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Can the CNF be the negation of the DNF?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Getting the CNF and DNF (Logic)Giving this formula in DNF and CNF propositional logicTurning CNF into DNFConverting DNF to CNF and vice versaNegation in Disjunctive Normal FormProve that any truth function $f$ can be represented by a formula $φ$ in cnf by negating a formula in dnfIs this the disjunctive and conjunctive nromal form for my porpositional formula F?Are my DNF and CNF for $A land (A lor C) implies (C lor B)$ correct?CNF quantifier eliminationIs it possible to get the CNF out of the DNF of this expression
$begingroup$
I've got the following question: Asking for the CNF of the following formula:
$$(lnot X lor Y)to bigl( Zlandlnot(lnot Yland X) bigr)$$
So the process should be like that:
$$lnot (lnot X lor Y)lor bigl( Zland (Ylor lnot X)bigr)$$
$$(Xland lnot Y)lor bigl((Zland Y)lor (Zland lnot X)bigr) $$
$$ (Xland lnot Y)lor (Zland Y)lor (Zland lnot X) $$
As we can see, It's nothing other than the DNF of the above formula, Can i say that the CNF is just the negation of this DNF?
The negation of this DNF looks like the following:
$$ (lnot Xlor Y)land (lnot Zlor lnot Y)land (lnot Zlor X) $$
Is it correct to say that this is the CNF of the above formula?
Thanks!!!
logic
asked Mar 27 at 19:48
D.RotnemerD.Rotnemer
14016
$endgroup$
add a comment |
$begingroup$
I've got the following question: Asking for the CNF of the following formula:
$$(lnot X lor Y)to bigl( Zlandlnot(lnot Yland X) bigr)$$
So the process should be like that:
$$lnot (lnot X lor Y)lor bigl( Zland (Ylor lnot X)bigr)$$
$$(Xland lnot Y)lor bigl((Zland Y)lor (Zland lnot X)bigr) $$
$$ (Xland lnot Y)lor (Zland Y)lor (Zland lnot X) $$
As we can see, It's nothing other than the DNF of the above formula, Can i say that the CNF is just the negation of this DNF?
The negation of this DNF looks like the following:
$$ (lnot Xlor Y)land (lnot Zlor lnot Y)land (lnot Zlor X) $$
Is it correct to say that this is the CNF of the above formula?
Thanks!!!
logic
asked Mar 27 at 19:48
D.RotnemerD.Rotnemer
14016
$endgroup$
$begingroup$
What you've found is a CNF, but not that of the above formula. It is a CNF of the negation.
$endgroup$
– user376343
Mar 27 at 20:25
$begingroup$
if you can use Karnaugh maps, do as follows: from K.map of the given formula you get easily the map of the negation, and also the DNF of the negation. Then you can use your trick to obtain the CNF it the given formula.
$endgroup$
– user376343
Mar 27 at 20:35
add a comment |
$begingroup$
I've got the following question: Asking for the CNF of the following formula:
$$(lnot X lor Y)to bigl( Zlandlnot(lnot Yland X) bigr)$$
So the process should be like that:
$$lnot (lnot X lor Y)lor bigl( Zland (Ylor lnot X)bigr)$$
$$(Xland lnot Y)lor bigl((Zland Y)lor (Zland lnot X)bigr) $$
$$ (Xland lnot Y)lor (Zland Y)lor (Zland lnot X) $$
As we can see, It's nothing other than the DNF of the above formula, Can i say that the CNF is just the negation of this DNF?
The negation of this DNF looks like the following:
$$ (lnot Xlor Y)land (lnot Zlor lnot Y)land (lnot Zlor X) $$
Is it correct to say that this is the CNF of the above formula?
Thanks!!!
logic
asked Mar 27 at 19:48
D.RotnemerD.Rotnemer
14016
$endgroup$
I've got the following question: Asking for the CNF of the following formula:
$$(lnot X lor Y)to bigl( Zlandlnot(lnot Yland X) bigr)$$
So the process should be like that:
$$lnot (lnot X lor Y)lor bigl( Zland (Ylor lnot X)bigr)$$
$$(Xland lnot Y)lor bigl((Zland Y)lor (Zland lnot X)bigr) $$
$$ (Xland lnot Y)lor (Zland Y)lor (Zland lnot X) $$
As we can see, It's nothing other than the DNF of the above formula, Can i say that the CNF is just the negation of this DNF?
The negation of this DNF looks like the following:
$$ (lnot Xlor Y)land (lnot Zlor lnot Y)land (lnot Zlor X) $$
Is it correct to say that this is the CNF of the above formula?
Thanks!!!
logic
logic
asked Mar 27 at 19:48
D.RotnemerD.Rotnemer
14016
asked Mar 27 at 19:48
D.RotnemerD.Rotnemer
14016
asked Mar 27 at 19:48
D.RotnemerD.Rotnemer
14016
asked Mar 27 at 19:48
D.RotnemerD.Rotnemer
14016
asked Mar 27 at 19:48
D.RotnemerD.Rotnemer
14016
14016
$begingroup$
What you've found is a CNF, but not that of the above formula. It is a CNF of the negation.
