Prove $p lor q Leftrightarrow (neg p) rightarrow q$, but with caveats.
$begingroup$
In this question, the professor asks us in parts i through iii to prove using truth tables that:
i. $neg (p lor q) Leftrightarrow (neg p) land (neg q)$
ii. $neg (p land q) Leftrightarrow (neg p) lor (neg q)$
iii. $neg (neg p) Leftrightarrow p$
Then in part iv. he changes the game and asks us to "use (i-iii) to derive (by the use of a succession of $Leftrightarrow$'s)" that $p lor q Leftrightarrow (neg p) rightarrow q$. That is, a truth table is now specifically not allowed.
I am just stuck. I get that statements i. through iii. are now "tools" that we can use to prove that $p lor q Leftrightarrow (neg p) rightarrow q$. I understand that $p lor q$ means one or both must be true, so if one is false (i.e. $neg p$), then the other must be true. But I don't see how I am supposed to prove that a statement is equivalent to an implication using i. through iii. if none of these statements i. through iii. involve an implication!
I tried submitting this combination of $Leftrightarrow$ statements and exposition:
"$p lor q Leftrightarrow neg (neg (p land q)) Leftrightarrow neg((neg p) land (neg q))$ where this last statement means that $p$ and $q$ cannot be simultaneously false, so if $p$ is false, then $q$ must be true,"
However, the professor said there's an easier way, and I should be able to prove the statement using ONLY $Leftrightarrow$ statements, and no exposition. I feel like I'm just missing something obvious. Or maybe he is assuming we're going to use outside knowledge and not JUST statements i. through iii.?
EDIT: In case anyone ever comes across this, the answer is that you are to use a combination of the rules i. through iii. as described, and also common, outside facts. So you have $(p lor q) equiv neg(neg p) lor q equiv neg p rightarrow q$.
discrete-mathematics logic proof-writing
asked Jan 20 at 18:46
1Teaches2Learn1Teaches2Learn
213
$endgroup$
|
show 1 more comment
$begingroup$
In this question, the professor asks us in parts i through iii to prove using truth tables that:
i. $neg (p lor q) Leftrightarrow (neg p) land (neg q)$
ii. $neg (p land q) Leftrightarrow (neg p) lor (neg q)$
iii. $neg (neg p) Leftrightarrow p$
Then in part iv. he changes the game and asks us to "use (i-iii) to derive (by the use of a succession of $Leftrightarrow$'s)" that $p lor q Leftrightarrow (neg p) rightarrow q$. That is, a truth table is now specifically not allowed.
I am just stuck. I get that statements i. through iii. are now "tools" that we can use to prove that $p lor q Leftrightarrow (neg p) rightarrow q$. I understand that $p lor q$ means one or both must be true, so if one is false (i.e. $neg p$), then the other must be true. But I don't see how I am supposed to prove that a statement is equivalent to an implication using i. through iii. if none of these statements i. through iii. involve an implication!
I tried submitting this combination of $Leftrightarrow$ statements and exposition:
"$p lor q Leftrightarrow neg (neg (p land q)) Leftrightarrow neg((neg p) land (neg q))$ where this last statement means that $p$ and $q$ cannot be simultaneously false, so if $p$ is false, then $q$ must be true,"
However, the professor said there's an easier way, and I should be able to prove the statement using ONLY $Leftrightarrow$ statements, and no exposition. I feel like I'm just missing something obvious. Or maybe he is assuming we're going to use outside knowledge and not JUST statements i. through iii.?
EDIT: In case anyone ever comes across this, the answer is that you are to use a combination of the rules i. through iii. as described, and also common, outside facts. So you have $(p lor q) equiv neg(neg p) lor q equiv neg p rightarrow q$.
discrete-mathematics logic proof-writing
asked Jan 20 at 18:46
1Teaches2Learn1Teaches2Learn
213
$endgroup$
3
$begingroup$
How do you define $pto q$?
$endgroup$
– John Douma
Jan 20 at 18:49
2
$begingroup$
You are right. With only (i)-(iii) you cannot, because there is nowhere the symbol $to$ in them... thus, it cannot "pop up" from nowhere.
$endgroup$
– Mauro ALLEGRANZA
Jan 20 at 18:50
$begingroup$
What is "exposition" ?
