How to convert $((Alor B)land ((Bleftrightarrow A)rightarrow C))lor(Crightarrow lnot A)$ into CNF.

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up vote
1
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I started off with:

$$((Alor B)land ((Bleftrightarrow A)rightarrow C))lor(Crightarrow lnot A)$$

And I've managed to get to:

$$((Alor B) land (((Bland lnot A) lor (A land lnot B)) lor C)) lor (lnot C lor lnot A)$$

How do I finish and get this into CNF?

logic propositional-calculus

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edited Aug 6 at 21:54

Shaun

7,41792972

asked Aug 6 at 21:08

random.string

113

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Don't you mean $Bwedgelnot A$?
  – Rushabh Mehta
  Aug 6 at 21:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Typo? That should be $((Alor B) land (((Blandlnot A) lor (A land lnot B)) lor C)) lor (lnot C lor lnot A)$
  – Graham Kemp
  Aug 6 at 21:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @GrahamKemp You're right, fixed the typo thanks
  – random.string
  Aug 6 at 21:20
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  *right sorry for being a grammar nazi
  – Rushabh Mehta
  Aug 6 at 21:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Since there are only 3 variables, have you considered making a truth table?
  – Ðlon Djurinsky
  Aug 6 at 21:22
  

  
  
  
  

 | 
show 2 more comments

up vote
1
down vote

favorite

I started off with:

$$((Alor B)land ((Bleftrightarrow A)rightarrow C))lor(Crightarrow lnot A)$$

And I've managed to get to:

$$((Alor B) land (((Bland lnot A) lor (A land lnot B)) lor C)) lor (lnot C lor lnot A)$$

How do I finish and get this into CNF?

logic propositional-calculus

share|cite|improve this question

edited Aug 6 at 21:54

Shaun

7,41792972

asked Aug 6 at 21:08

random.string

113

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Don't you mean $Bwedgelnot A$?
  – Rushabh Mehta
  Aug 6 at 21:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Typo? That should be $((Alor B) land (((Blandlnot A) lor (A land lnot B)) lor C)) lor (lnot C lor lnot A)$
  – Graham Kemp
  Aug 6 at 21:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @GrahamKemp You're right, fixed the typo thanks
  – random.string
  Aug 6 at 21:20
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  *right sorry for being a grammar nazi
  – Rushabh Mehta
  Aug 6 at 21:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Since there are only 3 variables, have you considered making a truth table?
  – Ðlon Djurinsky
  Aug 6 at 21:22
  

  
  
  
  

 | 
show 2 more comments

up vote
1
down vote

favorite

up vote
1
down vote

favorite

I started off with:

$$((Alor B)land ((Bleftrightarrow A)rightarrow C))lor(Crightarrow lnot A)$$

And I've managed to get to:

$$((Alor B) land (((Bland lnot A) lor (A land lnot B)) lor C)) lor (lnot C lor lnot A)$$

How do I finish and get this into CNF?

logic propositional-calculus

share|cite|improve this question

edited Aug 6 at 21:54

Shaun

7,41792972

asked Aug 6 at 21:08

random.string

113

I started off with:

$$((Alor B)land ((Bleftrightarrow A)rightarrow C))lor(Crightarrow lnot A)$$

And I've managed to get to:

$$((Alor B) land (((Bland lnot A) lor (A land lnot B)) lor C)) lor (lnot C lor lnot A)$$

How do I finish and get this into CNF?

logic propositional-calculus

share|cite|improve this question

edited Aug 6 at 21:54

Shaun

7,41792972

asked Aug 6 at 21:08

random.string

113

share|cite|improve this question

share|cite|improve this question

edited Aug 6 at 21:54

Shaun

7,41792972

edited Aug 6 at 21:54

Shaun

7,41792972

edited Aug 6 at 21:54

Shaun

7,41792972

7,41792972

asked Aug 6 at 21:08

random.string

113

asked Aug 6 at 21:08

random.string

113

asked Aug 6 at 21:08

random.string

113

113

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Don't you mean $Bwedgelnot A$?
  – Rushabh Mehta
  Aug 6 at 21:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Typo? That should be $((Alor B) land (((Blandlnot A) lor (A land lnot B)) lor C)) lor (lnot C lor lnot A)$
  – Graham Kemp
  Aug 6 at 21:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @GrahamKemp You're right, fixed the typo thanks
  – random.string
  Aug 6 at 21:20
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  *right sorry for being a grammar nazi
  – Rushabh Mehta
  Aug 6 at 21:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Since there are only 3 variables, have you considered making a truth table?
  – Ðlon Djurinsky
  Aug 6 at 21:22
  

