Duality discrete math problem
$begingroup$
This is the only answer I got wrong on my HW and the prof does not want to give us the correct answers before our midterm
The dual of a compound proposition that contains only the logical operators $lor$ , $land$ , and
$neg$ is the compound proposition obtained by replacing each $lor$ by $land$ , each $land$ by $lor$ , each $defT{{rm T}}defF{{rm F}}$
$T$ by $F$ , and each $F$ by $T$ . The dual of $s$ is denoted by $s^*$. Find the dual of these compound propositions.
a) $p lorneg q$
I got $neg p land q$
b) $p land (q lor (r land T))$
My answer was $neg p lor (neg q land r)$
c) $(p land neg q) lor (q land F)$
My answer was $(neg p lor q) land neg q$
I have tried googling the problem and cannot come up with anything on duals and our lectures are online and upon reviewing do not see anything. I am just confused and looking for a little guidance on what was incorrect with my answers.
propositional-calculus
edited Jul 21 '18 at 16:41
Rodrigo de Azevedo
13.2k41962
asked Oct 17 '13 at 9:59
KevinKevin
20115
$endgroup$
add a comment |
$begingroup$
This is the only answer I got wrong on my HW and the prof does not want to give us the correct answers before our midterm
The dual of a compound proposition that contains only the logical operators $lor$ , $land$ , and
$neg$ is the compound proposition obtained by replacing each $lor$ by $land$ , each $land$ by $lor$ , each $defT{{rm T}}defF{{rm F}}$
$T$ by $F$ , and each $F$ by $T$ . The dual of $s$ is denoted by $s^*$. Find the dual of these compound propositions.
a) $p lorneg q$
I got $neg p land q$
b) $p land (q lor (r land T))$
My answer was $neg p lor (neg q land r)$
c) $(p land neg q) lor (q land F)$
My answer was $(neg p lor q) land neg q$
I have tried googling the problem and cannot come up with anything on duals and our lectures are online and upon reviewing do not see anything. I am just confused and looking for a little guidance on what was incorrect with my answers.
propositional-calculus
edited Jul 21 '18 at 16:41
Rodrigo de Azevedo
13.2k41962
asked Oct 17 '13 at 9:59
KevinKevin
20115
$endgroup$
1
$begingroup$
This can't be "easy" if you didn't get it right. How about a slightly more descriptive title instead?
$endgroup$
– Asaf Karagila♦
Oct 17 '13 at 10:11
$begingroup$
Sorry about the misleading title. When I first did it I thought it was very simple and now that I see the answer it was just a simple mistake on my part.
$endgroup$
– Kevin
Oct 17 '13 at 11:03
add a comment |
$begingroup$
This is the only answer I got wrong on my HW and the prof does not want to give us the correct answers before our midterm
The dual of a compound proposition that contains only the logical operators $lor$ , $land$ , and
$neg$ is the compound proposition obtained by replacing each $lor$ by $land$ , each $land$ by $lor$ , each $defT{{rm T}}defF{{rm F}}$
$T$ by $F$ , and each $F$ by $T$ . The dual of $s$ is denoted by $s^*$. Find the dual of these compound propositions.
a) $p lorneg q$
I got $neg p land q$
b) $p land (q lor (r land T))$
My answer was $neg p lor (neg q land r)$
c) $(p land neg q) lor (q land F)$
My answer was $(neg p lor q) land neg q$
I have tried googling the problem and cannot come up with anything on duals and our lectures are online and upon reviewing do not see anything. I am just confused and looking for a little guidance on what was incorrect with my answers.
propositional-calculus
edited Jul 21 '18 at 16:41
Rodrigo de Azevedo
13.2k41962
asked Oct 17 '13 at 9:59
KevinKevin
20115
$endgroup$
This is the only answer I got wrong on my HW and the prof does not want to give us the correct answers before our midterm
The dual of a compound proposition that contains only the logical operators $lor$ , $land$ , and
$neg$ is the compound proposition obtained by replacing each $lor$ by $land$ , each $land$ by $lor$ , each $defT{{rm T}}defF{{rm F}}$
$T$ by $F$ , and each $F$ by $T$ . The dual of $s$ is denoted by $s^*$. Find the dual of these compound propositions.
