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Is $ forall x in emptyset. (P(x) land lnot P(x)) $ true?
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up vote
5
down vote
favorite
I have searched the forums but haven't found any answers to this question, so I thought I'd ask it myself.
Let $P(x)$ be an arbitrary predicate. Does the following statement evaluate to true or false?
$ forall x in emptyset. (P(x) land lnot P(x)) $
It should evaluate to true because there is no element, hence it is true for all elements.
But it should evaluate to false because $ P(x) land lnot P(x) $ is always false.
Does anyone have any ideas on how to answer this question?
logic predicate-logic
edited Jul 30 at 15:25
user21820
35.8k440136
asked Jul 30 at 9:51
Sven G.
285
add a comment |Â
up vote
5
down vote
favorite
I have searched the forums but haven't found any answers to this question, so I thought I'd ask it myself.
Let $P(x)$ be an arbitrary predicate. Does the following statement evaluate to true or false?
$ forall x in emptyset. (P(x) land lnot P(x)) $
It should evaluate to true because there is no element, hence it is true for all elements.
But it should evaluate to false because $ P(x) land lnot P(x) $ is always false.
Does anyone have any ideas on how to answer this question?
logic predicate-logic
edited Jul 30 at 15:25
user21820
35.8k440136
asked Jul 30 at 9:51
Sven G.
285
It might be easier to first convince yourself that "$forall xinemptyset(perp)$" is true, where "$perp$" is the symbol for falsity. This boils the problem down to its essence, namely that (false implies false) is true.
– Noah Schweber
Jul 30 at 17:44
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
I have searched the forums but haven't found any answers to this question, so I thought I'd ask it myself.
Let $P(x)$ be an arbitrary predicate. Does the following statement evaluate to true or false?
$ forall x in emptyset. (P(x) land lnot P(x)) $
It should evaluate to true because there is no element, hence it is true for all elements.
But it should evaluate to false because $ P(x) land lnot P(x) $ is always false.
Does anyone have any ideas on how to answer this question?
logic predicate-logic
edited Jul 30 at 15:25
user21820
35.8k440136
asked Jul 30 at 9:51
Sven G.
285
I have searched the forums but haven't found any answers to this question, so I thought I'd ask it myself.
Let $P(x)$ be an arbitrary predicate. Does the following statement evaluate to true or false?
$ forall x in emptyset. (P(x) land lnot P(x)) $
It should evaluate to true because there is no element, hence it is true for all elements.
But it should evaluate to false because $ P(x) land lnot P(x) $ is always false.
Does anyone have any ideas on how to answer this question?
logic predicate-logic
edited Jul 30 at 15:25
user21820
35.8k440136
asked Jul 30 at 9:51
Sven G.
285
edited Jul 30 at 15:25
user21820
35.8k440136
edited Jul 30 at 15:25
user21820
35.8k440136
edited Jul 30 at 15:25
user21820
35.8k440136
35.8k440136
asked Jul 30 at 9:51
Sven G.
285
asked Jul 30 at 9:51
Sven G.
285
asked Jul 30 at 9:51
Sven G.
285
285
It might be easier to first convince yourself that "$forall xinemptyset(perp)$" is true, where "$perp$" is the symbol for falsity. This boils the problem down to its essence, namely that (false implies false) is true.
– Noah Schweber
Jul 30 at 17:44
add a comment |Â
It might be easier to first convince yourself that "$forall xinemptyset(perp)$" is true, where "$perp$" is the symbol for falsity. This boils the problem down to its essence, namely that (false implies false) is true.
– Noah Schweber
Jul 30 at 17:44
It might be easier to first convince yourself that "$forall xinemptyset(perp)$" is true, where "$perp$" is the symbol for falsity. This boils the problem down to its essence, namely that (false implies false) is true.
– Noah Schweber
Jul 30 at 17:44
It might be easier to first convince yourself that "$forall xinemptyset(perp)$" is true, where "$perp$" is the symbol for falsity. This boils the problem down to its essence, namely that (false implies false) is true.
– Noah Schweber
Jul 30 at 17:44
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
15
down vote
accepted
The formula is equivalent to :
$forall x (x in emptyset to (P(x) land lnot P(x)))$.
Thus, it is TRUE,because $x in emptyset$ is always FALSE and the truth table for the conditional has that :
FALSE $to$ FALSE is TRUE.
