first that:



beginalign
((lnot C lor A) land lnot B) lor B &equiv (lnot C lor A lor B) land (lnot B lor B) &&textdistributivity law \
&equiv lnot C lor A lor B &&textidentity law (as $lnot B lor B$ is a tautology).
endalign



Then (the use of the associativity law below is always implicit),
beginalign
(A land B) lor ((lnot C lor A) land lnot B) lor B &equiv (A land B) lor (lnot C lor A lor B) &&textsubstitution \
& equiv (A land B) lor (A lor B) lor lnot C &&textcommutativity law \
&equiv ((A lor A lor B) land (B lor A lor B)) lor lnot C && textdistributivity law \
&equiv ((A lor B) land (A lor B)) lor lnot C && textidempotent law \
&equiv A lor B lor lnot C &&textidempotent law.
endalign






share|cite|improve this answer





















  • would the DNF be the same?
    – user10168997
    Aug 2 at 14:44










  • @user10168997 - Yes, of course. In this case, your starting formula $(A land B) lor ((lnot C lor A) land lnot B) lor B$ has a CNF and a DNF that coincide.
    – Taroccoesbrocco
    Aug 2 at 16:05











  • @user10168997 - Pay attention that a CNF of a given formula is not unique, and similarly for DNF. For instance, $A land (A lor B lor lnot C)$ is a CNF of $(A land B) lor ((lnot C lor A) land lnot B) lor B$ as well. So, properly speaking it is not clear what you refer to when you talk of the CNF of a given formula.
    – Taroccoesbrocco
    Aug 2 at 16:08















up vote
3
down vote













Yes, $A lor B lor lnot C$ is a conjunctive normal form (CNF) of $(A land B) lor ((lnot C lor A) land lnot B) lor B$.



A way to prove it is just to observe that $A lor B lor lnot C$ is a CNF and it is logically equivalent (you can check it by means of truth tables) to $(A land B) lor ((lnot C lor A) land lnot B) lor B$.



An alternative proof is to use logical equivalences as rewriting rules leading from $(A land B) lor ((lnot C lor A) land lnot B) lor B$ to $A lor B lor lnot C$. More precisely, observe first that:



beginalign
((lnot C lor A) land lnot B) lor B &equiv (lnot C lor A lor B) land (lnot B lor B) &&textdistributivity law \
&equiv lnot C lor A lor B &&textidentity law (as $lnot B lor B$ is a tautology).
endalign



Then (the use of the associativity law below is always implicit),
beginalign
(A land B) lor ((lnot C lor A) land lnot B) lor B &equiv (A land B) lor (lnot C lor A lor B) &&textsubstitution \
& equiv (A land B) lor (A lor B) lor lnot C &&textcommutativity law \
&equiv ((A lor A lor B) land (B lor A lor B)) lor lnot C && textdistributivity law \
&equiv ((A lor B) land (A lor B)) lor lnot C && textidempotent law \
&equiv A lor B lor lnot C &&textidempotent law.
endalign






share|cite|improve this answer





















  • would the DNF be the same?
    – user10168997
    Aug 2 at 14:44










  • @user10168997 - Yes, of course. In this case, your starting formula $(A land B) lor ((lnot C lor A) land lnot B) lor B$ has a CNF and a DNF that coincide.
    – Taroccoesbrocco
    Aug 2 at 16:05











  • @user10168997 - Pay attention that a CNF of a given formula is not unique, and similarly for DNF. For instance, $A land (A lor B lor lnot C)$ is a CNF of $(A land B) lor ((lnot C lor A) land lnot B) lor B$ as well. So, properly speaking it is not clear what you refer to when you talk of the CNF of a given formula.
    – Taroccoesbrocco
    Aug 2 at 16:08













up vote
3
down vote










up vote
3
down vote









Yes, $A lor B lor lnot C$ is a conjunctive normal form (CNF) of $(A land B) lor ((lnot C lor A) land lnot B) lor B$.



A way to prove it is just to observe that $A lor B lor lnot C$ is a CNF and it is logically equivalent (you can check it by means of truth tables) to $(A land B) lor ((lnot C lor A) land lnot B) lor B$.



