Mixing
How to solve tautology without using truth table?
$begingroup$
How do I determine that (¬a ∧ (a → b)) → ¬b is a tautology without a truth table. Ive tried switching a → b to ¬a or b but then I can't seem to do anything else to do after that.
discrete-mathematics
asked Jan 21 at 20:07
Aron JalbuenaAron Jalbuena
1
$endgroup$
add a comment |
$begingroup$
How do I determine that (¬a ∧ (a → b)) → ¬b is a tautology without a truth table. Ive tried switching a → b to ¬a or b but then I can't seem to do anything else to do after that.
discrete-mathematics
asked Jan 21 at 20:07
Aron JalbuenaAron Jalbuena
1
$endgroup$
add a comment |
$begingroup$
How do I determine that (¬a ∧ (a → b)) → ¬b is a tautology without a truth table. Ive tried switching a → b to ¬a or b but then I can't seem to do anything else to do after that.
discrete-mathematics
asked Jan 21 at 20:07
Aron JalbuenaAron Jalbuena
1
$endgroup$
How do I determine that (¬a ∧ (a → b)) → ¬b is a tautology without a truth table. Ive tried switching a → b to ¬a or b but then I can't seem to do anything else to do after that.
discrete-mathematics
discrete-mathematics
asked Jan 21 at 20:07
Aron JalbuenaAron Jalbuena
1
asked Jan 21 at 20:07
Aron JalbuenaAron Jalbuena
1
asked Jan 21 at 20:07
Aron JalbuenaAron Jalbuena
1
asked Jan 21 at 20:07
Aron JalbuenaAron Jalbuena
1
asked Jan 21 at 20:07
Aron JalbuenaAron Jalbuena
1
1
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
The statement you're trying to prove is false, so cannot be proven.
Take for example
a = It's Saturday
b = It's the weekend
Then your LHS suggests
1) it's not Saturday
and 2) if it's Saturday then it's the weekend
This does not imply that it's not the weekend! It could be Sunday.
answered Jan 21 at 20:15
bouncebackbounceback
40429
$endgroup$
add a comment |
$begingroup$
$neg a wedge (neg a vee b) $ can be rewritten as $(neg a vee neg a) wedge (neg a vee b)$. Now for the implication in question to be false, $neg b$ needs to be false, so $b$ is true. Assuming $b$ is true, we can get $(neg a vee neg a) wedge (neg a vee b)$ to be true when $a$ is false. So when $a$ is false and $b$ is true, we get that the overall implication in question is false. Thus, we find that your statement is not a tautology.
answered Jan 21 at 20:22
HyperionHyperion
646110
$endgroup$
add a comment |
$begingroup$
Since $ato b$ is equivalent to $neg alor b$, your statement is $neg atoneg b$ or equivalently $bto a$, which is clearly not a tautology.
answered Jan 21 at 20:32
J.G.J.G.
29.1k22845
$endgroup$
add a comment |
$begingroup$
Your formula must be wrong as it is showing not to be a tautology, are you sure you coped it correct?
answered Jan 21 at 20:37
Z.C97Z.C97
112
$endgroup$
$begingroup$
This is not a proper answer; it should be a comment under the question to request clarification from the OP.
$endgroup$
– YiFan
Jan 21 at 20:54
add a comment |
$begingroup$
There is a simple way of checking if an implication is a tautology. We have to establish that the given proposition is false.
If we have $Ato B $, then this is false only when $B$ is false and $A$ is true.
$ (¬a ∧ (a → b)) to ¬b;$can be re-interpreted in the above manner by assuming
$B$ is equivalent to $¬b$. We want to consider the case when B is false. If B is false, then b must be true.
$A$ is equivalent to $ (¬a ∧ (a → b))$. We want to prove that A is true.
We know that $b$ is true from $B$. So for $ato b$ is always true. Just by considering $a$ to be true, we get $T∧T$ which evaluates to True.
So we have the case where $Ato B$ is false when $a$ is true and $b$ is true. Hence we can prove by contradiction that $ (¬a ∧ (a → b)) to ¬b;$ is not a tautology.
answered Jan 22 at 6:13
SagnikSagnik
1377
$endgroup$
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The statement you're trying to prove is false, so cannot be proven.
Take for example
a = It's Saturday
b = It's the weekend
Then your LHS suggests
1) it's not Saturday
and 2) if it's Saturday then it's the weekend
This does not imply that it's not the weekend! It could be Sunday.
answered Jan 21 at 20:15
bouncebackbounceback
40429
$endgroup$
add a comment |
$begingroup$
The statement you're trying to prove is false, so cannot be proven.
Take for example
a = It's Saturday
b = It's the weekend
Then your LHS suggests
1) it's not Saturday
and 2) if it's Saturday then it's the weekend
This does not imply that it's not the weekend! It could be Sunday.
answered Jan 21 at 20:15
bouncebackbounceback
40429
$endgroup$
add a comment |
$begingroup$
The statement you're trying to prove is false, so cannot be proven.