  
  
  
  

 | 
show 2 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Don't you mean $Bwedgelnot A$?
  – Rushabh Mehta
  Aug 6 at 21:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Typo? That should be $((Alor B) land (((Blandlnot A) lor (A land lnot B)) lor C)) lor (lnot C lor lnot A)$
  – Graham Kemp
  Aug 6 at 21:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @GrahamKemp You're right, fixed the typo thanks
  – random.string
  Aug 6 at 21:20
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  *right sorry for being a grammar nazi
  – Rushabh Mehta
  Aug 6 at 21:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Since there are only 3 variables, have you considered making a truth table?
  – Ðlon Djurinsky
  Aug 6 at 21:22
  

  
  
  
  

Don't you mean $Bwedgelnot A$?
– Rushabh Mehta
Aug 6 at 21:18

Don't you mean $Bwedgelnot A$?
– Rushabh Mehta
Aug 6 at 21:18

Typo? That should be $((Alor B) land (((Blandlnot A) lor (A land lnot B)) lor C)) lor (lnot C lor lnot A)$
– Graham Kemp
Aug 6 at 21:18

Typo? That should be $((Alor B) land (((Blandlnot A) lor (A land lnot B)) lor C)) lor (lnot C lor lnot A)$
– Graham Kemp
Aug 6 at 21:18

@GrahamKemp You're right, fixed the typo thanks
– random.string
Aug 6 at 21:20

@GrahamKemp You're right, fixed the typo thanks
– random.string
Aug 6 at 21:20

*right sorry for being a grammar nazi
– Rushabh Mehta
Aug 6 at 21:21

*right sorry for being a grammar nazi
– Rushabh Mehta
Aug 6 at 21:21

Since there are only 3 variables, have you considered making a truth table?
– Ðlon Djurinsky
Aug 6 at 21:22

Since there are only 3 variables, have you considered making a truth table?
– Ðlon Djurinsky
Aug 6 at 21:22

 | 
show 2 more comments

2 Answers
2

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1
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Note that if $lnot (Awedge C)$, this expression is true, so we don't need to consider the other expression. Hence, we only need to consider the larger clause for the case that $(Awedge C)$ is true.

Hence, $(Avee B)$ is always true, so a true statement conjuncted with another statement is logically equivalent to that other statement.

So, so far we have

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

Next, note that under our assumption that $C$ is true, we know that the inner expression is a clause disjuncted into $C$, which means it is always true.

Hence, this expression can never be not true. The simplest CNF that could be made for this expression is $(Cveelnot C)$.

share|cite|improve this answer

edited Aug 6 at 21:29

answered Aug 6 at 21:24

Rushabh Mehta

1,060114

add a comment | 

up vote
0
down vote

Hint: $neg(Aleftrightarrow B)$ $equiv (Awedgeneg B)vee(neg Awedge B)\ equiv (Avee B)wedge(neg Aveeneg B) $

share|cite|improve this answer

answered Aug 6 at 21:23

Graham Kemp

80.1k43275

add a comment | 

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2 Answers
2

active

oldest

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2 Answers
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active

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votes

active

oldest

votes

up vote
1
down vote

Note that if $lnot (Awedge C)$, this expression is true, so we don't need to consider the other expression. Hence, we only need to consider the larger clause for the case that $(Awedge C)$ is true.

Hence, $(Avee B)$ is always true, so a true statement conjuncted with another statement is logically equivalent to that other statement.

So, so far we have

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

Next, note that under our assumption that $C$ is true, we know that the inner expression is a clause disjuncted into $C$, which means it is always true.