An alternative proof is to use logical equivalences as rewriting rules leading from $(A land B) lor ((lnot C lor A) land lnot B) lor B$ to $A lor B lor lnot C$. More precisely, observe first that:



beginalign
((lnot C lor A) land lnot B) lor B &equiv (lnot C lor A lor B) land (lnot B lor B) &&textdistributivity law \
&equiv lnot C lor A lor B &&textidentity law (as $lnot B lor B$ is a tautology).
endalign



Then (the use of the associativity law below is always implicit),
beginalign
(A land B) lor ((lnot C lor A) land lnot B) lor B &equiv (A land B) lor (lnot C lor A lor B) &&textsubstitution \
& equiv (A land B) lor (A lor B) lor lnot C &&textcommutativity law \
&equiv ((A lor A lor B) land (B lor A lor B)) lor lnot C && textdistributivity law \
&equiv ((A lor B) land (A lor B)) lor lnot C && textidempotent law \
&equiv A lor B lor lnot C &&textidempotent law.
endalign






share|cite|improve this answer













Yes, $A lor B lor lnot C$ is a conjunctive normal form (CNF) of $(A land B) lor ((lnot C lor A) land lnot B) lor B$.



A way to prove it is just to observe that $A lor B lor lnot C$ is a CNF and it is logically equivalent (you can check it by means of truth tables) to $(A land B) lor ((lnot C lor A) land lnot B) lor B$.



An alternative proof is to use logical equivalences as rewriting rules leading from $(A land B) lor ((lnot C lor A) land lnot B) lor B$ to $A lor B lor lnot C$. More precisely, observe first that:



beginalign
((lnot C lor A) land lnot B) lor B &equiv (lnot C lor A lor B) land (lnot B lor B) &&textdistributivity law \
&equiv lnot C lor A lor B &&textidentity law (as $lnot B lor B$ is a tautology).
endalign



Then (the use of the associativity law below is always implicit),
beginalign
(A land B) lor ((lnot C lor A) land lnot B) lor B &equiv (A land B) lor (lnot C lor A lor B) &&textsubstitution \
& equiv (A land B) lor (A lor B) lor lnot C &&textcommutativity law \
&equiv ((A lor A lor B) land (B lor A lor B)) lor lnot C && textdistributivity law \
&equiv ((A lor B) land (A lor B)) lor lnot C && textidempotent law \
&equiv A lor B lor lnot C &&textidempotent law.
endalign







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Aug 2 at 7:18









Taroccoesbrocco

3,31941330




3,31941330











  • would the DNF be the same?
    – user10168997
    Aug 2 at 14:44










  • @user10168997 - Yes, of course. In this case, your starting formula $(A land B) lor ((lnot C lor A) land lnot B) lor B$ has a CNF and a DNF that coincide.
    – Taroccoesbrocco
    Aug 2 at 16:05











  • @user10168997 - Pay attention that a CNF of a given formula is not unique, and similarly for DNF. For instance, $A land (A lor B lor lnot C)$ is a CNF of $(A land B) lor ((lnot C lor A) land lnot B) lor B$ as well. So, properly speaking it is not clear what you refer to when you talk of the CNF of a given formula.
    – Taroccoesbrocco
    Aug 2 at 16:08

















  • would the DNF be the same?
    – user10168997
    Aug 2 at 14:44










  • @user10168997 - Yes, of course. In this case, your starting formula $(A land B) lor ((lnot C lor A) land lnot B) lor B$ has a CNF and a DNF that coincide.
    – Taroccoesbrocco
    Aug 2 at 16:05











  • @user10168997 - Pay attention that a CNF of a given formula is not unique, and similarly for DNF. For instance, $A land (A lor B lor lnot C)$ is a CNF of $(A land B) lor ((lnot C lor A) land lnot B) lor B$ as well. So, properly speaking it is not clear what you refer to when you talk of the CNF of a given formula.
    – Taroccoesbrocco
    Aug 2 at 16:08
















would the DNF be the same?
– user10168997
Aug 2 at 14:44




would the DNF be the same?
– user10168997
Aug 2 at 14:44












@user10168997 - Yes, of course. In this case, your starting formula $(A land B) lor ((lnot C lor A) land lnot B) lor B$ has a CNF and a DNF that coincide.
– Taroccoesbrocco
Aug 2 at 16:05





@user10168997 - Yes, of course. In this case, your starting formula $(A land B) lor ((lnot C lor A) land lnot B) lor B$ has a CNF and a DNF that coincide.
– Taroccoesbrocco
Aug 2 at 16:05













@user10168997 - Pay attention that a CNF of a given formula is not unique, and similarly for DNF. For instance, $A land (A lor B lor lnot C)$ is a CNF of $(A land B) lor ((lnot C lor A) land lnot B) lor B$ as well. So, properly speaking it is not clear what you refer to when you talk of the CNF of a given formula.
– Taroccoesbrocco
Aug 2 at 16:08





@user10168997 - Pay attention that a CNF of a given formula is not unique, and similarly for DNF. For instance, $A land (A lor B lor lnot C)$ is a CNF of $(A land B) lor ((lnot C lor A) land lnot B) lor B$ as well. So, properly speaking it is not clear what you refer to when you talk of the CNF of a given formula.
– Taroccoesbrocco
Aug 2 at 16:08













 

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