Take for example
a = It's Saturday
b = It's the weekend
Then your LHS suggests
1) it's not Saturday
and 2) if it's Saturday then it's the weekend
This does not imply that it's not the weekend! It could be Sunday.
answered Jan 21 at 20:15
bouncebackbounceback
40429
$endgroup$
The statement you're trying to prove is false, so cannot be proven.
Take for example
a = It's Saturday
b = It's the weekend
Then your LHS suggests
1) it's not Saturday
and 2) if it's Saturday then it's the weekend
This does not imply that it's not the weekend! It could be Sunday.
answered Jan 21 at 20:15
bouncebackbounceback
40429
answered Jan 21 at 20:15
bouncebackbounceback
40429
answered Jan 21 at 20:15
bouncebackbounceback
40429
answered Jan 21 at 20:15
bouncebackbounceback
40429
40429
add a comment |
add a comment |
$begingroup$
$neg a wedge (neg a vee b) $ can be rewritten as $(neg a vee neg a) wedge (neg a vee b)$. Now for the implication in question to be false, $neg b$ needs to be false, so $b$ is true. Assuming $b$ is true, we can get $(neg a vee neg a) wedge (neg a vee b)$ to be true when $a$ is false. So when $a$ is false and $b$ is true, we get that the overall implication in question is false. Thus, we find that your statement is not a tautology.
answered Jan 21 at 20:22
HyperionHyperion
646110
$endgroup$
add a comment |
$begingroup$
$neg a wedge (neg a vee b) $ can be rewritten as $(neg a vee neg a) wedge (neg a vee b)$. Now for the implication in question to be false, $neg b$ needs to be false, so $b$ is true. Assuming $b$ is true, we can get $(neg a vee neg a) wedge (neg a vee b)$ to be true when $a$ is false. So when $a$ is false and $b$ is true, we get that the overall implication in question is false. Thus, we find that your statement is not a tautology.
answered Jan 21 at 20:22
HyperionHyperion
646110
$endgroup$
add a comment |
$begingroup$
$neg a wedge (neg a vee b) $ can be rewritten as $(neg a vee neg a) wedge (neg a vee b)$. Now for the implication in question to be false, $neg b$ needs to be false, so $b$ is true. Assuming $b$ is true, we can get $(neg a vee neg a) wedge (neg a vee b)$ to be true when $a$ is false. So when $a$ is false and $b$ is true, we get that the overall implication in question is false. Thus, we find that your statement is not a tautology.
answered Jan 21 at 20:22
HyperionHyperion
646110
$endgroup$
$neg a wedge (neg a vee b) $ can be rewritten as $(neg a vee neg a) wedge (neg a vee b)$. Now for the implication in question to be false, $neg b$ needs to be false, so $b$ is true. Assuming $b$ is true, we can get $(neg a vee neg a) wedge (neg a vee b)$ to be true when $a$ is false. So when $a$ is false and $b$ is true, we get that the overall implication in question is false. Thus, we find that your statement is not a tautology.
answered Jan 21 at 20:22
HyperionHyperion
646110
answered Jan 21 at 20:22
HyperionHyperion
646110
answered Jan 21 at 20:22
HyperionHyperion
646110
answered Jan 21 at 20:22
HyperionHyperion
646110
646110
add a comment |
add a comment |
$begingroup$
Since $ato b$ is equivalent to $neg alor b$, your statement is $neg atoneg b$ or equivalently $bto a$, which is clearly not a tautology.
answered Jan 21 at 20:32
J.G.J.G.
29.1k22845
$endgroup$
add a comment |
$begingroup$
Since $ato b$ is equivalent to $neg alor b$, your statement is $neg atoneg b$ or equivalently $bto a$, which is clearly not a tautology.
answered Jan 21 at 20:32
J.G.J.G.
29.1k22845
$endgroup$
add a comment |
$begingroup$
Since $ato b$ is equivalent to $neg alor b$, your statement is $neg atoneg b$ or equivalently $bto a$, which is clearly not a tautology.
answered Jan 21 at 20:32
J.G.J.G.
29.1k22845
$endgroup$
Since $ato b$ is equivalent to $neg alor b$, your statement is $neg atoneg b$ or equivalently $bto a$, which is clearly not a tautology.
answered Jan 21 at 20:32
J.G.J.G.
29.1k22845
answered Jan 21 at 20:32
J.G.J.G.
29.1k22845
answered Jan 21 at 20:32
J.G.J.G.
29.1k22845
answered Jan 21 at 20:32
J.G.J.G.
29.1k22845
29.1k22845
add a comment |
add a comment |
$begingroup$
Your formula must be wrong as it is showing not to be a tautology, are you sure you coped it correct?
answered Jan 21 at 20:37
Z.C97Z.C97
112
$endgroup$
$begingroup$
This is not a proper answer; it should be a comment under the question to request clarification from the OP.
$endgroup$
– YiFan
Jan 21 at 20:54
add a comment |
$begingroup$
Your formula must be wrong as it is showing not to be a tautology, are you sure you coped it correct?
answered Jan 21 at 20:37
Z.C97Z.C97
112
$endgroup$
$begingroup$
This is not a proper answer; it should be a comment under the question to request clarification from the OP.
