What does $forall x phi rightarrow psi$ mean?
$begingroup$
If I have a formula $forall x phi rightarrow psi$, how can I know if it means $(forall x phi) rightarrow psi$ or $forall x (phi rightarrow psi)$?
logic
asked Jan 20 at 18:34
roi_saumonroi_saumon
56938
$endgroup$
add a comment |
$begingroup$
If I have a formula $forall x phi rightarrow psi$, how can I know if it means $(forall x phi) rightarrow psi$ or $forall x (phi rightarrow psi)$?
logic
asked Jan 20 at 18:34
roi_saumonroi_saumon
56938
$endgroup$
1
$begingroup$
Optimality, only one of them makes sense, considering the number of free variables in each of $psi$ and $phi$. But it ought to be clarified.
$endgroup$
– Arthur
Jan 20 at 18:45
2
$begingroup$
Without explicit convention (see you textbook), usually the quantifiers applies as little as possible. Thus it is $(forall x phi) to psi$
$endgroup$
– Mauro ALLEGRANZA
Jan 20 at 18:46
add a comment |
$begingroup$
If I have a formula $forall x phi rightarrow psi$, how can I know if it means $(forall x phi) rightarrow psi$ or $forall x (phi rightarrow psi)$?
logic
asked Jan 20 at 18:34
roi_saumonroi_saumon
56938
$endgroup$
If I have a formula $forall x phi rightarrow psi$, how can I know if it means $(forall x phi) rightarrow psi$ or $forall x (phi rightarrow psi)$?
logic
logic
asked Jan 20 at 18:34
roi_saumonroi_saumon
56938
asked Jan 20 at 18:34
roi_saumonroi_saumon
56938
asked Jan 20 at 18:34
roi_saumonroi_saumon
56938
asked Jan 20 at 18:34
roi_saumonroi_saumon
56938
asked Jan 20 at 18:34
roi_saumonroi_saumon
56938
56938
1
$begingroup$
Optimality, only one of them makes sense, considering the number of free variables in each of $psi$ and $phi$. But it ought to be clarified.
$endgroup$
– Arthur
Jan 20 at 18:45
2
$begingroup$
Without explicit convention (see you textbook), usually the quantifiers applies as little as possible. Thus it is $(forall x phi) to psi$
$endgroup$
– Mauro ALLEGRANZA
Jan 20 at 18:46
add a comment |
1
$begingroup$
Optimality, only one of them makes sense, considering the number of free variables in each of $psi$ and $phi$. But it ought to be clarified.
$endgroup$
– Arthur
Jan 20 at 18:45
2
$begingroup$
Without explicit convention (see you textbook), usually the quantifiers applies as little as possible. Thus it is $(forall x phi) to psi$
$endgroup$
– Mauro ALLEGRANZA
Jan 20 at 18:46
1
1
$begingroup$
Optimality, only one of them makes sense, considering the number of free variables in each of $psi$ and $phi$. But it ought to be clarified.
$endgroup$
– Arthur
Jan 20 at 18:45
$begingroup$
Optimality, only one of them makes sense, considering the number of free variables in each of $psi$ and $phi$. But it ought to be clarified.
$endgroup$
– Arthur
Jan 20 at 18:45
2
2
$begingroup$
Without explicit convention (see you textbook), usually the quantifiers applies as little as possible. Thus it is $(forall x phi) to psi$
$endgroup$
– Mauro ALLEGRANZA
Jan 20 at 18:46
$begingroup$
Without explicit convention (see you textbook), usually the quantifiers applies as little as possible. Thus it is $(forall x phi) to psi$
$endgroup$
– Mauro ALLEGRANZA
Jan 20 at 18:46
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
According to how WFFs are formed, the former interpretation is correct, and the correct expression is
$$(forall x phi rightarrow psi)$$
and not
$$forall x(phi rightarrow psi)$$
Notice the parenthesis around the entire implication. You don't need parenthesis over $forall x phi$, as it is a WFF already.
I do think that
$$ (forall x phi) rightarrow psi$$
is more readable, though, even though it's not an official WFF.
answered Jan 20 at 18:52
MetricMetric
1,23649
$endgroup$
add a comment |
$begingroup$
It's a notational convention: quantifiers bind more tightly than Booleans, so the former interpretation is correct. In my opinion, though, it's a bit obnoxious to write it like that and parentheses really should be included.
Although sadly I've seen this deviated from occasionally - but I've never seen deviation from this convention in a textbook on logic, so it should serve you well.
