logical expressions and implication
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I am given 3 statements and I have to determine whether they implicate each other or not. however the use of $forall$ kind of confuses me.
the statements are the following ( forgive me if i'm addressing them in the wrong way) :
$(forall x alpha(x)) rightarrow (forall x beta(x)) $
$forall x (alpha(x) rightarrow beta(x))$
$exists x alpha(x) land lnot exists x beta(x) $
logic
asked Nov 25 at 15:43
lidor718
6
add a comment |
up vote
0
down vote
favorite
I am given 3 statements and I have to determine whether they implicate each other or not. however the use of $forall$ kind of confuses me.
the statements are the following ( forgive me if i'm addressing them in the wrong way) :
$(forall x alpha(x)) rightarrow (forall x beta(x)) $
$forall x (alpha(x) rightarrow beta(x))$
$exists x alpha(x) land lnot exists x beta(x) $
logic
asked Nov 25 at 15:43
lidor718
6
1
$forall$ means the relation should hold for every $x$ in the domain. I think you need to tell what is $alpha$ and $beta$ and write down how you tried the problem.
– zenith
Nov 25 at 15:54
That is it. I know what forall means, I just can't figure out if 1 implies 2 for example because of that forall there..
– lidor718
Nov 25 at 15:57
$(forall x alpha(x)) rightarrow (forall x beta(x))$ is the same as $(exists x neg alpha(x)) vee (forall x beta(x)) $. $forall x (alpha(x) rightarrow beta(x))$ is the same as $forall x (neg alpha(x) vee beta(x))$ $exists x alpha(x) wedge neg exists x beta(x)$ is the same as $forall x alpha(x) wedge forall x neg beta(x)$.
– mathnoob
Nov 25 at 16:06
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am given 3 statements and I have to determine whether they implicate each other or not. however the use of $forall$ kind of confuses me.
the statements are the following ( forgive me if i'm addressing them in the wrong way) :
$(forall x alpha(x)) rightarrow (forall x beta(x)) $
$forall x (alpha(x) rightarrow beta(x))$
$exists x alpha(x) land lnot exists x beta(x) $
logic
asked Nov 25 at 15:43
lidor718
6
I am given 3 statements and I have to determine whether they implicate each other or not. however the use of $forall$ kind of confuses me.
the statements are the following ( forgive me if i'm addressing them in the wrong way) :
$(forall x alpha(x)) rightarrow (forall x beta(x)) $
$forall x (alpha(x) rightarrow beta(x))$
$exists x alpha(x) land lnot exists x beta(x) $
logic
logic
asked Nov 25 at 15:43
lidor718
6
asked Nov 25 at 15:43
lidor718
6
asked Nov 25 at 15:43
lidor718
6
asked Nov 25 at 15:43
lidor718
6
asked Nov 25 at 15:43
lidor718
6
6
1
$forall$ means the relation should hold for every $x$ in the domain. I think you need to tell what is $alpha$ and $beta$ and write down how you tried the problem.
– zenith
Nov 25 at 15:54
That is it. I know what forall means, I just can't figure out if 1 implies 2 for example because of that forall there..
– lidor718
Nov 25 at 15:57
$(forall x alpha(x)) rightarrow (forall x beta(x))$ is the same as $(exists x neg alpha(x)) vee (forall x beta(x)) $. $forall x (alpha(x) rightarrow beta(x))$ is the same as $forall x (neg alpha(x) vee beta(x))$ $exists x alpha(x) wedge neg exists x beta(x)$ is the same as $forall x alpha(x) wedge forall x neg beta(x)$.
– mathnoob
Nov 25 at 16:06
add a comment |
1
$forall$ means the relation should hold for every $x$ in the domain. I think you need to tell what is $alpha$ and $beta$ and write down how you tried the problem.
– zenith
Nov 25 at 15:54
That is it. I know what forall means, I just can't figure out if 1 implies 2 for example because of that forall there..