a) $p lorneg q$
I got $neg p land q$
b) $p land (q lor (r land T))$
My answer was $neg p lor (neg q land r)$
c) $(p land neg q) lor (q land F)$
My answer was $(neg p lor q) land neg q$
I have tried googling the problem and cannot come up with anything on duals and our lectures are online and upon reviewing do not see anything. I am just confused and looking for a little guidance on what was incorrect with my answers.
propositional-calculus
propositional-calculus
edited Jul 21 '18 at 16:41
Rodrigo de Azevedo
13.2k41962
asked Oct 17 '13 at 9:59
KevinKevin
20115
edited Jul 21 '18 at 16:41
Rodrigo de Azevedo
13.2k41962
asked Oct 17 '13 at 9:59
KevinKevin
20115
edited Jul 21 '18 at 16:41
Rodrigo de Azevedo
13.2k41962
edited Jul 21 '18 at 16:41
Rodrigo de Azevedo
13.2k41962
edited Jul 21 '18 at 16:41
Rodrigo de Azevedo
13.2k41962
13.2k41962
asked Oct 17 '13 at 9:59
KevinKevin
20115
asked Oct 17 '13 at 9:59
KevinKevin
20115
asked Oct 17 '13 at 9:59
KevinKevin
20115
20115
1
$begingroup$
This can't be "easy" if you didn't get it right. How about a slightly more descriptive title instead?
$endgroup$
– Asaf Karagila♦
Oct 17 '13 at 10:11
$begingroup$
Sorry about the misleading title. When I first did it I thought it was very simple and now that I see the answer it was just a simple mistake on my part.
$endgroup$
– Kevin
Oct 17 '13 at 11:03
add a comment |
1
$begingroup$
This can't be "easy" if you didn't get it right. How about a slightly more descriptive title instead?
$endgroup$
– Asaf Karagila♦
Oct 17 '13 at 10:11
$begingroup$
Sorry about the misleading title. When I first did it I thought it was very simple and now that I see the answer it was just a simple mistake on my part.
$endgroup$
– Kevin
Oct 17 '13 at 11:03
1
1
$begingroup$
This can't be "easy" if you didn't get it right. How about a slightly more descriptive title instead?
$endgroup$
– Asaf Karagila♦
Oct 17 '13 at 10:11
$begingroup$
This can't be "easy" if you didn't get it right. How about a slightly more descriptive title instead?
$endgroup$
– Asaf Karagila♦
Oct 17 '13 at 10:11
$begingroup$
Sorry about the misleading title. When I first did it I thought it was very simple and now that I see the answer it was just a simple mistake on my part.
$endgroup$
– Kevin
Oct 17 '13 at 11:03
$begingroup$
Sorry about the misleading title. When I first did it I thought it was very simple and now that I see the answer it was just a simple mistake on my part.
$endgroup$
– Kevin
Oct 17 '13 at 11:03
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
You did more as you should. Forming the dual just wants you to replace $p$ by $neg p$ for each literal $p$, $lor$ by $land$ and vice versa and $T$ by $F$. You did more than that, in dualising (2), one obtains
$$ neg p lor bigl( neg q land (neg r lor F)bigr) $$
(you missed a $neg$ in front of $r$). We have of course $neg r lor F equiv neg r$, but this is not part of dualising. Same for (3), the dual proposition is
$$ (neg p lor q) land (neg q lor T)$$
edited Jul 21 '18 at 15:40
Parcly Taxel
44.8k1376111
answered Oct 17 '13 at 10:14
martinimartini
70.9k45992
$endgroup$
$begingroup$
Did you mean $neg r vee F equiv neg r $?
$endgroup$
– Ludolila
Oct 17 '13 at 10:34
$begingroup$
@Ludolila Yes, corrected, thx.