(For me, the obnoxiousness comes from the relative rarity. I wouldn't consider "$x^2-3x-4$" to be obnoxious, even though one could interpret it incorrectly if unfamiliar with PEMDAS. On the other hand, I would consider $xdiv ydiv z$ to be obnoxious. So there is an element of subjectivity to my assessment here.)
answered Jan 20 at 18:45
Noah SchweberNoah Schweber
125k10150287
$endgroup$
add a comment |
$begingroup$
It largely depends on whether $phi,$ and $psi$ are formulas including $x$, and even then, without parentheses indicating either interpretation you provide, then we must assume the former of your options.
Without parentheses displaying the scope of the universal quantifier for $x$, you should take $forall x phi rightarrow psi$ to mean $(forall x,(phi))to psi$ (where the quantifier only refers to the occurrence of $x in phi$).
Although, by convention, if all one is given is $forall x,phi to psi$, we can only take the quantifier to apply to $phi$, it is rather careless to write it this way. As you show in your question about what this might mean, readers shouldn't be left to guess one or the other; rather, the author should should write either $$(forall x(phi)) to psi,; text{ or else };forall x big(phi to psi)$$ depending on the scope of the quantifier on $x$, in order to help make themselves clear. Without parentheses, $forall x$ applies only to $phi$; however, if one intends for $forall x$ to bound all of $phi to psi$, then one must include all of the scope of the quantifier in parentheses.
answered Jan 20 at 18:46
jordan_glenjordan_glen
1
$endgroup$
$begingroup$
How can $forall x(phito psi)$ mean either?
$endgroup$
– Arthur
Jan 20 at 18:48
$begingroup$
I think I've made myself clear, now, @Arthur. Thanks for commenting.
$endgroup$
– jordan_glen
Jan 20 at 18:57
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
According to how WFFs are formed, the former interpretation is correct, and the correct expression is
$$(forall x phi rightarrow psi)$$
and not
$$forall x(phi rightarrow psi)$$
Notice the parenthesis around the entire implication. You don't need parenthesis over $forall x phi$, as it is a WFF already.
I do think that
$$ (forall x phi) rightarrow psi$$
is more readable, though, even though it's not an official WFF.
answered Jan 20 at 18:52
MetricMetric
1,23649
$endgroup$
add a comment |
$begingroup$
According to how WFFs are formed, the former interpretation is correct, and the correct expression is
$$(forall x phi rightarrow psi)$$
and not
$$forall x(phi rightarrow psi)$$
Notice the parenthesis around the entire implication. You don't need parenthesis over $forall x phi$, as it is a WFF already.
I do think that
$$ (forall x phi) rightarrow psi$$
is more readable, though, even though it's not an official WFF.
answered Jan 20 at 18:52
MetricMetric
1,23649
$endgroup$
add a comment |
$begingroup$
According to how WFFs are formed, the former interpretation is correct, and the correct expression is
$$(forall x phi rightarrow psi)$$
and not
$$forall x(phi rightarrow psi)$$
Notice the parenthesis around the entire implication. You don't need parenthesis over $forall x phi$, as it is a WFF already.
I do think that
$$ (forall x phi) rightarrow psi$$
is more readable, though, even though it's not an official WFF.
answered Jan 20 at 18:52
MetricMetric
1,23649
$endgroup$
According to how WFFs are formed, the former interpretation is correct, and the correct expression is
$$(forall x phi rightarrow psi)$$
and not
$$forall x(phi rightarrow psi)$$
Notice the parenthesis around the entire implication. You don't need parenthesis over $forall x phi$, as it is a WFF already.
I do think that
$$ (forall x phi) rightarrow psi$$
is more readable, though, even though it's not an official WFF.
answered Jan 20 at 18:52
MetricMetric
1,23649
answered Jan 20 at 18:52
MetricMetric
1,23649
answered Jan 20 at 18:52
MetricMetric
1,23649
answered Jan 20 at 18:52
MetricMetric
1,23649
1,23649
add a comment |
add a comment |
$begingroup$
It's a notational convention: quantifiers bind more tightly than Booleans, so the former interpretation is correct. In my opinion, though, it's a bit obnoxious to write it like that and parentheses really should be included.
Although sadly I've seen this deviated from occasionally - but I've never seen deviation from this convention in a textbook on logic, so it should serve you well.
(For me, the obnoxiousness comes from the relative rarity. I wouldn't consider "$x^2-3x-4$" to be obnoxious, even though one could interpret it incorrectly if unfamiliar with PEMDAS. On the other hand, I would consider $xdiv ydiv z$ to be obnoxious. So there is an element of subjectivity to my assessment here.)
answered Jan 20 at 18:45
Noah SchweberNoah Schweber
125k10150287
$endgroup$
add a comment |
$begingroup$
It's a notational convention: quantifiers bind more tightly than Booleans, so the former interpretation is correct. In my opinion, though, it's a bit obnoxious to write it like that and parentheses really should be included.