$endgroup$
– Mauro ALLEGRANZA
Jan 20 at 18:51
$begingroup$
Good question, "exposition" meaning "explanation in words, instead of just symbols."
$endgroup$
– 1Teaches2Learn
Jan 20 at 18:52
$begingroup$
Ok... $p lor q$ is "either $p$ or $q$ (inclusive or)" that means that at least one of them must be TRUE. If so, if $p$ is FALSE, then $q$ must necessarily be TRUE. And this is : from Left to Right...
$endgroup$
– Mauro ALLEGRANZA
Jan 20 at 18:55
|
show 1 more comment
$begingroup$
In this question, the professor asks us in parts i through iii to prove using truth tables that:
i. $neg (p lor q) Leftrightarrow (neg p) land (neg q)$
ii. $neg (p land q) Leftrightarrow (neg p) lor (neg q)$
iii. $neg (neg p) Leftrightarrow p$
Then in part iv. he changes the game and asks us to "use (i-iii) to derive (by the use of a succession of $Leftrightarrow$'s)" that $p lor q Leftrightarrow (neg p) rightarrow q$. That is, a truth table is now specifically not allowed.
I am just stuck. I get that statements i. through iii. are now "tools" that we can use to prove that $p lor q Leftrightarrow (neg p) rightarrow q$. I understand that $p lor q$ means one or both must be true, so if one is false (i.e. $neg p$), then the other must be true. But I don't see how I am supposed to prove that a statement is equivalent to an implication using i. through iii. if none of these statements i. through iii. involve an implication!
I tried submitting this combination of $Leftrightarrow$ statements and exposition:
"$p lor q Leftrightarrow neg (neg (p land q)) Leftrightarrow neg((neg p) land (neg q))$ where this last statement means that $p$ and $q$ cannot be simultaneously false, so if $p$ is false, then $q$ must be true,"
However, the professor said there's an easier way, and I should be able to prove the statement using ONLY $Leftrightarrow$ statements, and no exposition. I feel like I'm just missing something obvious. Or maybe he is assuming we're going to use outside knowledge and not JUST statements i. through iii.?
EDIT: In case anyone ever comes across this, the answer is that you are to use a combination of the rules i. through iii. as described, and also common, outside facts. So you have $(p lor q) equiv neg(neg p) lor q equiv neg p rightarrow q$.
discrete-mathematics logic proof-writing
asked Jan 20 at 18:46
1Teaches2Learn1Teaches2Learn
213
$endgroup$
In this question, the professor asks us in parts i through iii to prove using truth tables that:
i. $neg (p lor q) Leftrightarrow (neg p) land (neg q)$
ii. $neg (p land q) Leftrightarrow (neg p) lor (neg q)$
iii. $neg (neg p) Leftrightarrow p$
Then in part iv. he changes the game and asks us to "use (i-iii) to derive (by the use of a succession of $Leftrightarrow$'s)" that $p lor q Leftrightarrow (neg p) rightarrow q$. That is, a truth table is now specifically not allowed.
I am just stuck. I get that statements i. through iii. are now "tools" that we can use to prove that $p lor q Leftrightarrow (neg p) rightarrow q$. I understand that $p lor q$ means one or both must be true, so if one is false (i.e. $neg p$), then the other must be true. But I don't see how I am supposed to prove that a statement is equivalent to an implication using i. through iii. if none of these statements i. through iii. involve an implication!
I tried submitting this combination of $Leftrightarrow$ statements and exposition:
"$p lor q Leftrightarrow neg (neg (p land q)) Leftrightarrow neg((neg p) land (neg q))$ where this last statement means that $p$ and $q$ cannot be simultaneously false, so if $p$ is false, then $q$ must be true,"
However, the professor said there's an easier way, and I should be able to prove the statement using ONLY $Leftrightarrow$ statements, and no exposition. I feel like I'm just missing something obvious. Or maybe he is assuming we're going to use outside knowledge and not JUST statements i. through iii.?
