Finding $ alpha $ so that you get two tautologies
$begingroup$
I have this task to solve:
Find (if it exists) such a sentence formula α so that the following formulas are simultaneously
tautologies of the propositional calculus:
$$ (q rightarrow alpha) leftrightarrow (q rightarrow (p wedge r))$$
and
$$ (alpha rightarrow q) leftrightarrow (neg(p vee r) rightarrow q) $$
Generally I have idea to solve this task in use of table. So I created a table with 7 columns:
$p$,$q$,$r$,$(p wedge r)$, $(q rightarrow (p wedge r)$ $(q rightarrow alpha)$, $LS leftrightarrow RS $
and 8 rows with each case. In last column I get these values:
$alpha$, $neg alpha$, $1$, $1$, $neg alpha$, $neg alpha$, $1$, $1$
So if we consider tautology, both $alpha$ and $neg alpha$ should be $1$ so it can't be done. So $alpha$ like that does not exists and I don't have to consider second potential tautology. Have I right?
Eventually, there are other ways to solve tasks like that?
logic propositional-calculus
edited Dec 13 '18 at 20:28
Shaun
9,221113684
asked Dec 13 '18 at 20:23
VirtualUserVirtualUser
865114
$endgroup$
|
show 2 more comments
$begingroup$
I have this task to solve:
Find (if it exists) such a sentence formula α so that the following formulas are simultaneously
tautologies of the propositional calculus:
$$ (q rightarrow alpha) leftrightarrow (q rightarrow (p wedge r))$$
and
$$ (alpha rightarrow q) leftrightarrow (neg(p vee r) rightarrow q) $$
Generally I have idea to solve this task in use of table. So I created a table with 7 columns:
$p$,$q$,$r$,$(p wedge r)$, $(q rightarrow (p wedge r)$ $(q rightarrow alpha)$, $LS leftrightarrow RS $
and 8 rows with each case. In last column I get these values:
$alpha$, $neg alpha$, $1$, $1$, $neg alpha$, $neg alpha$, $1$, $1$
So if we consider tautology, both $alpha$ and $neg alpha$ should be $1$ so it can't be done. So $alpha$ like that does not exists and I don't have to consider second potential tautology. Have I right?
Eventually, there are other ways to solve tasks like that?
logic propositional-calculus
edited Dec 13 '18 at 20:28
Shaun
9,221113684
asked Dec 13 '18 at 20:23
VirtualUserVirtualUser
865114
$endgroup$
$begingroup$
Have you compared the values of alpha with the values of other columns? Remember, you're looking for a statement or formula, not necessarily a truth value
$endgroup$
– David Diaz
Dec 13 '18 at 20:27
$begingroup$
It should be tautology so $LS leftrightarrow RS $ must be true value. If I represent this by $alpha$ it should be true too. The same thing with $neg alpha $
$endgroup$
– VirtualUser
Dec 13 '18 at 20:30
1
$begingroup$
One could use the method of analytic tableaux.
$endgroup$
– Shaun
Dec 13 '18 at 20:30
$begingroup$
Ok, but my solution is good or not? Thanks for interesting link.
$endgroup$
– VirtualUser
Dec 13 '18 at 20:34
$begingroup$
@VirtualUser Your method of creating a truth-table is fine ... but your conclusion that it cannot be done is not. It can be done. Just see what value, if any, is forced on $alpha$ to make the truth-values come out right in each row. And then just generate an expression for that.
$endgroup$
– Bram28
Dec 13 '18 at 20:35
|
show 2 more comments
$begingroup$
I have this task to solve:
Find (if it exists) such a sentence formula α so that the following formulas are simultaneously
tautologies of the propositional calculus:
$$ (q rightarrow alpha) leftrightarrow (q rightarrow (p wedge r))$$
and
$$ (alpha rightarrow q) leftrightarrow (neg(p vee r) rightarrow q) $$
Generally I have idea to solve this task in use of table. So I created a table with 7 columns:
$p$,$q$,$r$,$(p wedge r)$, $(q rightarrow (p wedge r)$ $(q rightarrow alpha)$, $LS leftrightarrow RS $
and 8 rows with each case. In last column I get these values:
$alpha$, $neg alpha$, $1$, $1$, $neg alpha$, $neg alpha$, $1$, $1$
So if we consider tautology, both $alpha$ and $neg alpha$ should be $1$ so it can't be done. So $alpha$ like that does not exists and I don't have to consider second potential tautology. Have I right?