Although sadly I've seen this deviated from occasionally - but I've never seen deviation from this convention in a textbook on logic, so it should serve you well.
(For me, the obnoxiousness comes from the relative rarity. I wouldn't consider "$x^2-3x-4$" to be obnoxious, even though one could interpret it incorrectly if unfamiliar with PEMDAS. On the other hand, I would consider $xdiv ydiv z$ to be obnoxious. So there is an element of subjectivity to my assessment here.)
answered Jan 20 at 18:45
Noah SchweberNoah Schweber
125k10150287
$endgroup$
add a comment |
$begingroup$
It's a notational convention: quantifiers bind more tightly than Booleans, so the former interpretation is correct. In my opinion, though, it's a bit obnoxious to write it like that and parentheses really should be included.
Although sadly I've seen this deviated from occasionally - but I've never seen deviation from this convention in a textbook on logic, so it should serve you well.
(For me, the obnoxiousness comes from the relative rarity. I wouldn't consider "$x^2-3x-4$" to be obnoxious, even though one could interpret it incorrectly if unfamiliar with PEMDAS. On the other hand, I would consider $xdiv ydiv z$ to be obnoxious. So there is an element of subjectivity to my assessment here.)
answered Jan 20 at 18:45
Noah SchweberNoah Schweber
125k10150287
$endgroup$
It's a notational convention: quantifiers bind more tightly than Booleans, so the former interpretation is correct. In my opinion, though, it's a bit obnoxious to write it like that and parentheses really should be included.
Although sadly I've seen this deviated from occasionally - but I've never seen deviation from this convention in a textbook on logic, so it should serve you well.
(For me, the obnoxiousness comes from the relative rarity. I wouldn't consider "$x^2-3x-4$" to be obnoxious, even though one could interpret it incorrectly if unfamiliar with PEMDAS. On the other hand, I would consider $xdiv ydiv z$ to be obnoxious. So there is an element of subjectivity to my assessment here.)
answered Jan 20 at 18:45
Noah SchweberNoah Schweber
125k10150287
answered Jan 20 at 18:45
Noah SchweberNoah Schweber
125k10150287
answered Jan 20 at 18:45
Noah SchweberNoah Schweber
125k10150287
answered Jan 20 at 18:45
Noah SchweberNoah Schweber
125k10150287
125k10150287
add a comment |
add a comment |
$begingroup$
It largely depends on whether $phi,$ and $psi$ are formulas including $x$, and even then, without parentheses indicating either interpretation you provide, then we must assume the former of your options.
Without parentheses displaying the scope of the universal quantifier for $x$, you should take $forall x phi rightarrow psi$ to mean $(forall x,(phi))to psi$ (where the quantifier only refers to the occurrence of $x in phi$).
Although, by convention, if all one is given is $forall x,phi to psi$, we can only take the quantifier to apply to $phi$, it is rather careless to write it this way. As you show in your question about what this might mean, readers shouldn't be left to guess one or the other; rather, the author should should write either $$(forall x(phi)) to psi,; text{ or else };forall x big(phi to psi)$$ depending on the scope of the quantifier on $x$, in order to help make themselves clear. Without parentheses, $forall x$ applies only to $phi$; however, if one intends for $forall x$ to bound all of $phi to psi$, then one must include all of the scope of the quantifier in parentheses.
answered Jan 20 at 18:46
jordan_glenjordan_glen
1
$endgroup$
$begingroup$
How can $forall x(phito psi)$ mean either?
$endgroup$
– Arthur
Jan 20 at 18:48
$begingroup$
I think I've made myself clear, now, @Arthur. Thanks for commenting.
$endgroup$
– jordan_glen
Jan 20 at 18:57
add a comment |
$begingroup$
It largely depends on whether $phi,$ and $psi$ are formulas including $x$, and even then, without parentheses indicating either interpretation you provide, then we must assume the former of your options.
Without parentheses displaying the scope of the universal quantifier for $x$, you should take $forall x phi rightarrow psi$ to mean $(forall x,(phi))to psi$ (where the quantifier only refers to the occurrence of $x in phi$).