EDIT: In case anyone ever comes across this, the answer is that you are to use a combination of the rules i. through iii. as described, and also common, outside facts. So you have $(p lor q) equiv neg(neg p) lor q equiv neg p rightarrow q$.
discrete-mathematics logic proof-writing
discrete-mathematics logic proof-writing
asked Jan 20 at 18:46
1Teaches2Learn1Teaches2Learn
213
asked Jan 20 at 18:46
1Teaches2Learn1Teaches2Learn
213
edited Jan 27 at 8:54
1Teaches2Learn
asked Jan 20 at 18:46
1Teaches2Learn1Teaches2Learn
213
asked Jan 20 at 18:46
1Teaches2Learn1Teaches2Learn
213
asked Jan 20 at 18:46
1Teaches2Learn1Teaches2Learn
213
213
3
$begingroup$
How do you define $pto q$?
$endgroup$
– John Douma
Jan 20 at 18:49
2
$begingroup$
You are right. With only (i)-(iii) you cannot, because there is nowhere the symbol $to$ in them... thus, it cannot "pop up" from nowhere.
$endgroup$
– Mauro ALLEGRANZA
Jan 20 at 18:50
$begingroup$
What is "exposition" ?
$endgroup$
– Mauro ALLEGRANZA
Jan 20 at 18:51
$begingroup$
Good question, "exposition" meaning "explanation in words, instead of just symbols."
$endgroup$
– 1Teaches2Learn
Jan 20 at 18:52
$begingroup$
Ok... $p lor q$ is "either $p$ or $q$ (inclusive or)" that means that at least one of them must be TRUE. If so, if $p$ is FALSE, then $q$ must necessarily be TRUE. And this is : from Left to Right...
$endgroup$
– Mauro ALLEGRANZA
Jan 20 at 18:55
|
show 1 more comment
3
$begingroup$
How do you define $pto q$?
$endgroup$
– John Douma
Jan 20 at 18:49
2
$begingroup$
You are right. With only (i)-(iii) you cannot, because there is nowhere the symbol $to$ in them... thus, it cannot "pop up" from nowhere.
$endgroup$
– Mauro ALLEGRANZA
Jan 20 at 18:50
$begingroup$
What is "exposition" ?
$endgroup$
– Mauro ALLEGRANZA
Jan 20 at 18:51
$begingroup$
Good question, "exposition" meaning "explanation in words, instead of just symbols."
$endgroup$
– 1Teaches2Learn
Jan 20 at 18:52
$begingroup$
Ok... $p lor q$ is "either $p$ or $q$ (inclusive or)" that means that at least one of them must be TRUE. If so, if $p$ is FALSE, then $q$ must necessarily be TRUE. And this is : from Left to Right...
$endgroup$
– Mauro ALLEGRANZA
Jan 20 at 18:55
3
3
$begingroup$
How do you define $pto q$?
$endgroup$
– John Douma
Jan 20 at 18:49
$begingroup$
How do you define $pto q$?
$endgroup$
– John Douma
Jan 20 at 18:49
2
2
$begingroup$
You are right. With only (i)-(iii) you cannot, because there is nowhere the symbol $to$ in them... thus, it cannot "pop up" from nowhere.
$endgroup$
– Mauro ALLEGRANZA
Jan 20 at 18:50
$begingroup$
You are right. With only (i)-(iii) you cannot, because there is nowhere the symbol $to$ in them... thus, it cannot "pop up" from nowhere.
$endgroup$
– Mauro ALLEGRANZA
Jan 20 at 18:50
$begingroup$
What is "exposition" ?
$endgroup$
– Mauro ALLEGRANZA
Jan 20 at 18:51
$begingroup$
What is "exposition" ?
$endgroup$
– Mauro ALLEGRANZA
Jan 20 at 18:51
$begingroup$
Good question, "exposition" meaning "explanation in words, instead of just symbols."
$endgroup$
– 1Teaches2Learn
Jan 20 at 18:52
$begingroup$
Good question, "exposition" meaning "explanation in words, instead of just symbols."
$endgroup$
– 1Teaches2Learn
Jan 20 at 18:52
$begingroup$
Ok... $p lor q$ is "either $p$ or $q$ (inclusive or)" that means that at least one of them must be TRUE. If so, if $p$ is FALSE, then $q$ must necessarily be TRUE. And this is : from Left to Right...