$endgroup$
– martini
Oct 17 '13 at 10:35
$begingroup$
I see I thought I was supposed to reduce it by eliminating the T or F statement. Would part A be correct then? Thanks for the help!
$endgroup$
– Kevin
Oct 17 '13 at 11:02
add a comment |
$begingroup$
What you have done is wrong. Why you have negated all the proposition and the then find the dual?
Like dual of (p ∧ ¬q) is (p ∨ ¬q) not (¬p ∨ q).
Here is the definition of dual of a compound proposition:
The dual of a compound proposition that contains only the logical operators ∨, ∧, and ¬ is the compound proposition obtained by replacing each ∨ by ∧, each ∧ by ∨, each T by F, and each F by T. The dual of s is denoted by s∗. (Reference: Discrete Mathematics (7th Edition) Kenneth H. Rosen)
Example:
S =(p ∧ q)∨ (¬p ∨ q)∨ F
dual of S = (p ∨ q)∧ (¬p ∧ q)∧T
So the correct answer to your question is:
a) p∨¬q
dual: p∧¬q
b) p∧(q∨(r∧T))
dual: p∨(q∧(r∨F))
c) (p∧¬q)∨(q∧F)
dual: (p∨¬q)∧(q∨T)
answered Jan 18 at 16:49
Sumit RanjanSumit Ranjan
368
$endgroup$
add a comment |
$begingroup$
To obtain the dual of a formula , replace ∧ with V and T with F vice versa..
eg: dual of pVq is p∧q it is not ¬p∧¬q
reference: Discrete mathematical structures with applications to computer science by J.P Tremblay and R.Manohar.
answered Aug 18 '14 at 4:50
sowjanyasowjanya
1
$endgroup$
add a comment |
$begingroup$
When I first looked that this problem in Rosens, Discrete Mathematics and Its Applications, I was very confused. After a computer driven CTRL-F search, the word dual seemed to only be in the book twice. They Explain what a dual is inside the problem. If you're this far along in the assignment, then the real problem for the 3 part question is the way the word the question.
"The dual of a compound proposition that contains only the
logical operators ∨, ∧, and ¬ is the compound proposition
obtained by replacing each ∨ by ∧, each ∧ by ∨, each T
by F, and each F by T. The dual of s is denoted by s
∗."
The wording makes it seem as though the dual of an equation is found by switching
- V to ∧
- ∧ to V
- T to F
- F to T
- ¬q to q
but really, its just
- V to ∧
- ∧ to V
- T to F
- F to T
No need for changing the negations
They included the "extra" information on the negation because you can only find the dual of problems containing those symbols. So a problem containing a bi-conditional would not work. As far as we know this far in the book.
answered Sep 13 '16 at 22:01
Julian WeidnerJulian Weidner
1
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add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You did more as you should. Forming the dual just wants you to replace $p$ by $neg p$ for each literal $p$, $lor$ by $land$ and vice versa and $T$ by $F$. You did more than that, in dualising (2), one obtains
$$ neg p lor bigl( neg q land (neg r lor F)bigr) $$
(you missed a $neg$ in front of $r$). We have of course $neg r lor F equiv neg r$, but this is not part of dualising. Same for (3), the dual proposition is
$$ (neg p lor q) land (neg q lor T)$$
edited Jul 21 '18 at 15:40
Parcly Taxel
44.8k1376111
answered Oct 17 '13 at 10:14
martinimartini
70.9k45992
$endgroup$
$begingroup$
Did you mean $neg r vee F equiv neg r $?
$endgroup$
– Ludolila
Oct 17 '13 at 10:34
$begingroup$
@Ludolila Yes, corrected, thx.
$endgroup$
– martini
Oct 17 '13 at 10:35
$begingroup$
I see I thought I was supposed to reduce it by eliminating the T or F statement. Would part A be correct then? Thanks for the help!