See also Vacuous truth :
a statement that asserts that all members of the empty set have a certain property.
answered Jul 30 at 9:55
Mauro ALLEGRANZA
60.6k346105
2
Ex falsum sequitur quodlibet.
– drhab
Jul 30 at 9:58
add a comment |Â
up vote
4
down vote
It's a true statement.
Your statement is a universal statement, so to verify it, you can think of it as 'lining up' all objects to which the statement pertains, and checking whether they have the indicated property. So, for example, if your domain consists of two objects $a$ and $b$, then verifying the claim that all objects have property $P$ amounts to verifying that both $a$ and $b$ have property $P$.
Ok, but what if your domain is empty (as in your case)? Well, then the verification is trivial: Yes, 'all' (zero!) objects have the property $P$.
Here is another way to think about it: what would make a universal claim false? Well, the claim is false when you find some object that does not have property $P$ ... but if you find no such counterexamples, then the claim is true. Having no objects in your domain, however, means that there can be no counterexamples, so no matter what $P$ is, even if (as in your case) it expresses something impossible.
Consider this somewhat more concrete example:
"Every time I played the lottery, I won the jackpot!"
Huh! Am I a very lucky person? Well, consider the fact that actually I have never played the lottery ... does that make the statement true? Yes! All zero times that I played the lottery, I won the jackpot! Again, think of it as 'lining up' all points in time when I played the lottery and verifying that at all those times I won the jackpot. Well, given that there are no such points in time where I played the lottery, this verification is trivial: yes, it checks out!
Or, in terms of counterexamples: there is no point in time where i played the lottery and did not win the jackpot. So, the claim checks out and is therefore true.
Of course, it is also true that all zero times I played the lottery, I did not win the jackpot. So, all zero times that I played the lottery, I did win the jackpot as well as did not win the jackpot ... which is just like your statement.
answered Jul 30 at 11:01
Bram28
54.8k33881
add a comment |Â
up vote
0
down vote
$$forall x in emptyset. (P(x) land lnot P(x))$$
is true as you mentioned.
$$(P(x) land lnot P(x))$$ is also true for all $x$ in the empty set.
Thus there is no conflict.
answered Jul 30 at 10:11
Mohammad Riazi-Kermani
27.3k41851
add a comment |Â
up vote
0
down vote
Make use of the fact that for any propositions $A$ and $B$, we have $A implies [neg A implies B]$ regardless of whether $B$ is true or false. This tautology is the so-called principle of explosion.
Definition: $forall a: neg [ain emptyset]$
Suppose $xin emptyset$ (the initial premise).
From the definition of $emptyset$, we obtain $neg [xin emptyset]$.
Applying the principle of explosion to the initial premise, we have $neg [xin emptyset] implies P(x) land neg P(x)$.
Applying the rule of detachment, we have $P(x) land neg P(x)$.
Discharging the initial premise and generalizing on $x$, we obtain, as required:
$forall a:[ain emptyset implies P(a)land neg P(a)] $
answered Jul 30 at 13:12
Dan Christensen
7,99611730
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
15
down vote
accepted
The formula is equivalent to :
$forall x (x in emptyset to (P(x) land lnot P(x)))$.
Thus, it is TRUE,because $x in emptyset$ is always FALSE and the truth table for the conditional has that :
FALSE $to$ FALSE is TRUE.
See also Vacuous truth :
a statement that asserts that all members of the empty set have a certain property.
answered Jul 30 at 9:55
Mauro ALLEGRANZA
60.6k346105
2
Ex falsum sequitur quodlibet.
– drhab
Jul 30 at 9:58
add a comment |Â
up vote
15
down vote
accepted
The formula is equivalent to :
$forall x (x in emptyset to (P(x) land lnot P(x)))$.
Thus, it is TRUE,because $x in emptyset$ is always FALSE and the truth table for the conditional has that :
FALSE $to$ FALSE is TRUE.
See also Vacuous truth :
a statement that asserts that all members of the empty set have a certain property.
answered Jul 30 at 9:55
Mauro ALLEGRANZA
60.6k346105
2
Ex falsum sequitur quodlibet.
– drhab
Jul 30 at 9:58
add a comment |Â
up vote
15
down vote
accepted
up vote
15
down vote
accepted
The formula is equivalent to :
$forall x (x in emptyset to (P(x) land lnot P(x)))$.
Thus, it is TRUE,because $x in emptyset$ is always FALSE and the truth table for the conditional has that :
FALSE $to$ FALSE is TRUE.