Eventually, there are other ways to solve tasks like that?
logic propositional-calculus
edited Dec 13 '18 at 20:28
Shaun
9,221113684
asked Dec 13 '18 at 20:23
VirtualUserVirtualUser
865114
$endgroup$
I have this task to solve:
Find (if it exists) such a sentence formula α so that the following formulas are simultaneously
tautologies of the propositional calculus:
$$ (q rightarrow alpha) leftrightarrow (q rightarrow (p wedge r))$$
and
$$ (alpha rightarrow q) leftrightarrow (neg(p vee r) rightarrow q) $$
Generally I have idea to solve this task in use of table. So I created a table with 7 columns:
$p$,$q$,$r$,$(p wedge r)$, $(q rightarrow (p wedge r)$ $(q rightarrow alpha)$, $LS leftrightarrow RS $
and 8 rows with each case. In last column I get these values:
$alpha$, $neg alpha$, $1$, $1$, $neg alpha$, $neg alpha$, $1$, $1$
So if we consider tautology, both $alpha$ and $neg alpha$ should be $1$ so it can't be done. So $alpha$ like that does not exists and I don't have to consider second potential tautology. Have I right?
Eventually, there are other ways to solve tasks like that?
logic propositional-calculus
logic propositional-calculus
edited Dec 13 '18 at 20:28
Shaun
9,221113684
asked Dec 13 '18 at 20:23
VirtualUserVirtualUser
865114
edited Dec 13 '18 at 20:28
Shaun
9,221113684
asked Dec 13 '18 at 20:23
VirtualUserVirtualUser
865114
edited Dec 13 '18 at 20:28
Shaun
9,221113684
edited Dec 13 '18 at 20:28
Shaun
9,221113684
edited Dec 13 '18 at 20:28
Shaun
9,221113684
9,221113684
asked Dec 13 '18 at 20:23
VirtualUserVirtualUser
865114
asked Dec 13 '18 at 20:23
VirtualUserVirtualUser
865114
asked Dec 13 '18 at 20:23
VirtualUserVirtualUser
865114
865114
$begingroup$
Have you compared the values of alpha with the values of other columns? Remember, you're looking for a statement or formula, not necessarily a truth value
$endgroup$
– David Diaz
Dec 13 '18 at 20:27
$begingroup$
It should be tautology so $LS leftrightarrow RS $ must be true value. If I represent this by $alpha$ it should be true too. The same thing with $neg alpha $
$endgroup$
– VirtualUser
Dec 13 '18 at 20:30
1
$begingroup$
One could use the method of analytic tableaux.
$endgroup$
– Shaun
Dec 13 '18 at 20:30
$begingroup$
Ok, but my solution is good or not? Thanks for interesting link.
$endgroup$
– VirtualUser
Dec 13 '18 at 20:34
$begingroup$
@VirtualUser Your method of creating a truth-table is fine ... but your conclusion that it cannot be done is not. It can be done. Just see what value, if any, is forced on $alpha$ to make the truth-values come out right in each row. And then just generate an expression for that.
$endgroup$
– Bram28
Dec 13 '18 at 20:35
|
show 2 more comments
$begingroup$
Have you compared the values of alpha with the values of other columns? Remember, you're looking for a statement or formula, not necessarily a truth value
$endgroup$
– David Diaz
Dec 13 '18 at 20:27
$begingroup$
It should be tautology so $LS leftrightarrow RS $ must be true value. If I represent this by $alpha$ it should be true too. The same thing with $neg alpha $
$endgroup$
– VirtualUser
Dec 13 '18 at 20:30
1
$begingroup$
One could use the method of analytic tableaux.