$endgroup$
– YiFan
Jan 21 at 20:54
add a comment |
$begingroup$
Your formula must be wrong as it is showing not to be a tautology, are you sure you coped it correct?
answered Jan 21 at 20:37
Z.C97Z.C97
112
$endgroup$
Your formula must be wrong as it is showing not to be a tautology, are you sure you coped it correct?
answered Jan 21 at 20:37
Z.C97Z.C97
112
answered Jan 21 at 20:37
Z.C97Z.C97
112
answered Jan 21 at 20:37
Z.C97Z.C97
112
answered Jan 21 at 20:37
Z.C97Z.C97
112
112
$begingroup$
This is not a proper answer; it should be a comment under the question to request clarification from the OP.
$endgroup$
– YiFan
Jan 21 at 20:54
add a comment |
$begingroup$
This is not a proper answer; it should be a comment under the question to request clarification from the OP.
$endgroup$
– YiFan
Jan 21 at 20:54
$begingroup$
This is not a proper answer; it should be a comment under the question to request clarification from the OP.
$endgroup$
– YiFan
Jan 21 at 20:54
$begingroup$
This is not a proper answer; it should be a comment under the question to request clarification from the OP.
$endgroup$
– YiFan
Jan 21 at 20:54
add a comment |
$begingroup$
There is a simple way of checking if an implication is a tautology. We have to establish that the given proposition is false.
If we have $Ato B $, then this is false only when $B$ is false and $A$ is true.
$ (¬a ∧ (a → b)) to ¬b;$can be re-interpreted in the above manner by assuming
$B$ is equivalent to $¬b$. We want to consider the case when B is false. If B is false, then b must be true.
$A$ is equivalent to $ (¬a ∧ (a → b))$. We want to prove that A is true.
We know that $b$ is true from $B$. So for $ato b$ is always true. Just by considering $a$ to be true, we get $T∧T$ which evaluates to True.
So we have the case where $Ato B$ is false when $a$ is true and $b$ is true. Hence we can prove by contradiction that $ (¬a ∧ (a → b)) to ¬b;$ is not a tautology.
answered Jan 22 at 6:13
SagnikSagnik
1377
$endgroup$
add a comment |
$begingroup$
There is a simple way of checking if an implication is a tautology. We have to establish that the given proposition is false.
If we have $Ato B $, then this is false only when $B$ is false and $A$ is true.
$ (¬a ∧ (a → b)) to ¬b;$can be re-interpreted in the above manner by assuming
$B$ is equivalent to $¬b$. We want to consider the case when B is false. If B is false, then b must be true.
$A$ is equivalent to $ (¬a ∧ (a → b))$. We want to prove that A is true.
We know that $b$ is true from $B$. So for $ato b$ is always true. Just by considering $a$ to be true, we get $T∧T$ which evaluates to True.
So we have the case where $Ato B$ is false when $a$ is true and $b$ is true. Hence we can prove by contradiction that $ (¬a ∧ (a → b)) to ¬b;$ is not a tautology.
answered Jan 22 at 6:13
SagnikSagnik
1377
$endgroup$
add a comment |
$begingroup$
There is a simple way of checking if an implication is a tautology. We have to establish that the given proposition is false.
If we have $Ato B $, then this is false only when $B$ is false and $A$ is true.
$ (¬a ∧ (a → b)) to ¬b;$can be re-interpreted in the above manner by assuming
$B$ is equivalent to $¬b$. We want to consider the case when B is false. If B is false, then b must be true.
$A$ is equivalent to $ (¬a ∧ (a → b))$. We want to prove that A is true.
We know that $b$ is true from $B$. So for $ato b$ is always true. Just by considering $a$ to be true, we get $T∧T$ which evaluates to True.
So we have the case where $Ato B$ is false when $a$ is true and $b$ is true. Hence we can prove by contradiction that $ (¬a ∧ (a → b)) to ¬b;$ is not a tautology.
answered Jan 22 at 6:13
SagnikSagnik
1377
$endgroup$
There is a simple way of checking if an implication is a tautology. We have to establish that the given proposition is false.
If we have $Ato B $, then this is false only when $B$ is false and $A$ is true.
$ (¬a ∧ (a → b)) to ¬b;$can be re-interpreted in the above manner by assuming
$B$ is equivalent to $¬b$. We want to consider the case when B is false. If B is false, then b must be true.
$A$ is equivalent to $ (¬a ∧ (a → b))$. We want to prove that A is true.
We know that $b$ is true from $B$. So for $ato b$ is always true. Just by considering $a$ to be true, we get $T∧T$ which evaluates to True.
So we have the case where $Ato B$ is false when $a$ is true and $b$ is true. Hence we can prove by contradiction that $ (¬a ∧ (a → b)) to ¬b;$ is not a tautology.
answered Jan 22 at 6:13
SagnikSagnik
1377
answered Jan 22 at 6:13
SagnikSagnik
1377
answered Jan 22 at 6:13
SagnikSagnik
1377
answered Jan 22 at 6:13
SagnikSagnik
1377
1377
add a comment |
add a comment |
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