$endgroup$
– Kevin
Oct 17 '13 at 11:02
add a comment |
$begingroup$
You did more as you should. Forming the dual just wants you to replace $p$ by $neg p$ for each literal $p$, $lor$ by $land$ and vice versa and $T$ by $F$. You did more than that, in dualising (2), one obtains
$$ neg p lor bigl( neg q land (neg r lor F)bigr) $$
(you missed a $neg$ in front of $r$). We have of course $neg r lor F equiv neg r$, but this is not part of dualising. Same for (3), the dual proposition is
$$ (neg p lor q) land (neg q lor T)$$
edited Jul 21 '18 at 15:40
Parcly Taxel
44.8k1376111
answered Oct 17 '13 at 10:14
martinimartini
70.9k45992
$endgroup$
$begingroup$
Did you mean $neg r vee F equiv neg r $?
$endgroup$
– Ludolila
Oct 17 '13 at 10:34
$begingroup$
@Ludolila Yes, corrected, thx.
$endgroup$
– martini
Oct 17 '13 at 10:35
$begingroup$
I see I thought I was supposed to reduce it by eliminating the T or F statement. Would part A be correct then? Thanks for the help!
$endgroup$
– Kevin
Oct 17 '13 at 11:02
add a comment |
$begingroup$
You did more as you should. Forming the dual just wants you to replace $p$ by $neg p$ for each literal $p$, $lor$ by $land$ and vice versa and $T$ by $F$. You did more than that, in dualising (2), one obtains
$$ neg p lor bigl( neg q land (neg r lor F)bigr) $$
(you missed a $neg$ in front of $r$). We have of course $neg r lor F equiv neg r$, but this is not part of dualising. Same for (3), the dual proposition is
$$ (neg p lor q) land (neg q lor T)$$
edited Jul 21 '18 at 15:40
Parcly Taxel
44.8k1376111
answered Oct 17 '13 at 10:14
martinimartini
70.9k45992
$endgroup$
You did more as you should. Forming the dual just wants you to replace $p$ by $neg p$ for each literal $p$, $lor$ by $land$ and vice versa and $T$ by $F$. You did more than that, in dualising (2), one obtains
$$ neg p lor bigl( neg q land (neg r lor F)bigr) $$
(you missed a $neg$ in front of $r$). We have of course $neg r lor F equiv neg r$, but this is not part of dualising. Same for (3), the dual proposition is
$$ (neg p lor q) land (neg q lor T)$$
edited Jul 21 '18 at 15:40
Parcly Taxel
44.8k1376111
answered Oct 17 '13 at 10:14
martinimartini
70.9k45992
edited Jul 21 '18 at 15:40
Parcly Taxel
44.8k1376111
edited Jul 21 '18 at 15:40
Parcly Taxel
44.8k1376111
edited Jul 21 '18 at 15:40
Parcly Taxel
44.8k1376111
44.8k1376111
answered Oct 17 '13 at 10:14
martinimartini
70.9k45992
answered Oct 17 '13 at 10:14
martinimartini
70.9k45992
answered Oct 17 '13 at 10:14
martinimartini
70.9k45992
70.9k45992
$begingroup$
Did you mean $neg r vee F equiv neg r $?
$endgroup$
– Ludolila
Oct 17 '13 at 10:34
$begingroup$
@Ludolila Yes, corrected, thx.
$endgroup$
– martini
Oct 17 '13 at 10:35
$begingroup$
I see I thought I was supposed to reduce it by eliminating the T or F statement. Would part A be correct then? Thanks for the help!
$endgroup$
– Kevin
Oct 17 '13 at 11:02
add a comment |
$begingroup$
Did you mean $neg r vee F equiv neg r $?
$endgroup$
– Ludolila
Oct 17 '13 at 10:34
$begingroup$
@Ludolila Yes, corrected, thx.
$endgroup$
– martini
Oct 17 '13 at 10:35
$begingroup$
I see I thought I was supposed to reduce it by eliminating the T or F statement. Would part A be correct then? Thanks for the help!
$endgroup$
– Kevin
Oct 17 '13 at 11:02
$begingroup$
Did you mean $neg r vee F equiv neg r $?