– lidor718
Nov 25 at 15:57
$(forall x alpha(x)) rightarrow (forall x beta(x))$ is the same as $(exists x neg alpha(x)) vee (forall x beta(x)) $. $forall x (alpha(x) rightarrow beta(x))$ is the same as $forall x (neg alpha(x) vee beta(x))$ $exists x alpha(x) wedge neg exists x beta(x)$ is the same as $forall x alpha(x) wedge forall x neg beta(x)$.
– mathnoob
Nov 25 at 16:06
1
1
$forall$ means the relation should hold for every $x$ in the domain. I think you need to tell what is $alpha$ and $beta$ and write down how you tried the problem.
– zenith
Nov 25 at 15:54
$forall$ means the relation should hold for every $x$ in the domain. I think you need to tell what is $alpha$ and $beta$ and write down how you tried the problem.
– zenith
Nov 25 at 15:54
That is it. I know what forall means, I just can't figure out if 1 implies 2 for example because of that forall there..
– lidor718
Nov 25 at 15:57
That is it. I know what forall means, I just can't figure out if 1 implies 2 for example because of that forall there..
– lidor718
Nov 25 at 15:57
$(forall x alpha(x)) rightarrow (forall x beta(x))$ is the same as $(exists x neg alpha(x)) vee (forall x beta(x)) $. $forall x (alpha(x) rightarrow beta(x))$ is the same as $forall x (neg alpha(x) vee beta(x))$ $exists x alpha(x) wedge neg exists x beta(x)$ is the same as $forall x alpha(x) wedge forall x neg beta(x)$.
– mathnoob
Nov 25 at 16:06
$(forall x alpha(x)) rightarrow (forall x beta(x))$ is the same as $(exists x neg alpha(x)) vee (forall x beta(x)) $. $forall x (alpha(x) rightarrow beta(x))$ is the same as $forall x (neg alpha(x) vee beta(x))$ $exists x alpha(x) wedge neg exists x beta(x)$ is the same as $forall x alpha(x) wedge forall x neg beta(x)$.
– mathnoob
Nov 25 at 16:06
add a comment |
1 Answer
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If 2. holds in some model, this means that for all $x$ in the model, if we know $alpha(x)$ holds for that $x$ then we can conclude $beta(x)$ holds for that $x$ as well.
So if we know or assume that for all $x$ in the model $alpha(x)$ holds, so we
assume $forall x alpha(x)$, then applying 2. at each $x$ of the model, we also know that $beta(x)$ holds for all $x$, so $forall x beta(x)$. So the implication
$$(forall x alpha(x)) rightarrow (forall x beta(x))$$
is valid in that model.
Conclusion: 2 implies 1 in all models.
If we have a model for 3. then it cannot be a model for 1 anymore: for if all elements would satisfy $alpha$ (and there is at least one, from satisfying 3.) then all elements would satisfy $beta$ too, and this is also not the case when 3. is satisfied.
Conclusion: 3 implies not 1, or equivalent, 1 implies "not 3".
answered Nov 25 at 16:02
Henno Brandsma
103k345112
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
If 2. holds in some model, this means that for all $x$ in the model, if we know $alpha(x)$ holds for that $x$ then we can conclude $beta(x)$ holds for that $x$ as well.
So if we know or assume that for all $x$ in the model $alpha(x)$ holds, so we
assume $forall x alpha(x)$, then applying 2. at each $x$ of the model, we also know that $beta(x)$ holds for all $x$, so $forall x beta(x)$. So the implication
$$(forall x alpha(x)) rightarrow (forall x beta(x))$$
is valid in that model.
Conclusion: 2 implies 1 in all models.
If we have a model for 3. then it cannot be a model for 1 anymore: for if all elements would satisfy $alpha$ (and there is at least one, from satisfying 3.) then all elements would satisfy $beta$ too, and this is also not the case when 3. is satisfied.