$endgroup$
– Shaun
Dec 13 '18 at 20:30
$begingroup$
Ok, but my solution is good or not? Thanks for interesting link.
$endgroup$
– VirtualUser
Dec 13 '18 at 20:34
$begingroup$
@VirtualUser Your method of creating a truth-table is fine ... but your conclusion that it cannot be done is not. It can be done. Just see what value, if any, is forced on $alpha$ to make the truth-values come out right in each row. And then just generate an expression for that.
$endgroup$
– Bram28
Dec 13 '18 at 20:35
$begingroup$
Have you compared the values of alpha with the values of other columns? Remember, you're looking for a statement or formula, not necessarily a truth value
$endgroup$
– David Diaz
Dec 13 '18 at 20:27
$begingroup$
Have you compared the values of alpha with the values of other columns? Remember, you're looking for a statement or formula, not necessarily a truth value
$endgroup$
– David Diaz
Dec 13 '18 at 20:27
$begingroup$
It should be tautology so $LS leftrightarrow RS $ must be true value. If I represent this by $alpha$ it should be true too. The same thing with $neg alpha $
$endgroup$
– VirtualUser
Dec 13 '18 at 20:30
$begingroup$
It should be tautology so $LS leftrightarrow RS $ must be true value. If I represent this by $alpha$ it should be true too. The same thing with $neg alpha $
$endgroup$
– VirtualUser
Dec 13 '18 at 20:30
1
1
$begingroup$
One could use the method of analytic tableaux.
$endgroup$
– Shaun
Dec 13 '18 at 20:30
$begingroup$
One could use the method of analytic tableaux.
$endgroup$
– Shaun
Dec 13 '18 at 20:30
$begingroup$
Ok, but my solution is good or not? Thanks for interesting link.
$endgroup$
– VirtualUser
Dec 13 '18 at 20:34
$begingroup$
Ok, but my solution is good or not? Thanks for interesting link.
$endgroup$
– VirtualUser
Dec 13 '18 at 20:34
$begingroup$
@VirtualUser Your method of creating a truth-table is fine ... but your conclusion that it cannot be done is not. It can be done. Just see what value, if any, is forced on $alpha$ to make the truth-values come out right in each row. And then just generate an expression for that.
$endgroup$
– Bram28
Dec 13 '18 at 20:35
$begingroup$
@VirtualUser Your method of creating a truth-table is fine ... but your conclusion that it cannot be done is not. It can be done. Just see what value, if any, is forced on $alpha$ to make the truth-values come out right in each row. And then just generate an expression for that.
$endgroup$
– Bram28
Dec 13 '18 at 20:35
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
You'll eventually get to a correct answer by working out a truth-table (and, as a general method, that is to be preferred over what I'm about to do below), but here is a more direct approach by careful inspection of the statements involved.
First, note that the first conditional holds True whenever $q$ is False, so nothing is required of $alpha$ in the case where $q$ is False. So the only interesting case is where $q$ is True. And in that case, $alpha$ needs to have the same truth-value as $p land r$ in order for the biconditional to be True, i.e. just when $p$ and $r$ are both True.
Likewise, you can immediately see that the second biconditional is True whenever $q$ is True, so here the only interesting case is where $q$ is False, and in that case, the biconditional is True just when $alpha$ has the same value as $neg (p lor r)$, i.e. when both $p$ and $r$ are False.
So, the only cases where $alpha$ should be True are where $q$, $p$, and $r$ are all True, or when they are all False:
$$alpha = (q land p land r) lor (neg q land neg p land neg r)$$
answered Dec 13 '18 at 20:52
Bram28Bram28
62.6k44793
$endgroup$
$begingroup$
Thanks for your reply I think that I have understood this problem, it is interesting alternative for table.
$endgroup$
– VirtualUser
Dec 13 '18 at 20:59
1
$begingroup$
@VirtualUser Yes, the statements were 'nice' enough that I was able to derive it this way pretty quickly ... but the Truth-table method will be more general.