$endgroup$
– Ludolila
Oct 17 '13 at 10:34
$begingroup$
Did you mean $neg r vee F equiv neg r $?
$endgroup$
– Ludolila
Oct 17 '13 at 10:34
$begingroup$
@Ludolila Yes, corrected, thx.
$endgroup$
– martini
Oct 17 '13 at 10:35
$begingroup$
@Ludolila Yes, corrected, thx.
$endgroup$
– martini
Oct 17 '13 at 10:35
$begingroup$
I see I thought I was supposed to reduce it by eliminating the T or F statement. Would part A be correct then? Thanks for the help!
$endgroup$
– Kevin
Oct 17 '13 at 11:02
$begingroup$
I see I thought I was supposed to reduce it by eliminating the T or F statement. Would part A be correct then? Thanks for the help!
$endgroup$
– Kevin
Oct 17 '13 at 11:02
add a comment |
$begingroup$
What you have done is wrong. Why you have negated all the proposition and the then find the dual?
Like dual of (p ∧ ¬q) is (p ∨ ¬q) not (¬p ∨ q).
Here is the definition of dual of a compound proposition:
The dual of a compound proposition that contains only the logical operators ∨, ∧, and ¬ is the compound proposition obtained by replacing each ∨ by ∧, each ∧ by ∨, each T by F, and each F by T. The dual of s is denoted by s∗. (Reference: Discrete Mathematics (7th Edition) Kenneth H. Rosen)
Example:
S =(p ∧ q)∨ (¬p ∨ q)∨ F
dual of S = (p ∨ q)∧ (¬p ∧ q)∧T
So the correct answer to your question is:
a) p∨¬q
dual: p∧¬q
b) p∧(q∨(r∧T))
dual: p∨(q∧(r∨F))
c) (p∧¬q)∨(q∧F)
dual: (p∨¬q)∧(q∨T)
answered Jan 18 at 16:49
Sumit RanjanSumit Ranjan
368
$endgroup$
add a comment |
$begingroup$
What you have done is wrong. Why you have negated all the proposition and the then find the dual?
Like dual of (p ∧ ¬q) is (p ∨ ¬q) not (¬p ∨ q).
Here is the definition of dual of a compound proposition:
The dual of a compound proposition that contains only the logical operators ∨, ∧, and ¬ is the compound proposition obtained by replacing each ∨ by ∧, each ∧ by ∨, each T by F, and each F by T. The dual of s is denoted by s∗. (Reference: Discrete Mathematics (7th Edition) Kenneth H. Rosen)
Example:
S =(p ∧ q)∨ (¬p ∨ q)∨ F
dual of S = (p ∨ q)∧ (¬p ∧ q)∧T
So the correct answer to your question is:
a) p∨¬q
dual: p∧¬q
b) p∧(q∨(r∧T))
dual: p∨(q∧(r∨F))
c) (p∧¬q)∨(q∧F)
dual: (p∨¬q)∧(q∨T)
answered Jan 18 at 16:49
Sumit RanjanSumit Ranjan
368
$endgroup$
add a comment |
$begingroup$
What you have done is wrong. Why you have negated all the proposition and the then find the dual?
Like dual of (p ∧ ¬q) is (p ∨ ¬q) not (¬p ∨ q).
Here is the definition of dual of a compound proposition:
The dual of a compound proposition that contains only the logical operators ∨, ∧, and ¬ is the compound proposition obtained by replacing each ∨ by ∧, each ∧ by ∨, each T by F, and each F by T. The dual of s is denoted by s∗. (Reference: Discrete Mathematics (7th Edition) Kenneth H. Rosen)
Example:
S =(p ∧ q)∨ (¬p ∨ q)∨ F
dual of S = (p ∨ q)∧ (¬p ∧ q)∧T
So the correct answer to your question is:
a) p∨¬q
dual: p∧¬q
b) p∧(q∨(r∧T))
dual: p∨(q∧(r∨F))
c) (p∧¬q)∨(q∧F)
dual: (p∨¬q)∧(q∨T)
answered Jan 18 at 16:49
Sumit RanjanSumit Ranjan
368
$endgroup$
What you have done is wrong. Why you have negated all the proposition and the then find the dual?
