Dual of each of these compound propositions.
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I'm doing my Discrete Structure assignment. The question asked to find the dual of each of these compound propositions. Note that T denotes a tautology and F denotes a contradiction. The questions are:
a) $p∧¬q∧¬r$
b) $(p∧q∧r)∨s$
c) $(p∨F)∧(q∨T)$
My answers:
a) $¬p∨q∨r$
b) $(¬p∨¬q∨¬r)∧¬s$
c) $(¬p∧T)∨(¬q∧F)$
Did I got it right?
Thank you!
discrete-mathematics logic proof-verification
asked Feb 17 '16 at 2:17
Krin YongvongphaiboonKrin Yongvongphaiboon
64
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add a comment |
$begingroup$
I'm doing my Discrete Structure assignment. The question asked to find the dual of each of these compound propositions. Note that T denotes a tautology and F denotes a contradiction. The questions are:
a) $p∧¬q∧¬r$
b) $(p∧q∧r)∨s$
c) $(p∨F)∧(q∨T)$
My answers:
a) $¬p∨q∨r$
b) $(¬p∨¬q∨¬r)∧¬s$
c) $(¬p∧T)∨(¬q∧F)$
Did I got it right?
Thank you!
discrete-mathematics logic proof-verification
asked Feb 17 '16 at 2:17
Krin YongvongphaiboonKrin Yongvongphaiboon
64
$endgroup$
$begingroup$
What is your question?
$endgroup$
– Yeah..
Feb 17 '16 at 2:22
$begingroup$
@Yeah. The question appears to be : "Did I get it right?"
$endgroup$
– Graham Kemp
Feb 17 '16 at 2:23
$begingroup$
@GrahamKemp YES! :P
$endgroup$
– Krin Yongvongphaiboon
Feb 17 '16 at 2:24
add a comment |
$begingroup$
I'm doing my Discrete Structure assignment. The question asked to find the dual of each of these compound propositions. Note that T denotes a tautology and F denotes a contradiction. The questions are:
a) $p∧¬q∧¬r$
b) $(p∧q∧r)∨s$
c) $(p∨F)∧(q∨T)$
My answers:
a) $¬p∨q∨r$
b) $(¬p∨¬q∨¬r)∧¬s$
c) $(¬p∧T)∨(¬q∧F)$
Did I got it right?
Thank you!
discrete-mathematics logic proof-verification
asked Feb 17 '16 at 2:17
Krin YongvongphaiboonKrin Yongvongphaiboon
64
$endgroup$
I'm doing my Discrete Structure assignment. The question asked to find the dual of each of these compound propositions. Note that T denotes a tautology and F denotes a contradiction. The questions are:
a) $p∧¬q∧¬r$
b) $(p∧q∧r)∨s$
c) $(p∨F)∧(q∨T)$
My answers:
a) $¬p∨q∨r$
b) $(¬p∨¬q∨¬r)∧¬s$
c) $(¬p∧T)∨(¬q∧F)$
Did I got it right?
Thank you!
discrete-mathematics logic proof-verification
discrete-mathematics logic proof-verification
asked Feb 17 '16 at 2:17
Krin YongvongphaiboonKrin Yongvongphaiboon
64
asked Feb 17 '16 at 2:17
Krin YongvongphaiboonKrin Yongvongphaiboon
64
edited Feb 17 '16 at 2:25
Krin Yongvongphaiboon
asked Feb 17 '16 at 2:17
Krin YongvongphaiboonKrin Yongvongphaiboon
64
asked Feb 17 '16 at 2:17
Krin YongvongphaiboonKrin Yongvongphaiboon
64
asked Feb 17 '16 at 2:17
Krin YongvongphaiboonKrin Yongvongphaiboon
64
64
$begingroup$
What is your question?
$endgroup$
– Yeah..
Feb 17 '16 at 2:22
$begingroup$
@Yeah. The question appears to be : "Did I get it right?"
$endgroup$
– Graham Kemp
Feb 17 '16 at 2:23
$begingroup$
@GrahamKemp YES! :P
$endgroup$
– Krin Yongvongphaiboon
Feb 17 '16 at 2:24
add a comment |
$begingroup$
What is your question?
$endgroup$
– Yeah..
Feb 17 '16 at 2:22
$begingroup$
@Yeah. The question appears to be : "Did I get it right?"
$endgroup$
– Graham Kemp
Feb 17 '16 at 2:23
$begingroup$
@GrahamKemp YES! :P
$endgroup$
– Krin Yongvongphaiboon
Feb 17 '16 at 2:24
$begingroup$
What is your question?
$endgroup$
– Yeah..
Feb 17 '16 at 2:22
$begingroup$
What is your question?
$endgroup$
– Yeah..
Feb 17 '16 at 2:22
$begingroup$
@Yeah. The question appears to be : "Did I get it right?"
$endgroup$
– Graham Kemp
Feb 17 '16 at 2:23
$begingroup$
@Yeah. The question appears to be : "Did I get it right?"
