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

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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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  • 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














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





















  • 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












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













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












share|cite|improve this question




share|cite|improve this question








edited Aug 6 at 21:54









Shaun

7,41792972




7,41792972









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
















  • 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










2 Answers
2






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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 $(Cveelnot C)$.






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






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      2 Answers
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      2 Answers
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      up vote
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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



























        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

























          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















          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















          share|cite|improve this answer



          share|cite|improve this answer








          edited Aug 6 at 21:29


























          answered Aug 6 at 21:24









          Rushabh Mehta

          1,060114




          1,060114




















              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

























                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























                  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













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







                  share|cite|improve this answer













                  share|cite|improve this answer



                  share|cite|improve this answer











                  answered Aug 6 at 21:23









                  Graham Kemp

                  80.1k43275




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