Like dual of (p ∧ ¬q) is (p ∨ ¬q) not (¬p ∨ q).
Here is the definition of dual of a compound proposition:
The dual of a compound proposition that contains only the logical operators ∨, ∧, and ¬ is the compound proposition obtained by replacing each ∨ by ∧, each ∧ by ∨, each T by F, and each F by T. The dual of s is denoted by s∗. (Reference: Discrete Mathematics (7th Edition) Kenneth H. Rosen)
Example:
S =(p ∧ q)∨ (¬p ∨ q)∨ F
dual of S = (p ∨ q)∧ (¬p ∧ q)∧T
So the correct answer to your question is:
a) p∨¬q
dual: p∧¬q
b) p∧(q∨(r∧T))
dual: p∨(q∧(r∨F))
c) (p∧¬q)∨(q∧F)
dual: (p∨¬q)∧(q∨T)
answered Jan 18 at 16:49
Sumit RanjanSumit Ranjan
368
edited Jan 18 at 16:54
answered Jan 18 at 16:49
Sumit RanjanSumit Ranjan
368
answered Jan 18 at 16:49
Sumit RanjanSumit Ranjan
368
answered Jan 18 at 16:49
Sumit RanjanSumit Ranjan
368
368
add a comment |
add a comment |
$begingroup$
To obtain the dual of a formula , replace ∧ with V and T with F vice versa..
eg: dual of pVq is p∧q it is not ¬p∧¬q
reference: Discrete mathematical structures with applications to computer science by J.P Tremblay and R.Manohar.
answered Aug 18 '14 at 4:50
sowjanyasowjanya
1
$endgroup$
add a comment |
$begingroup$
To obtain the dual of a formula , replace ∧ with V and T with F vice versa..
eg: dual of pVq is p∧q it is not ¬p∧¬q
reference: Discrete mathematical structures with applications to computer science by J.P Tremblay and R.Manohar.
answered Aug 18 '14 at 4:50
sowjanyasowjanya
1
$endgroup$
add a comment |
$begingroup$
To obtain the dual of a formula , replace ∧ with V and T with F vice versa..
eg: dual of pVq is p∧q it is not ¬p∧¬q
reference: Discrete mathematical structures with applications to computer science by J.P Tremblay and R.Manohar.
answered Aug 18 '14 at 4:50
sowjanyasowjanya
1
$endgroup$
To obtain the dual of a formula , replace ∧ with V and T with F vice versa..
eg: dual of pVq is p∧q it is not ¬p∧¬q
reference: Discrete mathematical structures with applications to computer science by J.P Tremblay and R.Manohar.
answered Aug 18 '14 at 4:50
sowjanyasowjanya
1
answered Aug 18 '14 at 4:50
sowjanyasowjanya
1
answered Aug 18 '14 at 4:50
sowjanyasowjanya
1
answered Aug 18 '14 at 4:50
sowjanyasowjanya
1
1
add a comment |
add a comment |
$begingroup$
When I first looked that this problem in Rosens, Discrete Mathematics and Its Applications, I was very confused. After a computer driven CTRL-F search, the word dual seemed to only be in the book twice. They Explain what a dual is inside the problem. If you're this far along in the assignment, then the real problem for the 3 part question is the way the word the question.
"The dual of a compound proposition that contains only the
logical operators ∨, ∧, and ¬ is the compound proposition
obtained by replacing each ∨ by ∧, each ∧ by ∨, each T
by F, and each F by T. The dual of s is denoted by s
∗."
