How is the order of statements gives a different answer? [on hold]Why are the different ways to write a universal statements equivalent?Can the logical equivalence of two different statements to the same proposition imply that the three of them are false?Determining the equivalence of different subsets, unions and intersections?How many different answer keys are possible?Are provable statements at the bottom of the arithmetical hierarchy?Existential quantifier before/after the implication sign gives different contrapositives.The $n$th statement in a list of $100$ statements is “Exactly $n$ of the statements in this list are false”.The Logic of False StatementsHow to prove two statements are equivalent and give a counterexample if they're not.How many true statements are there in following statements?
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How is the order of statements gives a different answer? [on hold]
Why are the different ways to write a universal statements equivalent?Can the logical equivalence of two different statements to the same proposition imply that the three of them are false?Determining the equivalence of different subsets, unions and intersections?How many different answer keys are possible?Are provable statements at the bottom of the arithmetical hierarchy?Existential quantifier before/after the implication sign gives different contrapositives.The $n$th statement in a list of $100$ statements is “Exactly $n$ of the statements in this list are false”.The Logic of False StatementsHow to prove two statements are equivalent and give a counterexample if they're not.How many true statements are there in following statements?
$begingroup$
arent both statements equivalent? so both should be false?
discrete-mathematics logic proof-theory
edited yesterday
Arthur
117k7116200
asked yesterday
user10798572user10798572
24
$endgroup$
put on hold as off-topic by José Carlos Santos, Eevee Trainer, John Omielan, Jyrki Lahtonen, Peter Foreman yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Eevee Trainer, John Omielan, Jyrki Lahtonen, Peter Foreman
add a comment |
$begingroup$
arent both statements equivalent? so both should be false?
discrete-mathematics logic proof-theory
edited yesterday
Arthur
117k7116200
asked yesterday
user10798572user10798572
24
$endgroup$
put on hold as off-topic by José Carlos Santos, Eevee Trainer, John Omielan, Jyrki Lahtonen, Peter Foreman yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Eevee Trainer, John Omielan, Jyrki Lahtonen, Peter Foreman
$begingroup$
In general, the order of a "for all" and a "there exists" is very important, and affects the meaning of the statement. If $exists$ comes first, in something like $exists xtext s.t. forall y ldots$, the same $x$ must work simultaneously for all $y$ (so $x$ cannot depend on $y$). However, if the $forall$ comes first, in something like $forall y, exists xtext s.t. ldots$, the $x$ is allowed to depend on $y$, i.e. we are allowed to have different $x$ for different $y$. Thus your two statements are not equivalent.
$endgroup$
– Minus One-Twelfth
yesterday
$begingroup$
" If ∃ comes first, in something like ∃𝑥 s.t. ∀𝑦…, the same 𝑥 must work simultaneously for all 𝑦 (so 𝑥 cannot depend on 𝑦)." then shouldnt it be written as !∃𝑥 meaning a unique x
$endgroup$
– user10798572
yesterday
1
$begingroup$
When writing $exists xforall y$ we say nothing about how many $x$ are such that they work for all $y$ simultaneously; there is at least one, but may just as well be many. With $exists!xforall y$, there is exactly one $x$ that works for all $y$ simultaneously.
$endgroup$
– Arthur
yesterday
add a comment |
$begingroup$
arent both statements equivalent? so both should be false?
discrete-mathematics logic proof-theory
edited yesterday
Arthur
117k7116200
asked yesterday
user10798572user10798572
24
$endgroup$
arent both statements equivalent? so both should be false?