Conclusion: 3 implies not 1, or equivalent, 1 implies "not 3".
answered Nov 25 at 16:02
Henno Brandsma
103k345112
add a comment |
up vote
1
down vote
If 2. holds in some model, this means that for all $x$ in the model, if we know $alpha(x)$ holds for that $x$ then we can conclude $beta(x)$ holds for that $x$ as well.
So if we know or assume that for all $x$ in the model $alpha(x)$ holds, so we
assume $forall x alpha(x)$, then applying 2. at each $x$ of the model, we also know that $beta(x)$ holds for all $x$, so $forall x beta(x)$. So the implication
$$(forall x alpha(x)) rightarrow (forall x beta(x))$$
is valid in that model.
Conclusion: 2 implies 1 in all models.
If we have a model for 3. then it cannot be a model for 1 anymore: for if all elements would satisfy $alpha$ (and there is at least one, from satisfying 3.) then all elements would satisfy $beta$ too, and this is also not the case when 3. is satisfied.
Conclusion: 3 implies not 1, or equivalent, 1 implies "not 3".
answered Nov 25 at 16:02
Henno Brandsma
103k345112
add a comment |
up vote
1
down vote
up vote
1
down vote
If 2. holds in some model, this means that for all $x$ in the model, if we know $alpha(x)$ holds for that $x$ then we can conclude $beta(x)$ holds for that $x$ as well.
So if we know or assume that for all $x$ in the model $alpha(x)$ holds, so we
assume $forall x alpha(x)$, then applying 2. at each $x$ of the model, we also know that $beta(x)$ holds for all $x$, so $forall x beta(x)$. So the implication
$$(forall x alpha(x)) rightarrow (forall x beta(x))$$
is valid in that model.
Conclusion: 2 implies 1 in all models.
If we have a model for 3. then it cannot be a model for 1 anymore: for if all elements would satisfy $alpha$ (and there is at least one, from satisfying 3.) then all elements would satisfy $beta$ too, and this is also not the case when 3. is satisfied.
Conclusion: 3 implies not 1, or equivalent, 1 implies "not 3".
answered Nov 25 at 16:02
Henno Brandsma
103k345112
If 2. holds in some model, this means that for all $x$ in the model, if we know $alpha(x)$ holds for that $x$ then we can conclude $beta(x)$ holds for that $x$ as well.
So if we know or assume that for all $x$ in the model $alpha(x)$ holds, so we
assume $forall x alpha(x)$, then applying 2. at each $x$ of the model, we also know that $beta(x)$ holds for all $x$, so $forall x beta(x)$. So the implication
$$(forall x alpha(x)) rightarrow (forall x beta(x))$$
is valid in that model.
Conclusion: 2 implies 1 in all models.
If we have a model for 3. then it cannot be a model for 1 anymore: for if all elements would satisfy $alpha$ (and there is at least one, from satisfying 3.) then all elements would satisfy $beta$ too, and this is also not the case when 3. is satisfied.
Conclusion: 3 implies not 1, or equivalent, 1 implies "not 3".
answered Nov 25 at 16:02
Henno Brandsma
103k345112
answered Nov 25 at 16:02
Henno Brandsma
103k345112
answered Nov 25 at 16:02
Henno Brandsma
103k345112
answered Nov 25 at 16:02
Henno Brandsma
103k345112
103k345112
add a comment |
add a comment |
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1
$forall$ means the relation should hold for every $x$ in the domain. I think you need to tell what is $alpha$ and $beta$ and write down how you tried the problem.
– zenith
Nov 25 at 15:54
That is it. I know what forall means, I just can't figure out if 1 implies 2 for example because of that forall there..
– lidor718
Nov 25 at 15:57
$(forall x alpha(x)) rightarrow (forall x beta(x))$ is the same as $(exists x neg alpha(x)) vee (forall x beta(x)) $. $forall x (alpha(x) rightarrow beta(x))$ is the same as $forall x (neg alpha(x) vee beta(x))$ $exists x alpha(x) wedge neg exists x beta(x)$ is the same as $forall x alpha(x) wedge forall x neg beta(x)$.
– mathnoob
Nov 25 at 16:06