The wording makes it seem as though the dual of an equation is found by switching
- V to ∧
- ∧ to V
- T to F
- F to T
- ¬q to q
but really, its just
- V to ∧
- ∧ to V
- T to F
- F to T
No need for changing the negations
They included the "extra" information on the negation because you can only find the dual of problems containing those symbols. So a problem containing a bi-conditional would not work. As far as we know this far in the book.
answered Sep 13 '16 at 22:01
Julian WeidnerJulian Weidner
1
$endgroup$
add a comment |
$begingroup$
When I first looked that this problem in Rosens, Discrete Mathematics and Its Applications, I was very confused. After a computer driven CTRL-F search, the word dual seemed to only be in the book twice. They Explain what a dual is inside the problem. If you're this far along in the assignment, then the real problem for the 3 part question is the way the word the question.
"The dual of a compound proposition that contains only the
logical operators ∨, ∧, and ¬ is the compound proposition
obtained by replacing each ∨ by ∧, each ∧ by ∨, each T
by F, and each F by T. The dual of s is denoted by s
∗."
The wording makes it seem as though the dual of an equation is found by switching
- V to ∧
- ∧ to V
- T to F
- F to T
- ¬q to q
but really, its just
- V to ∧
- ∧ to V
- T to F
- F to T
No need for changing the negations
They included the "extra" information on the negation because you can only find the dual of problems containing those symbols. So a problem containing a bi-conditional would not work. As far as we know this far in the book.
answered Sep 13 '16 at 22:01
Julian WeidnerJulian Weidner
1
$endgroup$
add a comment |
$begingroup$
When I first looked that this problem in Rosens, Discrete Mathematics and Its Applications, I was very confused. After a computer driven CTRL-F search, the word dual seemed to only be in the book twice. They Explain what a dual is inside the problem. If you're this far along in the assignment, then the real problem for the 3 part question is the way the word the question.
"The dual of a compound proposition that contains only the
logical operators ∨, ∧, and ¬ is the compound proposition
obtained by replacing each ∨ by ∧, each ∧ by ∨, each T
by F, and each F by T. The dual of s is denoted by s
∗."
The wording makes it seem as though the dual of an equation is found by switching
- V to ∧
- ∧ to V
- T to F
- F to T
- ¬q to q
but really, its just
- V to ∧
- ∧ to V
- T to F
- F to T
No need for changing the negations
They included the "extra" information on the negation because you can only find the dual of problems containing those symbols. So a problem containing a bi-conditional would not work. As far as we know this far in the book.
answered Sep 13 '16 at 22:01
Julian WeidnerJulian Weidner
1
$endgroup$
When I first looked that this problem in Rosens, Discrete Mathematics and Its Applications, I was very confused. After a computer driven CTRL-F search, the word dual seemed to only be in the book twice. They Explain what a dual is inside the problem. If you're this far along in the assignment, then the real problem for the 3 part question is the way the word the question.
"The dual of a compound proposition that contains only the
logical operators ∨, ∧, and ¬ is the compound proposition
obtained by replacing each ∨ by ∧, each ∧ by ∨, each T
by F, and each F by T. The dual of s is denoted by s
∗."
The wording makes it seem as though the dual of an equation is found by switching
- V to ∧
- ∧ to V
- T to F
- F to T
- ¬q to q
but really, its just
- V to ∧
- ∧ to V
- T to F
- F to T
No need for changing the negations
They included the "extra" information on the negation because you can only find the dual of problems containing those symbols. So a problem containing a bi-conditional would not work. As far as we know this far in the book.
answered Sep 13 '16 at 22:01
Julian WeidnerJulian Weidner
1
answered Sep 13 '16 at 22:01
Julian WeidnerJulian Weidner
1
answered Sep 13 '16 at 22:01
Julian WeidnerJulian Weidner
1
answered Sep 13 '16 at 22:01
Julian WeidnerJulian Weidner
1
1
add a comment |
add a comment |
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$begingroup$
This can't be "easy" if you didn't get it right. How about a slightly more descriptive title instead?
$endgroup$
– Asaf Karagila♦
Oct 17 '13 at 10:11
$begingroup$
Sorry about the misleading title. When I first did it I thought it was very simple and now that I see the answer it was just a simple mistake on my part.
$endgroup$
– Kevin
Oct 17 '13 at 11:03