discrete-mathematics logic proof-theory
discrete-mathematics logic proof-theory
edited yesterday
Arthur
117k7116200
asked yesterday
user10798572user10798572
24
edited yesterday
Arthur
117k7116200
asked yesterday
user10798572user10798572
24
edited yesterday
Arthur
117k7116200
edited yesterday
Arthur
117k7116200
edited yesterday
Arthur
117k7116200
117k7116200
asked yesterday
user10798572user10798572
24
asked yesterday
user10798572user10798572
24
asked yesterday
user10798572user10798572
24
24
put on hold as off-topic by José Carlos Santos, Eevee Trainer, John Omielan, Jyrki Lahtonen, Peter Foreman yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Eevee Trainer, John Omielan, Jyrki Lahtonen, Peter Foreman
put on hold as off-topic by José Carlos Santos, Eevee Trainer, John Omielan, Jyrki Lahtonen, Peter Foreman yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Eevee Trainer, John Omielan, Jyrki Lahtonen, Peter Foreman
$begingroup$
In general, the order of a "for all" and a "there exists" is very important, and affects the meaning of the statement. If $exists$ comes first, in something like $exists xtext s.t. forall y ldots$, the same $x$ must work simultaneously for all $y$ (so $x$ cannot depend on $y$). However, if the $forall$ comes first, in something like $forall y, exists xtext s.t. ldots$, the $x$ is allowed to depend on $y$, i.e. we are allowed to have different $x$ for different $y$. Thus your two statements are not equivalent.
$endgroup$
– Minus One-Twelfth
yesterday
$begingroup$
" If ∃ comes first, in something like ∃𝑥 s.t. ∀𝑦…, the same 𝑥 must work simultaneously for all 𝑦 (so 𝑥 cannot depend on 𝑦)." then shouldnt it be written as !∃𝑥 meaning a unique x
$endgroup$
– user10798572
yesterday
1
$begingroup$
When writing $exists xforall y$ we say nothing about how many $x$ are such that they work for all $y$ simultaneously; there is at least one, but may just as well be many. With $exists!xforall y$, there is exactly one $x$ that works for all $y$ simultaneously.
$endgroup$
– Arthur
yesterday
add a comment |
$begingroup$
In general, the order of a "for all" and a "there exists" is very important, and affects the meaning of the statement. If $exists$ comes first, in something like $exists xtext s.t. forall y ldots$, the same $x$ must work simultaneously for all $y$ (so $x$ cannot depend on $y$). However, if the $forall$ comes first, in something like $forall y, exists xtext s.t. ldots$, the $x$ is allowed to depend on $y$, i.e. we are allowed to have different $x$ for different $y$. Thus your two statements are not equivalent.
$endgroup$
– Minus One-Twelfth
yesterday
$begingroup$
" If ∃ comes first, in something like ∃𝑥 s.t. ∀𝑦…, the same 𝑥 must work simultaneously for all 𝑦 (so 𝑥 cannot depend on 𝑦)." then shouldnt it be written as !∃𝑥 meaning a unique x
$endgroup$
– user10798572
yesterday
1
$begingroup$
When writing $exists xforall y$ we say nothing about how many $x$ are such that they work for all $y$ simultaneously; there is at least one, but may just as well be many. With $exists!xforall y$, there is exactly one $x$ that works for all $y$ simultaneously.
$endgroup$
– Arthur
yesterday
$begingroup$
In general, the order of a "for all" and a "there exists" is very important, and affects the meaning of the statement. If $exists$ comes first, in something like $exists xtext s.t. forall y ldots$, the same $x$ must work simultaneously for all $y$ (so $x$ cannot depend on $y$). However, if the $forall$ comes first, in something like $forall y, exists xtext s.t. ldots$, the $x$ is allowed to depend on $y$, i.e. we are allowed to have different $x$ for different $y$. Thus your two statements are not equivalent.
$endgroup$
– Minus One-Twelfth
yesterday
$begingroup$
In general, the order of a "for all" and a "there exists" is very important, and affects the meaning of the statement. If $exists$ comes first, in something like $exists xtext s.t. forall y ldots$, the same $x$ must work simultaneously for all $y$ (so $x$ cannot depend on $y$). However, if the $forall$ comes first, in something like $forall y, exists xtext s.t. ldots$, the $x$ is allowed to depend on $y$, i.e. we are allowed to have different $x$ for different $y$. Thus your two statements are not equivalent.