See also Vacuous truth :
a statement that asserts that all members of the empty set have a certain property.
answered Jul 30 at 9:55
Mauro ALLEGRANZA
60.6k346105
The formula is equivalent to :
$forall x (x in emptyset to (P(x) land lnot P(x)))$.
Thus, it is TRUE,because $x in emptyset$ is always FALSE and the truth table for the conditional has that :
FALSE $to$ FALSE is TRUE.
See also Vacuous truth :
a statement that asserts that all members of the empty set have a certain property.
answered Jul 30 at 9:55
Mauro ALLEGRANZA
60.6k346105
edited Jul 30 at 10:03
answered Jul 30 at 9:55
Mauro ALLEGRANZA
60.6k346105
answered Jul 30 at 9:55
Mauro ALLEGRANZA
60.6k346105
answered Jul 30 at 9:55
Mauro ALLEGRANZA
60.6k346105
60.6k346105
2
Ex falsum sequitur quodlibet.
– drhab
Jul 30 at 9:58
add a comment |Â
2
Ex falsum sequitur quodlibet.
– drhab
Jul 30 at 9:58
2
2
Ex falsum sequitur quodlibet.
– drhab
Jul 30 at 9:58
Ex falsum sequitur quodlibet.
– drhab
Jul 30 at 9:58
add a comment |Â
up vote
4
down vote
It's a true statement.
Your statement is a universal statement, so to verify it, you can think of it as 'lining up' all objects to which the statement pertains, and checking whether they have the indicated property. So, for example, if your domain consists of two objects $a$ and $b$, then verifying the claim that all objects have property $P$ amounts to verifying that both $a$ and $b$ have property $P$.
Ok, but what if your domain is empty (as in your case)? Well, then the verification is trivial: Yes, 'all' (zero!) objects have the property $P$.
Here is another way to think about it: what would make a universal claim false? Well, the claim is false when you find some object that does not have property $P$ ... but if you find no such counterexamples, then the claim is true. Having no objects in your domain, however, means that there can be no counterexamples, so no matter what $P$ is, even if (as in your case) it expresses something impossible.
Consider this somewhat more concrete example:
"Every time I played the lottery, I won the jackpot!"
Huh! Am I a very lucky person? Well, consider the fact that actually I have never played the lottery ... does that make the statement true? Yes! All zero times that I played the lottery, I won the jackpot! Again, think of it as 'lining up' all points in time when I played the lottery and verifying that at all those times I won the jackpot. Well, given that there are no such points in time where I played the lottery, this verification is trivial: yes, it checks out!
Or, in terms of counterexamples: there is no point in time where i played the lottery and did not win the jackpot. So, the claim checks out and is therefore true.
Of course, it is also true that all zero times I played the lottery, I did not win the jackpot. So, all zero times that I played the lottery, I did win the jackpot as well as did not win the jackpot ... which is just like your statement.
answered Jul 30 at 11:01
Bram28
54.8k33881
add a comment |Â
up vote
4
down vote
It's a true statement.
Your statement is a universal statement, so to verify it, you can think of it as 'lining up' all objects to which the statement pertains, and checking whether they have the indicated property. So, for example, if your domain consists of two objects $a$ and $b$, then verifying the claim that all objects have property $P$ amounts to verifying that both $a$ and $b$ have property $P$.
Ok, but what if your domain is empty (as in your case)? Well, then the verification is trivial: Yes, 'all' (zero!) objects have the property $P$.
Here is another way to think about it: what would make a universal claim false? Well, the claim is false when you find some object that does not have property $P$ ... but if you find no such counterexamples, then the claim is true. Having no objects in your domain, however, means that there can be no counterexamples, so no matter what $P$ is, even if (as in your case) it expresses something impossible.
Consider this somewhat more concrete example:
"Every time I played the lottery, I won the jackpot!"
Huh! Am I a very lucky person? Well, consider the fact that actually I have never played the lottery ... does that make the statement true? Yes! All zero times that I played the lottery, I won the jackpot! Again, think of it as 'lining up' all points in time when I played the lottery and verifying that at all those times I won the jackpot. Well, given that there are no such points in time where I played the lottery, this verification is trivial: yes, it checks out!
Or, in terms of counterexamples: there is no point in time where i played the lottery and did not win the jackpot. So, the claim checks out and is therefore true.