$endgroup$
– Bram28
Dec 13 '18 at 21:13
add a comment |
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$begingroup$
You'll eventually get to a correct answer by working out a truth-table (and, as a general method, that is to be preferred over what I'm about to do below), but here is a more direct approach by careful inspection of the statements involved.
First, note that the first conditional holds True whenever $q$ is False, so nothing is required of $alpha$ in the case where $q$ is False. So the only interesting case is where $q$ is True. And in that case, $alpha$ needs to have the same truth-value as $p land r$ in order for the biconditional to be True, i.e. just when $p$ and $r$ are both True.
Likewise, you can immediately see that the second biconditional is True whenever $q$ is True, so here the only interesting case is where $q$ is False, and in that case, the biconditional is True just when $alpha$ has the same value as $neg (p lor r)$, i.e. when both $p$ and $r$ are False.
So, the only cases where $alpha$ should be True are where $q$, $p$, and $r$ are all True, or when they are all False:
$$alpha = (q land p land r) lor (neg q land neg p land neg r)$$
answered Dec 13 '18 at 20:52
Bram28Bram28
62.6k44793
$endgroup$
$begingroup$
Thanks for your reply I think that I have understood this problem, it is interesting alternative for table.
$endgroup$
– VirtualUser
Dec 13 '18 at 20:59
1
$begingroup$
@VirtualUser Yes, the statements were 'nice' enough that I was able to derive it this way pretty quickly ... but the Truth-table method will be more general.
$endgroup$
– Bram28
Dec 13 '18 at 21:13
add a comment |
$begingroup$
You'll eventually get to a correct answer by working out a truth-table (and, as a general method, that is to be preferred over what I'm about to do below), but here is a more direct approach by careful inspection of the statements involved.
First, note that the first conditional holds True whenever $q$ is False, so nothing is required of $alpha$ in the case where $q$ is False. So the only interesting case is where $q$ is True. And in that case, $alpha$ needs to have the same truth-value as $p land r$ in order for the biconditional to be True, i.e. just when $p$ and $r$ are both True.
Likewise, you can immediately see that the second biconditional is True whenever $q$ is True, so here the only interesting case is where $q$ is False, and in that case, the biconditional is True just when $alpha$ has the same value as $neg (p lor r)$, i.e. when both $p$ and $r$ are False.
So, the only cases where $alpha$ should be True are where $q$, $p$, and $r$ are all True, or when they are all False:
$$alpha = (q land p land r) lor (neg q land neg p land neg r)$$
answered Dec 13 '18 at 20:52
Bram28Bram28
62.6k44793
$endgroup$
$begingroup$
Thanks for your reply I think that I have understood this problem, it is interesting alternative for table.
$endgroup$
– VirtualUser
Dec 13 '18 at 20:59
1
$begingroup$
@VirtualUser Yes, the statements were 'nice' enough that I was able to derive it this way pretty quickly ... but the Truth-table method will be more general.
$endgroup$
– Bram28
Dec 13 '18 at 21:13
add a comment |
$begingroup$
You'll eventually get to a correct answer by working out a truth-table (and, as a general method, that is to be preferred over what I'm about to do below), but here is a more direct approach by careful inspection of the statements involved.
First, note that the first conditional holds True whenever $q$ is False, so nothing is required of $alpha$ in the case where $q$ is False. So the only interesting case is where $q$ is True. And in that case, $alpha$ needs to have the same truth-value as $p land r$ in order for the biconditional to be True, i.e. just when $p$ and $r$ are both True.
Likewise, you can immediately see that the second biconditional is True whenever $q$ is True, so here the only interesting case is where $q$ is False, and in that case, the biconditional is True just when $alpha$ has the same value as $neg (p lor r)$, i.e. when both $p$ and $r$ are False.