$endgroup$
– Graham Kemp
Feb 17 '16 at 2:23
$begingroup$
@GrahamKemp YES! :P
$endgroup$
– Krin Yongvongphaiboon
Feb 17 '16 at 2:24
$begingroup$
@GrahamKemp YES! :P
$endgroup$
– Krin Yongvongphaiboon
Feb 17 '16 at 2:24
add a comment |
2 Answers
2
active
oldest
votes
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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
S* =(p ∨ q)∧ (¬p ∧ q)∧T (dual of S is denoted by S*)
So the correct answer to your question is:
a) p∧¬q∧¬r
b) (p∧q∧r)∨s
c) (p∨F)∧(q∨T)
answers:
a) p∨¬q∨¬r
b) (p∨q∨r)∧s
c) (p∧T)∨(q∧F)
answered Jan 18 at 16:39
Sumit RanjanSumit Ranjan
315
$endgroup$
add a comment |
$begingroup$
Yes. All three check: okay.
You've negated all the propositions and dualled all the operators.
That is what you needed to do.
$endgroup$
$begingroup$
Thank you for confirming :)
$endgroup$
– Krin Yongvongphaiboon
Feb 17 '16 at 2:25
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
S* =(p ∨ q)∧ (¬p ∧ q)∧T (dual of S is denoted by S*)
So the correct answer to your question is:
a) p∧¬q∧¬r
b) (p∧q∧r)∨s
c) (p∨F)∧(q∨T)
answers:
a) p∨¬q∨¬r
b) (p∨q∨r)∧s
c) (p∧T)∨(q∧F)
answered Jan 18 at 16:39
Sumit RanjanSumit Ranjan
315
$endgroup$
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
S* =(p ∨ q)∧ (¬p ∧ q)∧T (dual of S is denoted by S*)
So the correct answer to your question is:
a) p∧¬q∧¬r
b) (p∧q∧r)∨s
c) (p∨F)∧(q∨T)
answers:
a) p∨¬q∨¬r
b) (p∨q∨r)∧s
c) (p∧T)∨(q∧F)
answered Jan 18 at 16:39
Sumit RanjanSumit Ranjan
315
$endgroup$
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
S* =(p ∨ q)∧ (¬p ∧ q)∧T (dual of S is denoted by S*)
So the correct answer to your question is:
a) p∧¬q∧¬r
b) (p∧q∧r)∨s
c) (p∨F)∧(q∨T)
answers:
a) p∨¬q∨¬r
b) (p∨q∨r)∧s
c) (p∧T)∨(q∧F)
answered Jan 18 at 16:39
Sumit RanjanSumit Ranjan
315
$endgroup$
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
S* =(p ∨ q)∧ (¬p ∧ q)∧T (dual of S is denoted by S*)
So the correct answer to your question is:
a) p∧¬q∧¬r
b) (p∧q∧r)∨s
c) (p∨F)∧(q∨T)
answers:
a) p∨¬q∨¬r
b) (p∨q∨r)∧s
c) (p∧T)∨(q∧F)
answered Jan 18 at 16:39
Sumit RanjanSumit Ranjan
315
edited Jan 18 at 16:57
answered Jan 18 at 16:39
Sumit RanjanSumit Ranjan
315
answered Jan 18 at 16:39
Sumit RanjanSumit Ranjan
315
answered Jan 18 at 16:39
Sumit RanjanSumit Ranjan
315
315
add a comment |
add a comment |
$begingroup$
Yes. All three check: okay.
You've negated all the propositions and dualled all the operators.
That is what you needed to do.
$endgroup$
$begingroup$
Thank you for confirming :)
$endgroup$
– Krin Yongvongphaiboon
Feb 17 '16 at 2:25
add a comment |
$begingroup$
Yes. All three check: okay.
You've negated all the propositions and dualled all the operators.
That is what you needed to do.
$endgroup$
$begingroup$
Thank you for confirming :)
$endgroup$
– Krin Yongvongphaiboon
Feb 17 '16 at 2:25
add a comment |
$begingroup$
Yes. All three check: okay.
You've negated all the propositions and dualled all the operators.
That is what you needed to do.
$endgroup$
Yes. All three check: okay.
You've negated all the propositions and dualled all the operators.
That is what you needed to do.
answered Feb 17 '16 at 2:21
community wiki
Graham Kemp
$begingroup$
Thank you for confirming :)
$endgroup$
– Krin Yongvongphaiboon
Feb 17 '16 at 2:25
add a comment |
$begingroup$
Thank you for confirming :)
$endgroup$
– Krin Yongvongphaiboon
Feb 17 '16 at 2:25
$begingroup$
Thank you for confirming :)
$endgroup$
– Krin Yongvongphaiboon
Feb 17 '16 at 2:25
$begingroup$
Thank you for confirming :)
$endgroup$
– Krin Yongvongphaiboon
Feb 17 '16 at 2:25
add a comment |
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$begingroup$
What is your question?
$endgroup$
– Yeah..
Feb 17 '16 at 2:22
$begingroup$
@Yeah. The question appears to be : "Did I get it right?"
$endgroup$
– Graham Kemp
Feb 17 '16 at 2:23
$begingroup$
@GrahamKemp YES! :P
$endgroup$
– Krin Yongvongphaiboon
Feb 17 '16 at 2:24