$endgroup$
– user376343
Mar 27 at 20:25
$begingroup$
if you can use Karnaugh maps, do as follows: from K.map of the given formula you get easily the map of the negation, and also the DNF of the negation. Then you can use your trick to obtain the CNF it the given formula.
$endgroup$
– user376343
Mar 27 at 20:35
add a comment |
$begingroup$
What you've found is a CNF, but not that of the above formula. It is a CNF of the negation.
$endgroup$
– user376343
Mar 27 at 20:25
$begingroup$
if you can use Karnaugh maps, do as follows: from K.map of the given formula you get easily the map of the negation, and also the DNF of the negation. Then you can use your trick to obtain the CNF it the given formula.
$endgroup$
– user376343
Mar 27 at 20:35
$begingroup$
What you've found is a CNF, but not that of the above formula. It is a CNF of the negation.
$endgroup$
– user376343
Mar 27 at 20:25
$begingroup$
What you've found is a CNF, but not that of the above formula. It is a CNF of the negation.
$endgroup$
– user376343
Mar 27 at 20:25
$begingroup$
if you can use Karnaugh maps, do as follows: from K.map of the given formula you get easily the map of the negation, and also the DNF of the negation. Then you can use your trick to obtain the CNF it the given formula.
$endgroup$
– user376343
Mar 27 at 20:35
$begingroup$
if you can use Karnaugh maps, do as follows: from K.map of the given formula you get easily the map of the negation, and also the DNF of the negation. Then you can use your trick to obtain the CNF it the given formula.
$endgroup$
– user376343
Mar 27 at 20:35
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No, you cannot say that a CNF is the negation of the DNF, because it certainly is not so.
The normal forms of a statement are equivalent to that statement, and hence to each other.
$(Xlandlnot Y)lor (Zland Y)lor (Zlandlnot X)$ is a DNF for the statement, as is the simpler $(Xlandlnot Y)lor Z$. (Use distribution, commutation, de Morgan's, and absorption.)
Thus a CNF for that statement would be $(Xlor Z)land(lnot Ylor Z)$
answered Mar 27 at 22:40
Graham KempGraham Kemp
88.1k43579
$endgroup$
add a comment |
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$begingroup$
No, you cannot say that a CNF is the negation of the DNF, because it certainly is not so.
The normal forms of a statement are equivalent to that statement, and hence to each other.
$(Xlandlnot Y)lor (Zland Y)lor (Zlandlnot X)$ is a DNF for the statement, as is the simpler $(Xlandlnot Y)lor Z$. (Use distribution, commutation, de Morgan's, and absorption.)
Thus a CNF for that statement would be $(Xlor Z)land(lnot Ylor Z)$
answered Mar 27 at 22:40
Graham KempGraham Kemp
88.1k43579
$endgroup$
add a comment |
$begingroup$
No, you cannot say that a CNF is the negation of the DNF, because it certainly is not so.
The normal forms of a statement are equivalent to that statement, and hence to each other.
$(Xlandlnot Y)lor (Zland Y)lor (Zlandlnot X)$ is a DNF for the statement, as is the simpler $(Xlandlnot Y)lor Z$. (Use distribution, commutation, de Morgan's, and absorption.)
Thus a CNF for that statement would be $(Xlor Z)land(lnot Ylor Z)$
answered Mar 27 at 22:40
Graham KempGraham Kemp
88.1k43579
$endgroup$
add a comment |
$begingroup$
No, you cannot say that a CNF is the negation of the DNF, because it certainly is not so.
The normal forms of a statement are equivalent to that statement, and hence to each other.
$(Xlandlnot Y)lor (Zland Y)lor (Zlandlnot X)$ is a DNF for the statement, as is the simpler $(Xlandlnot Y)lor Z$. (Use distribution, commutation, de Morgan's, and absorption.)
Thus a CNF for that statement would be $(Xlor Z)land(lnot Ylor Z)$
answered Mar 27 at 22:40
Graham KempGraham Kemp
88.1k43579
$endgroup$
No, you cannot say that a CNF is the negation of the DNF, because it certainly is not so.
The normal forms of a statement are equivalent to that statement, and hence to each other.
$(Xlandlnot Y)lor (Zland Y)lor (Zlandlnot X)$ is a DNF for the statement, as is the simpler $(Xlandlnot Y)lor Z$. (Use distribution, commutation, de Morgan's, and absorption.)
Thus a CNF for that statement would be $(Xlor Z)land(lnot Ylor Z)$
answered Mar 27 at 22:40
Graham KempGraham Kemp
88.1k43579
answered Mar 27 at 22:40
Graham KempGraham Kemp
88.1k43579
answered Mar 27 at 22:40
Graham KempGraham Kemp
88.1k43579
answered Mar 27 at 22:40
Graham KempGraham Kemp
88.1k43579
88.1k43579
add a comment |
add a comment |
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$begingroup$
What you've found is a CNF, but not that of the above formula. It is a CNF of the negation.
$endgroup$
– user376343
Mar 27 at 20:25
$begingroup$
if you can use Karnaugh maps, do as follows: from K.map of the given formula you get easily the map of the negation, and also the DNF of the negation. Then you can use your trick to obtain the CNF it the given formula.
$endgroup$
– user376343
Mar 27 at 20:35