$endgroup$
– Minus One-Twelfth
yesterday
$begingroup$
" If ∃ comes first, in something like ∃𝑥 s.t. ∀𝑦…, the same 𝑥 must work simultaneously for all 𝑦 (so 𝑥 cannot depend on 𝑦)." then shouldnt it be written as !∃𝑥 meaning a unique x
$endgroup$
– user10798572
yesterday
$begingroup$
" If ∃ comes first, in something like ∃𝑥 s.t. ∀𝑦…, the same 𝑥 must work simultaneously for all 𝑦 (so 𝑥 cannot depend on 𝑦)." then shouldnt it be written as !∃𝑥 meaning a unique x
$endgroup$
– user10798572
yesterday
1
1
$begingroup$
When writing $exists xforall y$ we say nothing about how many $x$ are such that they work for all $y$ simultaneously; there is at least one, but may just as well be many. With $exists!xforall y$, there is exactly one $x$ that works for all $y$ simultaneously.
$endgroup$
– Arthur
yesterday
$begingroup$
When writing $exists xforall y$ we say nothing about how many $x$ are such that they work for all $y$ simultaneously; there is at least one, but may just as well be many. With $exists!xforall y$, there is exactly one $x$ that works for all $y$ simultaneously.
$endgroup$
– Arthur
yesterday
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
No. Both statements are not equivalent. In the first one, it says there is an $x$ ( a real number) for each and every real number $y$ such that $xy=1$. This is clearly untrue. For example, for $y=5$, the only $x$ that will yield $xy=1$ will be $frac15$. But this $x=frac15$ is not the multiplicative inverse of any other real number.
In the second statement, it says for every $y$ there exists an $x$ such that $xy=1$. This is different in that it says that the $x$ for each $y$ does not have to be the same real number as opposed to what was implied by the first statement. And, of course, this is true. Because, for any $y=z$, the multiplicative inverse would be $frac1z$.
edited yesterday
Peter Foreman
3,4421216
answered yesterday
A.AsadA.Asad
48919
$endgroup$
$begingroup$
Also, the second statement is only true if we are talking about all real numbers excluding 0. 0 does not have a multiplicative inverse.
$endgroup$
– A.Asad
yesterday
1
$begingroup$
true, $mathbbR^ast$ is the set of non-zero reals.
$endgroup$
– Henno Brandsma
yesterday
$begingroup$
Thank you! I was not familiar with the terminology
$endgroup$
– A.Asad
yesterday
$begingroup$
For any unary ring $R$, $R^ast$ is the set of multiplicative units. This is a commonly used special case.
$endgroup$
– Henno Brandsma
yesterday
$begingroup$
for the first statement, it doesnt say a unique solution !E, so i can change the X for every Y making it true? right
$endgroup$
– user10798572
yesterday
|
show 1 more comment
$begingroup$
No, what you showed is correct
$$exists x in mathbbR^ast: forall y in mathbbR^ast: xy=1tagi$$
is false; there is no $x$ that works for all $y$, but
$$forall y in mathbbR^ast: exists x in mathbbR^ast: xy=1tagii$$
is true, as for every $y$ we can find an $x$.
This is how the semantics of these statements is defined. Read up on it..
There are hidden brackets here, really:
$$exists x in mathbbR^ast: left(forall y in mathbbR^ast: xy=1right)tagi'$$
e.g. and the $x$ is a "constant" (bound variable) within the bracketed statement. So the statement holds if we can find an $x$ such that for that specific $x$ the inner statement holds, and for that, for each $y$ the statement $xy=1$ should hold, which it does not. Etc.
answered yesterday
Henno BrandsmaHenno Brandsma
112k348120
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No. Both statements are not equivalent. In the first one, it says there is an $x$ ( a real number) for each and every real number $y$ such that $xy=1$. This is clearly untrue. For example, for $y=5$, the only $x$ that will yield $xy=1$ will be $frac15$. But this $x=frac15$ is not the multiplicative inverse of any other real number.