Of course, it is also true that all zero times I played the lottery, I did not win the jackpot. So, all zero times that I played the lottery, I did win the jackpot as well as did not win the jackpot ... which is just like your statement.
answered Jul 30 at 11:01
Bram28
54.8k33881
add a comment |Â
up vote
4
down vote
up vote
4
down vote
It's a true statement.
Your statement is a universal statement, so to verify it, you can think of it as 'lining up' all objects to which the statement pertains, and checking whether they have the indicated property. So, for example, if your domain consists of two objects $a$ and $b$, then verifying the claim that all objects have property $P$ amounts to verifying that both $a$ and $b$ have property $P$.
Ok, but what if your domain is empty (as in your case)? Well, then the verification is trivial: Yes, 'all' (zero!) objects have the property $P$.
Here is another way to think about it: what would make a universal claim false? Well, the claim is false when you find some object that does not have property $P$ ... but if you find no such counterexamples, then the claim is true. Having no objects in your domain, however, means that there can be no counterexamples, so no matter what $P$ is, even if (as in your case) it expresses something impossible.
Consider this somewhat more concrete example:
"Every time I played the lottery, I won the jackpot!"
Huh! Am I a very lucky person? Well, consider the fact that actually I have never played the lottery ... does that make the statement true? Yes! All zero times that I played the lottery, I won the jackpot! Again, think of it as 'lining up' all points in time when I played the lottery and verifying that at all those times I won the jackpot. Well, given that there are no such points in time where I played the lottery, this verification is trivial: yes, it checks out!
Or, in terms of counterexamples: there is no point in time where i played the lottery and did not win the jackpot. So, the claim checks out and is therefore true.
Of course, it is also true that all zero times I played the lottery, I did not win the jackpot. So, all zero times that I played the lottery, I did win the jackpot as well as did not win the jackpot ... which is just like your statement.
answered Jul 30 at 11:01
Bram28
54.8k33881
It's a true statement.
Your statement is a universal statement, so to verify it, you can think of it as 'lining up' all objects to which the statement pertains, and checking whether they have the indicated property. So, for example, if your domain consists of two objects $a$ and $b$, then verifying the claim that all objects have property $P$ amounts to verifying that both $a$ and $b$ have property $P$.
Ok, but what if your domain is empty (as in your case)? Well, then the verification is trivial: Yes, 'all' (zero!) objects have the property $P$.
Here is another way to think about it: what would make a universal claim false? Well, the claim is false when you find some object that does not have property $P$ ... but if you find no such counterexamples, then the claim is true. Having no objects in your domain, however, means that there can be no counterexamples, so no matter what $P$ is, even if (as in your case) it expresses something impossible.
Consider this somewhat more concrete example:
"Every time I played the lottery, I won the jackpot!"
Huh! Am I a very lucky person? Well, consider the fact that actually I have never played the lottery ... does that make the statement true? Yes! All zero times that I played the lottery, I won the jackpot! Again, think of it as 'lining up' all points in time when I played the lottery and verifying that at all those times I won the jackpot. Well, given that there are no such points in time where I played the lottery, this verification is trivial: yes, it checks out!
Or, in terms of counterexamples: there is no point in time where i played the lottery and did not win the jackpot. So, the claim checks out and is therefore true.
Of course, it is also true that all zero times I played the lottery, I did not win the jackpot. So, all zero times that I played the lottery, I did win the jackpot as well as did not win the jackpot ... which is just like your statement.
answered Jul 30 at 11:01
Bram28
54.8k33881
edited Jul 30 at 11:09
answered Jul 30 at 11:01
Bram28
54.8k33881
answered Jul 30 at 11:01
Bram28
54.8k33881
answered Jul 30 at 11:01
Bram28
54.8k33881
54.8k33881
add a comment |Â
add a comment |Â
up vote
0
down vote
$$forall x in emptyset. (P(x) land lnot P(x))$$
is true as you mentioned.
$$(P(x) land lnot P(x))$$ is also true for all $x$ in the empty set.
Thus there is no conflict.
answered Jul 30 at 10:11
Mohammad Riazi-Kermani
27.3k41851
add a comment |Â
up vote
0
down vote
$$forall x in emptyset. (P(x) land lnot P(x))$$
is true as you mentioned.
$$(P(x) land lnot P(x))$$ is also true for all $x$ in the empty set.
Thus there is no conflict.
answered Jul 30 at 10:11
Mohammad Riazi-Kermani
27.3k41851
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$$forall x in emptyset. (P(x) land lnot P(x))$$
is true as you mentioned.
$$(P(x) land lnot P(x))$$ is also true for all $x$ in the empty set.