Hence, this expression can never be not true. The simplest CNF that could be made for this expression is $(Cveelnot C)$.

share|cite|improve this answer

edited Aug 6 at 21:29

answered Aug 6 at 21:24

Rushabh Mehta

1,060114

add a comment | 

up vote
1
down vote

Note that if $lnot (Awedge C)$, this expression is true, so we don't need to consider the other expression. Hence, we only need to consider the larger clause for the case that $(Awedge C)$ is true.

Hence, $(Avee B)$ is always true, so a true statement conjuncted with another statement is logically equivalent to that other statement.

So, so far we have

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

Next, note that under our assumption that $C$ is true, we know that the inner expression is a clause disjuncted into $C$, which means it is always true.

Hence, this expression can never be not true. The simplest CNF that could be made for this expression is $(Cveelnot C)$.

share|cite|improve this answer

edited Aug 6 at 21:29

answered Aug 6 at 21:24

Rushabh Mehta

1,060114

add a comment | 

up vote
1
down vote

up vote
1
down vote

Note that if $lnot (Awedge C)$, this expression is true, so we don't need to consider the other expression. Hence, we only need to consider the larger clause for the case that $(Awedge C)$ is true.

Hence, $(Avee B)$ is always true, so a true statement conjuncted with another statement is logically equivalent to that other statement.

So, so far we have

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

Next, note that under our assumption that $C$ is true, we know that the inner expression is a clause disjuncted into $C$, which means it is always true.

Hence, this expression can never be not true. The simplest CNF that could be made for this expression is $(Cveelnot C)$.

share|cite|improve this answer

edited Aug 6 at 21:29

answered Aug 6 at 21:24

Rushabh Mehta

1,060114

Note that if $lnot (Awedge C)$, this expression is true, so we don't need to consider the other expression. Hence, we only need to consider the larger clause for the case that $(Awedge C)$ is true.

Hence, $(Avee B)$ is always true, so a true statement conjuncted with another statement is logically equivalent to that other statement.

So, so far we have

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

Next, note that under our assumption that $C$ is true, we know that the inner expression is a clause disjuncted into $C$, which means it is always true.

Hence, this expression can never be not true. The simplest CNF that could be made for this expression is $(Cveelnot C)$.

share|cite|improve this answer

edited Aug 6 at 21:29

answered Aug 6 at 21:24

Rushabh Mehta

1,060114

share|cite|improve this answer

share|cite|improve this answer

edited Aug 6 at 21:29

edited Aug 6 at 21:29

edited Aug 6 at 21:29

answered Aug 6 at 21:24

Rushabh Mehta

1,060114

answered Aug 6 at 21:24

Rushabh Mehta

1,060114

answered Aug 6 at 21:24

Rushabh Mehta

1,060114

1,060114

add a comment | 

add a comment | 

up vote
0
down vote

Hint: $neg(Aleftrightarrow B)$ $equiv (Awedgeneg B)vee(neg Awedge B)\ equiv (Avee B)wedge(neg Aveeneg B) $

share|cite|improve this answer

answered Aug 6 at 21:23

Graham Kemp

80.1k43275

add a comment | 

up vote
0
down vote

Hint: $neg(Aleftrightarrow B)$ $equiv (Awedgeneg B)vee(neg Awedge B)\ equiv (Avee B)wedge(neg Aveeneg B) $

share|cite|improve this answer

answered Aug 6 at 21:23

Graham Kemp

80.1k43275

add a comment | 

up vote
0
down vote

up vote
0
down vote

Hint: $neg(Aleftrightarrow B)$ $equiv (Awedgeneg B)vee(neg Awedge B)\ equiv (Avee B)wedge(neg Aveeneg B) $

share|cite|improve this answer

answered Aug 6 at 21:23

Graham Kemp

80.1k43275

Hint: $neg(Aleftrightarrow B)$ $equiv (Awedgeneg B)vee(neg Awedge B)\ equiv (Avee B)wedge(neg Aveeneg B) $

share|cite|improve this answer

answered Aug 6 at 21:23

Graham Kemp

80.1k43275

share|cite|improve this answer

share|cite|improve this answer

answered Aug 6 at 21:23

Graham Kemp

80.1k43275

answered Aug 6 at 21:23

Graham Kemp

80.1k43275

answered Aug 6 at 21:23

Graham Kemp

80.1k43275

80.1k43275

add a comment | 

add a comment | 

 

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