$endgroup$
– Mauro ALLEGRANZA
Jan 20 at 18:55
$begingroup$
Ok... $p lor q$ is "either $p$ or $q$ (inclusive or)" that means that at least one of them must be TRUE. If so, if $p$ is FALSE, then $q$ must necessarily be TRUE. And this is : from Left to Right...
$endgroup$
– Mauro ALLEGRANZA
Jan 20 at 18:55
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
You're absolutely right. With no rule involving $rightarrow$ in (i)-(iii) this is impossible.
But, typically it is given that $p rightarrow q Leftrightarrow neg p lor q$
With that, I am sure you can do it!
And ask your professor about this ....
answered Jan 20 at 19:04
Bram28Bram28
63.1k44793
$endgroup$
$begingroup$
Thank you so much! This was my first time asking a question, and I was not expecting help to come so quickly! I love this site. (Thanks also to Mauro Allegranza.)
$endgroup$
– 1Teaches2Learn
Jan 20 at 19:30
$begingroup$
@1Teaches2Learn Ha ha, yes, I remember my first post ... and almost instantaneous reply! :)
$endgroup$
– Bram28
Jan 20 at 19:45
$begingroup$
Thanks again! I did end up getting this question correct. I have submitted an upvote for you, but it apparently will not go through until I get enough "karma" or whatever this site calls it! So I'm off to earn more of it, whatever it's called :P
$endgroup$
– 1Teaches2Learn
Jan 27 at 8:55
add a comment |
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1 Answer
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$begingroup$
You're absolutely right. With no rule involving $rightarrow$ in (i)-(iii) this is impossible.
But, typically it is given that $p rightarrow q Leftrightarrow neg p lor q$
With that, I am sure you can do it!
And ask your professor about this ....
answered Jan 20 at 19:04
Bram28Bram28
63.1k44793
$endgroup$
$begingroup$
Thank you so much! This was my first time asking a question, and I was not expecting help to come so quickly! I love this site. (Thanks also to Mauro Allegranza.)
$endgroup$
– 1Teaches2Learn
Jan 20 at 19:30
$begingroup$
@1Teaches2Learn Ha ha, yes, I remember my first post ... and almost instantaneous reply! :)
$endgroup$
– Bram28
Jan 20 at 19:45
$begingroup$
Thanks again! I did end up getting this question correct. I have submitted an upvote for you, but it apparently will not go through until I get enough "karma" or whatever this site calls it! So I'm off to earn more of it, whatever it's called :P
$endgroup$
– 1Teaches2Learn
Jan 27 at 8:55
add a comment |
$begingroup$
You're absolutely right. With no rule involving $rightarrow$ in (i)-(iii) this is impossible.
But, typically it is given that $p rightarrow q Leftrightarrow neg p lor q$
With that, I am sure you can do it!
And ask your professor about this ....
answered Jan 20 at 19:04
Bram28Bram28
63.1k44793
$endgroup$
$begingroup$
Thank you so much! This was my first time asking a question, and I was not expecting help to come so quickly! I love this site. (Thanks also to Mauro Allegranza.)
$endgroup$
– 1Teaches2Learn
Jan 20 at 19:30
$begingroup$
@1Teaches2Learn Ha ha, yes, I remember my first post ... and almost instantaneous reply! :)
$endgroup$
– Bram28
Jan 20 at 19:45
$begingroup$
Thanks again! I did end up getting this question correct. I have submitted an upvote for you, but it apparently will not go through until I get enough "karma" or whatever this site calls it! So I'm off to earn more of it, whatever it's called :P
$endgroup$
– 1Teaches2Learn
Jan 27 at 8:55
add a comment |
$begingroup$
You're absolutely right. With no rule involving $rightarrow$ in (i)-(iii) this is impossible.
But, typically it is given that $p rightarrow q Leftrightarrow neg p lor q$
With that, I am sure you can do it!
And ask your professor about this ....
answered Jan 20 at 19:04
Bram28Bram28
63.1k44793
$endgroup$
You're absolutely right. With no rule involving $rightarrow$ in (i)-(iii) this is impossible.
But, typically it is given that $p rightarrow q Leftrightarrow neg p lor q$
With that, I am sure you can do it!