In the second statement, it says for every $y$ there exists an $x$ such that $xy=1$. This is different in that it says that the $x$ for each $y$ does not have to be the same real number as opposed to what was implied by the first statement. And, of course, this is true. Because, for any $y=z$, the multiplicative inverse would be $frac1z$.
edited yesterday
Peter Foreman
3,4421216
answered yesterday
A.AsadA.Asad
48919
$endgroup$
$begingroup$
Also, the second statement is only true if we are talking about all real numbers excluding 0. 0 does not have a multiplicative inverse.
$endgroup$
– A.Asad
yesterday
1
$begingroup$
true, $mathbbR^ast$ is the set of non-zero reals.
$endgroup$
– Henno Brandsma
yesterday
$begingroup$
Thank you! I was not familiar with the terminology
$endgroup$
– A.Asad
yesterday
$begingroup$
For any unary ring $R$, $R^ast$ is the set of multiplicative units. This is a commonly used special case.
$endgroup$
– Henno Brandsma
yesterday
$begingroup$
for the first statement, it doesnt say a unique solution !E, so i can change the X for every Y making it true? right
$endgroup$
– user10798572
yesterday
|
show 1 more comment
$begingroup$
No. Both statements are not equivalent. In the first one, it says there is an $x$ ( a real number) for each and every real number $y$ such that $xy=1$. This is clearly untrue. For example, for $y=5$, the only $x$ that will yield $xy=1$ will be $frac15$. But this $x=frac15$ is not the multiplicative inverse of any other real number.
In the second statement, it says for every $y$ there exists an $x$ such that $xy=1$. This is different in that it says that the $x$ for each $y$ does not have to be the same real number as opposed to what was implied by the first statement. And, of course, this is true. Because, for any $y=z$, the multiplicative inverse would be $frac1z$.
edited yesterday
Peter Foreman
3,4421216
answered yesterday
A.AsadA.Asad
48919
$endgroup$
$begingroup$
Also, the second statement is only true if we are talking about all real numbers excluding 0. 0 does not have a multiplicative inverse.
$endgroup$
– A.Asad
yesterday
1
$begingroup$
true, $mathbbR^ast$ is the set of non-zero reals.
$endgroup$
– Henno Brandsma
yesterday
$begingroup$
Thank you! I was not familiar with the terminology
$endgroup$
– A.Asad
yesterday
$begingroup$
For any unary ring $R$, $R^ast$ is the set of multiplicative units. This is a commonly used special case.
$endgroup$
– Henno Brandsma
yesterday
$begingroup$
for the first statement, it doesnt say a unique solution !E, so i can change the X for every Y making it true? right
$endgroup$
– user10798572
yesterday
|
show 1 more comment
$begingroup$
No. Both statements are not equivalent. In the first one, it says there is an $x$ ( a real number) for each and every real number $y$ such that $xy=1$. This is clearly untrue. For example, for $y=5$, the only $x$ that will yield $xy=1$ will be $frac15$. But this $x=frac15$ is not the multiplicative inverse of any other real number.
In the second statement, it says for every $y$ there exists an $x$ such that $xy=1$. This is different in that it says that the $x$ for each $y$ does not have to be the same real number as opposed to what was implied by the first statement. And, of course, this is true. Because, for any $y=z$, the multiplicative inverse would be $frac1z$.
edited yesterday
Peter Foreman
3,4421216
answered yesterday
A.AsadA.Asad
48919
$endgroup$
No. Both statements are not equivalent. In the first one, it says there is an $x$ ( a real number) for each and every real number $y$ such that $xy=1$. This is clearly untrue. For example, for $y=5$, the only $x$ that will yield $xy=1$ will be $frac15$. But this $x=frac15$ is not the multiplicative inverse of any other real number.
In the second statement, it says for every $y$ there exists an $x$ such that $xy=1$. This is different in that it says that the $x$ for each $y$ does not have to be the same real number as opposed to what was implied by the first statement. And, of course, this is true. Because, for any $y=z$, the multiplicative inverse would be $frac1z$.
edited yesterday
Peter Foreman
3,4421216
answered yesterday
A.AsadA.Asad
48919
edited yesterday
Peter Foreman
3,4421216
edited yesterday
Peter Foreman
3,4421216
edited yesterday
Peter Foreman
3,4421216
3,4421216
answered yesterday
A.AsadA.Asad
48919
answered yesterday
A.AsadA.Asad
48919
answered yesterday
A.AsadA.Asad
48919
48919
$begingroup$
Also, the second statement is only true if we are talking about all real numbers excluding 0. 0 does not have a multiplicative inverse.