Although, by convention, if all one is given is $forall x,phi to psi$, we can only take the quantifier to apply to $phi$, it is rather careless to write it this way. As you show in your question about what this might mean, readers shouldn't be left to guess one or the other; rather, the author should should write either $$(forall x(phi)) to psi,; text{ or else };forall x big(phi to psi)$$ depending on the scope of the quantifier on $x$, in order to help make themselves clear. Without parentheses, $forall x$ applies only to $phi$; however, if one intends for $forall x$ to bound all of $phi to psi$, then one must include all of the scope of the quantifier in parentheses.
answered Jan 20 at 18:46
jordan_glenjordan_glen
1
$endgroup$
$begingroup$
How can $forall x(phito psi)$ mean either?
$endgroup$
– Arthur
Jan 20 at 18:48
$begingroup$
I think I've made myself clear, now, @Arthur. Thanks for commenting.
$endgroup$
– jordan_glen
Jan 20 at 18:57
add a comment |
$begingroup$
It largely depends on whether $phi,$ and $psi$ are formulas including $x$, and even then, without parentheses indicating either interpretation you provide, then we must assume the former of your options.
Without parentheses displaying the scope of the universal quantifier for $x$, you should take $forall x phi rightarrow psi$ to mean $(forall x,(phi))to psi$ (where the quantifier only refers to the occurrence of $x in phi$).
Although, by convention, if all one is given is $forall x,phi to psi$, we can only take the quantifier to apply to $phi$, it is rather careless to write it this way. As you show in your question about what this might mean, readers shouldn't be left to guess one or the other; rather, the author should should write either $$(forall x(phi)) to psi,; text{ or else };forall x big(phi to psi)$$ depending on the scope of the quantifier on $x$, in order to help make themselves clear. Without parentheses, $forall x$ applies only to $phi$; however, if one intends for $forall x$ to bound all of $phi to psi$, then one must include all of the scope of the quantifier in parentheses.
answered Jan 20 at 18:46
jordan_glenjordan_glen
1
$endgroup$
It largely depends on whether $phi,$ and $psi$ are formulas including $x$, and even then, without parentheses indicating either interpretation you provide, then we must assume the former of your options.
Without parentheses displaying the scope of the universal quantifier for $x$, you should take $forall x phi rightarrow psi$ to mean $(forall x,(phi))to psi$ (where the quantifier only refers to the occurrence of $x in phi$).
Although, by convention, if all one is given is $forall x,phi to psi$, we can only take the quantifier to apply to $phi$, it is rather careless to write it this way. As you show in your question about what this might mean, readers shouldn't be left to guess one or the other; rather, the author should should write either $$(forall x(phi)) to psi,; text{ or else };forall x big(phi to psi)$$ depending on the scope of the quantifier on $x$, in order to help make themselves clear. Without parentheses, $forall x$ applies only to $phi$; however, if one intends for $forall x$ to bound all of $phi to psi$, then one must include all of the scope of the quantifier in parentheses.
answered Jan 20 at 18:46
jordan_glenjordan_glen
1
edited Jan 20 at 19:07
answered Jan 20 at 18:46
jordan_glenjordan_glen
1
answered Jan 20 at 18:46
jordan_glenjordan_glen
1
answered Jan 20 at 18:46
jordan_glenjordan_glen
1
1
$begingroup$
How can $forall x(phito psi)$ mean either?
$endgroup$
– Arthur
Jan 20 at 18:48
$begingroup$
I think I've made myself clear, now, @Arthur. Thanks for commenting.
$endgroup$
– jordan_glen
Jan 20 at 18:57
add a comment |
$begingroup$
How can $forall x(phito psi)$ mean either?
$endgroup$
– Arthur
Jan 20 at 18:48
$begingroup$
I think I've made myself clear, now, @Arthur. Thanks for commenting.
$endgroup$
– jordan_glen
Jan 20 at 18:57
$begingroup$
How can $forall x(phito psi)$ mean either?
$endgroup$
– Arthur
Jan 20 at 18:48
$begingroup$
How can $forall x(phito psi)$ mean either?
$endgroup$
– Arthur
Jan 20 at 18:48
$begingroup$
I think I've made myself clear, now, @Arthur. Thanks for commenting.
$endgroup$
– jordan_glen
Jan 20 at 18:57
$begingroup$
I think I've made myself clear, now, @Arthur. Thanks for commenting.
$endgroup$
– jordan_glen
Jan 20 at 18:57
add a comment |
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1
$begingroup$
Optimality, only one of them makes sense, considering the number of free variables in each of $psi$ and $phi$. But it ought to be clarified.
$endgroup$
– Arthur
Jan 20 at 18:45
2
$begingroup$
Without explicit convention (see you textbook), usually the quantifiers applies as little as possible. Thus it is $(forall x phi) to psi$
$endgroup$
– Mauro ALLEGRANZA
Jan 20 at 18:46