And ask your professor about this ....
answered Jan 20 at 19:04
Bram28Bram28
63.1k44793
answered Jan 20 at 19:04
Bram28Bram28
63.1k44793
answered Jan 20 at 19:04
Bram28Bram28
63.1k44793
answered Jan 20 at 19:04
Bram28Bram28
63.1k44793
63.1k44793
$begingroup$
Thank you so much! This was my first time asking a question, and I was not expecting help to come so quickly! I love this site. (Thanks also to Mauro Allegranza.)
$endgroup$
– 1Teaches2Learn
Jan 20 at 19:30
$begingroup$
@1Teaches2Learn Ha ha, yes, I remember my first post ... and almost instantaneous reply! :)
$endgroup$
– Bram28
Jan 20 at 19:45
$begingroup$
Thanks again! I did end up getting this question correct. I have submitted an upvote for you, but it apparently will not go through until I get enough "karma" or whatever this site calls it! So I'm off to earn more of it, whatever it's called :P
$endgroup$
– 1Teaches2Learn
Jan 27 at 8:55
add a comment |
$begingroup$
Thank you so much! This was my first time asking a question, and I was not expecting help to come so quickly! I love this site. (Thanks also to Mauro Allegranza.)
$endgroup$
– 1Teaches2Learn
Jan 20 at 19:30
$begingroup$
@1Teaches2Learn Ha ha, yes, I remember my first post ... and almost instantaneous reply! :)
$endgroup$
– Bram28
Jan 20 at 19:45
$begingroup$
Thanks again! I did end up getting this question correct. I have submitted an upvote for you, but it apparently will not go through until I get enough "karma" or whatever this site calls it! So I'm off to earn more of it, whatever it's called :P
$endgroup$
– 1Teaches2Learn
Jan 27 at 8:55
$begingroup$
Thank you so much! This was my first time asking a question, and I was not expecting help to come so quickly! I love this site. (Thanks also to Mauro Allegranza.)
$endgroup$
– 1Teaches2Learn
Jan 20 at 19:30
$begingroup$
Thank you so much! This was my first time asking a question, and I was not expecting help to come so quickly! I love this site. (Thanks also to Mauro Allegranza.)
$endgroup$
– 1Teaches2Learn
Jan 20 at 19:30
$begingroup$
@1Teaches2Learn Ha ha, yes, I remember my first post ... and almost instantaneous reply! :)
$endgroup$
– Bram28
Jan 20 at 19:45
$begingroup$
@1Teaches2Learn Ha ha, yes, I remember my first post ... and almost instantaneous reply! :)
$endgroup$
– Bram28
Jan 20 at 19:45
$begingroup$
Thanks again! I did end up getting this question correct. I have submitted an upvote for you, but it apparently will not go through until I get enough "karma" or whatever this site calls it! So I'm off to earn more of it, whatever it's called :P
$endgroup$
– 1Teaches2Learn
Jan 27 at 8:55
$begingroup$
Thanks again! I did end up getting this question correct. I have submitted an upvote for you, but it apparently will not go through until I get enough "karma" or whatever this site calls it! So I'm off to earn more of it, whatever it's called :P
$endgroup$
– 1Teaches2Learn
Jan 27 at 8:55
add a comment |
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3
$begingroup$
How do you define $pto q$?
$endgroup$
– John Douma
Jan 20 at 18:49
2
$begingroup$
You are right. With only (i)-(iii) you cannot, because there is nowhere the symbol $to$ in them... thus, it cannot "pop up" from nowhere.
$endgroup$
– Mauro ALLEGRANZA
Jan 20 at 18:50
$begingroup$
What is "exposition" ?
$endgroup$
– Mauro ALLEGRANZA
Jan 20 at 18:51
$begingroup$
Good question, "exposition" meaning "explanation in words, instead of just symbols."
$endgroup$
– 1Teaches2Learn
Jan 20 at 18:52
$begingroup$
Ok... $p lor q$ is "either $p$ or $q$ (inclusive or)" that means that at least one of them must be TRUE. If so, if $p$ is FALSE, then $q$ must necessarily be TRUE. And this is : from Left to Right...
$endgroup$
– Mauro ALLEGRANZA
Jan 20 at 18:55