Thus there is no conflict.
answered Jul 30 at 10:11
Mohammad Riazi-Kermani
27.3k41851
$$forall x in emptyset. (P(x) land lnot P(x))$$
is true as you mentioned.
$$(P(x) land lnot P(x))$$ is also true for all $x$ in the empty set.
Thus there is no conflict.
answered Jul 30 at 10:11
Mohammad Riazi-Kermani
27.3k41851
answered Jul 30 at 10:11
Mohammad Riazi-Kermani
27.3k41851
answered Jul 30 at 10:11
Mohammad Riazi-Kermani
27.3k41851
answered Jul 30 at 10:11
Mohammad Riazi-Kermani
27.3k41851
27.3k41851
add a comment |Â
add a comment |Â
up vote
0
down vote
Make use of the fact that for any propositions $A$ and $B$, we have $A implies [neg A implies B]$ regardless of whether $B$ is true or false. This tautology is the so-called principle of explosion.
Definition: $forall a: neg [ain emptyset]$
Suppose $xin emptyset$ (the initial premise).
From the definition of $emptyset$, we obtain $neg [xin emptyset]$.
Applying the principle of explosion to the initial premise, we have $neg [xin emptyset] implies P(x) land neg P(x)$.
Applying the rule of detachment, we have $P(x) land neg P(x)$.
Discharging the initial premise and generalizing on $x$, we obtain, as required:
$forall a:[ain emptyset implies P(a)land neg P(a)] $
answered Jul 30 at 13:12
Dan Christensen
7,99611730
add a comment |Â
up vote
0
down vote
Make use of the fact that for any propositions $A$ and $B$, we have $A implies [neg A implies B]$ regardless of whether $B$ is true or false. This tautology is the so-called principle of explosion.
Definition: $forall a: neg [ain emptyset]$
Suppose $xin emptyset$ (the initial premise).
From the definition of $emptyset$, we obtain $neg [xin emptyset]$.
Applying the principle of explosion to the initial premise, we have $neg [xin emptyset] implies P(x) land neg P(x)$.
Applying the rule of detachment, we have $P(x) land neg P(x)$.
Discharging the initial premise and generalizing on $x$, we obtain, as required:
$forall a:[ain emptyset implies P(a)land neg P(a)] $
answered Jul 30 at 13:12
Dan Christensen
7,99611730
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Make use of the fact that for any propositions $A$ and $B$, we have $A implies [neg A implies B]$ regardless of whether $B$ is true or false. This tautology is the so-called principle of explosion.
Definition: $forall a: neg [ain emptyset]$
Suppose $xin emptyset$ (the initial premise).
From the definition of $emptyset$, we obtain $neg [xin emptyset]$.
Applying the principle of explosion to the initial premise, we have $neg [xin emptyset] implies P(x) land neg P(x)$.
Applying the rule of detachment, we have $P(x) land neg P(x)$.
Discharging the initial premise and generalizing on $x$, we obtain, as required:
$forall a:[ain emptyset implies P(a)land neg P(a)] $
answered Jul 30 at 13:12
Dan Christensen
7,99611730
Make use of the fact that for any propositions $A$ and $B$, we have $A implies [neg A implies B]$ regardless of whether $B$ is true or false. This tautology is the so-called principle of explosion.
Definition: $forall a: neg [ain emptyset]$
Suppose $xin emptyset$ (the initial premise).
From the definition of $emptyset$, we obtain $neg [xin emptyset]$.
Applying the principle of explosion to the initial premise, we have $neg [xin emptyset] implies P(x) land neg P(x)$.
Applying the rule of detachment, we have $P(x) land neg P(x)$.
Discharging the initial premise and generalizing on $x$, we obtain, as required:
$forall a:[ain emptyset implies P(a)land neg P(a)] $
answered Jul 30 at 13:12
Dan Christensen
7,99611730
edited Jul 30 at 13:18
answered Jul 30 at 13:12
Dan Christensen
7,99611730
answered Jul 30 at 13:12
Dan Christensen
7,99611730
answered Jul 30 at 13:12
Dan Christensen
7,99611730
7,99611730
add a comment |Â
add a comment |Â
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It might be easier to first convince yourself that "$forall xinemptyset(perp)$" is true, where "$perp$" is the symbol for falsity. This boils the problem down to its essence, namely that (false implies false) is true.
– Noah Schweber
Jul 30 at 17:44