So, the only cases where $alpha$ should be True are where $q$, $p$, and $r$ are all True, or when they are all False:
$$alpha = (q land p land r) lor (neg q land neg p land neg r)$$
answered Dec 13 '18 at 20:52
Bram28Bram28
62.6k44793
$endgroup$
You'll eventually get to a correct answer by working out a truth-table (and, as a general method, that is to be preferred over what I'm about to do below), but here is a more direct approach by careful inspection of the statements involved.
First, note that the first conditional holds True whenever $q$ is False, so nothing is required of $alpha$ in the case where $q$ is False. So the only interesting case is where $q$ is True. And in that case, $alpha$ needs to have the same truth-value as $p land r$ in order for the biconditional to be True, i.e. just when $p$ and $r$ are both True.
Likewise, you can immediately see that the second biconditional is True whenever $q$ is True, so here the only interesting case is where $q$ is False, and in that case, the biconditional is True just when $alpha$ has the same value as $neg (p lor r)$, i.e. when both $p$ and $r$ are False.
So, the only cases where $alpha$ should be True are where $q$, $p$, and $r$ are all True, or when they are all False:
$$alpha = (q land p land r) lor (neg q land neg p land neg r)$$
answered Dec 13 '18 at 20:52
Bram28Bram28
62.6k44793
answered Dec 13 '18 at 20:52
Bram28Bram28
62.6k44793
answered Dec 13 '18 at 20:52
Bram28Bram28
62.6k44793
answered Dec 13 '18 at 20:52
Bram28Bram28
62.6k44793
62.6k44793
$begingroup$
Thanks for your reply I think that I have understood this problem, it is interesting alternative for table.
$endgroup$
– VirtualUser
Dec 13 '18 at 20:59
1
$begingroup$
@VirtualUser Yes, the statements were 'nice' enough that I was able to derive it this way pretty quickly ... but the Truth-table method will be more general.
$endgroup$
– Bram28
Dec 13 '18 at 21:13
add a comment |
$begingroup$
Thanks for your reply I think that I have understood this problem, it is interesting alternative for table.
$endgroup$
– VirtualUser
Dec 13 '18 at 20:59
1
$begingroup$
@VirtualUser Yes, the statements were 'nice' enough that I was able to derive it this way pretty quickly ... but the Truth-table method will be more general.
$endgroup$
– Bram28
Dec 13 '18 at 21:13
$begingroup$
Thanks for your reply I think that I have understood this problem, it is interesting alternative for table.
$endgroup$
– VirtualUser
Dec 13 '18 at 20:59
$begingroup$
Thanks for your reply I think that I have understood this problem, it is interesting alternative for table.
$endgroup$
– VirtualUser
Dec 13 '18 at 20:59
1
1
$begingroup$
@VirtualUser Yes, the statements were 'nice' enough that I was able to derive it this way pretty quickly ... but the Truth-table method will be more general.
$endgroup$
– Bram28
Dec 13 '18 at 21:13
$begingroup$
@VirtualUser Yes, the statements were 'nice' enough that I was able to derive it this way pretty quickly ... but the Truth-table method will be more general.
$endgroup$
– Bram28
Dec 13 '18 at 21:13
add a comment |
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$begingroup$
Have you compared the values of alpha with the values of other columns? Remember, you're looking for a statement or formula, not necessarily a truth value
$endgroup$
– David Diaz
Dec 13 '18 at 20:27
$begingroup$
It should be tautology so $LS leftrightarrow RS $ must be true value. If I represent this by $alpha$ it should be true too. The same thing with $neg alpha $
$endgroup$
– VirtualUser
Dec 13 '18 at 20:30
1
$begingroup$
One could use the method of analytic tableaux.
$endgroup$
– Shaun
Dec 13 '18 at 20:30
$begingroup$
Ok, but my solution is good or not? Thanks for interesting link.
$endgroup$
– VirtualUser
Dec 13 '18 at 20:34
$begingroup$
@VirtualUser Your method of creating a truth-table is fine ... but your conclusion that it cannot be done is not. It can be done. Just see what value, if any, is forced on $alpha$ to make the truth-values come out right in each row. And then just generate an expression for that.
$endgroup$
– Bram28
Dec 13 '18 at 20:35