$endgroup$
– A.Asad
yesterday
1
$begingroup$
true, $mathbbR^ast$ is the set of non-zero reals.
$endgroup$
– Henno Brandsma
yesterday
$begingroup$
Thank you! I was not familiar with the terminology
$endgroup$
– A.Asad
yesterday
$begingroup$
For any unary ring $R$, $R^ast$ is the set of multiplicative units. This is a commonly used special case.
$endgroup$
– Henno Brandsma
yesterday
$begingroup$
for the first statement, it doesnt say a unique solution !E, so i can change the X for every Y making it true? right
$endgroup$
– user10798572
yesterday
|
show 1 more comment
$begingroup$
Also, the second statement is only true if we are talking about all real numbers excluding 0. 0 does not have a multiplicative inverse.
$endgroup$
– A.Asad
yesterday
1
$begingroup$
true, $mathbbR^ast$ is the set of non-zero reals.
$endgroup$
– Henno Brandsma
yesterday
$begingroup$
Thank you! I was not familiar with the terminology
$endgroup$
– A.Asad
yesterday
$begingroup$
For any unary ring $R$, $R^ast$ is the set of multiplicative units. This is a commonly used special case.
$endgroup$
– Henno Brandsma
yesterday
$begingroup$
for the first statement, it doesnt say a unique solution !E, so i can change the X for every Y making it true? right
$endgroup$
– user10798572
yesterday
$begingroup$
Also, the second statement is only true if we are talking about all real numbers excluding 0. 0 does not have a multiplicative inverse.
$endgroup$
– A.Asad
yesterday
$begingroup$
Also, the second statement is only true if we are talking about all real numbers excluding 0. 0 does not have a multiplicative inverse.
$endgroup$
– A.Asad
yesterday
1
1
$begingroup$
true, $mathbbR^ast$ is the set of non-zero reals.
$endgroup$
– Henno Brandsma
yesterday
$begingroup$
true, $mathbbR^ast$ is the set of non-zero reals.
$endgroup$
– Henno Brandsma
yesterday
$begingroup$
Thank you! I was not familiar with the terminology
$endgroup$
– A.Asad
yesterday
$begingroup$
Thank you! I was not familiar with the terminology
$endgroup$
– A.Asad
yesterday
$begingroup$
For any unary ring $R$, $R^ast$ is the set of multiplicative units. This is a commonly used special case.
$endgroup$
– Henno Brandsma
yesterday
$begingroup$
For any unary ring $R$, $R^ast$ is the set of multiplicative units. This is a commonly used special case.
$endgroup$
– Henno Brandsma
yesterday
$begingroup$
for the first statement, it doesnt say a unique solution !E, so i can change the X for every Y making it true? right
$endgroup$
– user10798572
yesterday
$begingroup$
for the first statement, it doesnt say a unique solution !E, so i can change the X for every Y making it true? right
$endgroup$
– user10798572
yesterday
|
show 1 more comment
$begingroup$
No, what you showed is correct
$$exists x in mathbbR^ast: forall y in mathbbR^ast: xy=1tagi$$
is false; there is no $x$ that works for all $y$, but
$$forall y in mathbbR^ast: exists x in mathbbR^ast: xy=1tagii$$
is true, as for every $y$ we can find an $x$.
This is how the semantics of these statements is defined. Read up on it..
There are hidden brackets here, really:
$$exists x in mathbbR^ast: left(forall y in mathbbR^ast: xy=1right)tagi'$$
e.g. and the $x$ is a "constant" (bound variable) within the bracketed statement. So the statement holds if we can find an $x$ such that for that specific $x$ the inner statement holds, and for that, for each $y$ the statement $xy=1$ should hold, which it does not. Etc.
answered yesterday
Henno BrandsmaHenno Brandsma
112k348120
$endgroup$
add a comment |
$begingroup$
No, what you showed is correct
$$exists x in mathbbR^ast: forall y in mathbbR^ast: xy=1tagi$$
is false; there is no $x$ that works for all $y$, but
$$forall y in mathbbR^ast: exists x in mathbbR^ast: xy=1tagii$$
is true, as for every $y$ we can find an $x$.
This is how the semantics of these statements is defined. Read up on it..
There are hidden brackets here, really:
$$exists x in mathbbR^ast: left(forall y in mathbbR^ast: xy=1right)tagi'$$
e.g. and the $x$ is a "constant" (bound variable) within the bracketed statement. So the statement holds if we can find an $x$ such that for that specific $x$ the inner statement holds, and for that, for each $y$ the statement $xy=1$ should hold, which it does not. Etc.
answered yesterday
Henno BrandsmaHenno Brandsma
112k348120
$endgroup$
add a comment |
$begingroup$
No, what you showed is correct
$$exists x in mathbbR^ast: forall y in mathbbR^ast: xy=1tagi$$
is false; there is no $x$ that works for all $y$, but
$$forall y in mathbbR^ast: exists x in mathbbR^ast: xy=1tagii$$
is true, as for every $y$ we can find an $x$.
This is how the semantics of these statements is defined. Read up on it..
There are hidden brackets here, really:
$$exists x in mathbbR^ast: left(forall y in mathbbR^ast: xy=1right)tagi'$$
e.g. and the $x$ is a "constant" (bound variable) within the bracketed statement. So the statement holds if we can find an $x$ such that for that specific $x$ the inner statement holds, and for that, for each $y$ the statement $xy=1$ should hold, which it does not. Etc.
answered yesterday
Henno BrandsmaHenno Brandsma
112k348120
$endgroup$
No, what you showed is correct
$$exists x in mathbbR^ast: forall y in mathbbR^ast: xy=1tagi$$
is false; there is no $x$ that works for all $y$, but
$$forall y in mathbbR^ast: exists x in mathbbR^ast: xy=1tagii$$
is true, as for every $y$ we can find an $x$.
This is how the semantics of these statements is defined. Read up on it..
There are hidden brackets here, really:
$$exists x in mathbbR^ast: left(forall y in mathbbR^ast: xy=1right)tagi'$$
e.g. and the $x$ is a "constant" (bound variable) within the bracketed statement. So the statement holds if we can find an $x$ such that for that specific $x$ the inner statement holds, and for that, for each $y$ the statement $xy=1$ should hold, which it does not. Etc.
answered yesterday
Henno BrandsmaHenno Brandsma
112k348120
edited yesterday
answered yesterday
Henno BrandsmaHenno Brandsma
112k348120
answered yesterday
Henno BrandsmaHenno Brandsma
112k348120
answered yesterday
Henno BrandsmaHenno Brandsma
112k348120
112k348120
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In general, the order of a "for all" and a "there exists" is very important, and affects the meaning of the statement. If $exists$ comes first, in something like $exists xtext s.t. forall y ldots$, the same $x$ must work simultaneously for all $y$ (so $x$ cannot depend on $y$). However, if the $forall$ comes first, in something like $forall y, exists xtext s.t. ldots$, the $x$ is allowed to depend on $y$, i.e. we are allowed to have different $x$ for different $y$. Thus your two statements are not equivalent.
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– Minus One-Twelfth
yesterday
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" If ∃ comes first, in something like ∃𝑥 s.t. ∀𝑦…, the same 𝑥 must work simultaneously for all 𝑦 (so 𝑥 cannot depend on 𝑦)." then shouldnt it be written as !∃𝑥 meaning a unique x
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– user10798572
yesterday
1
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When writing $exists xforall y$ we say nothing about how many $x$ are such that they work for all $y$ simultaneously; there is at least one, but may just as well be many. With $exists!xforall y$, there is exactly one $x$ that works for all $y$ simultaneously.
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– Arthur
yesterday