Mixing
Express Negation in Simple English: There is a student in this class who has chatted with exactly one other...
$begingroup$
Am I correct in the following:
If the domain is all students, and C(x,y) is the predicate of x having chatted with y. Then the sentence
There is a student in this class who has chatted with exactly one other student
Can be represented as
$$exists xexists y(yne xland forall z(zne xto(z=yleftrightarrow C(x,z))))$$
Which is logically equivalent to:
$$exists x exists y[(xneq y) landforall z ([z=x] lor[(zneq y) lor C(x,y)] land [lnot C(x,y) lor (z = y)])]$$
The negation of which would be:
$$forall x forall y [(x=y) lor exists z([zneq x] land [(z =y) land lnot C(x,y)] lor[C(x,y) land(z neq y)])]$$
That is also logically equivalent to:
$$forall x forall y [(x = y) lor exists z([z neq x] land [C(x,y) leftrightarrow (z neq y)])]$$
Which can be translated to: All students have spoken with at least one other student or themselves?
discrete-mathematics logic first-order-logic predicate-logic logic-translation
edited Jan 17 at 20:48
Bram28
63.1k44793
asked Jan 16 at 6:33
ElliottElliott
424
$endgroup$
add a comment |
$begingroup$
Am I correct in the following:
If the domain is all students, and C(x,y) is the predicate of x having chatted with y. Then the sentence
There is a student in this class who has chatted with exactly one other student
Can be represented as
$$exists xexists y(yne xland forall z(zne xto(z=yleftrightarrow C(x,z))))$$
Which is logically equivalent to:
$$exists x exists y[(xneq y) landforall z ([z=x] lor[(zneq y) lor C(x,y)] land [lnot C(x,y) lor (z = y)])]$$
The negation of which would be:
$$forall x forall y [(x=y) lor exists z([zneq x] land [(z =y) land lnot C(x,y)] lor[C(x,y) land(z neq y)])]$$
That is also logically equivalent to:
$$forall x forall y [(x = y) lor exists z([z neq x] land [C(x,y) leftrightarrow (z neq y)])]$$
Which can be translated to: All students have spoken with at least one other student or themselves?
discrete-mathematics logic first-order-logic predicate-logic logic-translation
edited Jan 17 at 20:48
Bram28
63.1k44793
asked Jan 16 at 6:33
ElliottElliott
424
$endgroup$
add a comment |
$begingroup$
Am I correct in the following:
If the domain is all students, and C(x,y) is the predicate of x having chatted with y. Then the sentence
There is a student in this class who has chatted with exactly one other student
Can be represented as
$$exists xexists y(yne xland forall z(zne xto(z=yleftrightarrow C(x,z))))$$
Which is logically equivalent to:
$$exists x exists y[(xneq y) landforall z ([z=x] lor[(zneq y) lor C(x,y)] land [lnot C(x,y) lor (z = y)])]$$
The negation of which would be:
$$forall x forall y [(x=y) lor exists z([zneq x] land [(z =y) land lnot C(x,y)] lor[C(x,y) land(z neq y)])]$$
That is also logically equivalent to:
$$forall x forall y [(x = y) lor exists z([z neq x] land [C(x,y) leftrightarrow (z neq y)])]$$
Which can be translated to: All students have spoken with at least one other student or themselves?
discrete-mathematics logic first-order-logic predicate-logic logic-translation
edited Jan 17 at 20:48
Bram28
63.1k44793
asked Jan 16 at 6:33
ElliottElliott
424
$endgroup$
Am I correct in the following:
If the domain is all students, and C(x,y) is the predicate of x having chatted with y. Then the sentence
There is a student in this class who has chatted with exactly one other student
Can be represented as
$$exists xexists y(yne xland forall z(zne xto(z=yleftrightarrow C(x,z))))$$
Which is logically equivalent to:
$$exists x exists y[(xneq y) landforall z ([z=x] lor[(zneq y) lor C(x,y)] land [lnot C(x,y) lor (z = y)])]$$
The negation of which would be:
$$forall x forall y [(x=y) lor exists z([zneq x] land [(z =y) land lnot C(x,y)] lor[C(x,y) land(z neq y)])]$$
That is also logically equivalent to:
$$forall x forall y [(x = y) lor exists z([z neq x] land [C(x,y) leftrightarrow (z neq y)])]$$
Which can be translated to: All students have spoken with at least one other student or themselves?
discrete-mathematics logic first-order-logic predicate-logic logic-translation
discrete-mathematics logic first-order-logic predicate-logic logic-translation
edited Jan 17 at 20:48
Bram28
63.1k44793
asked Jan 16 at 6:33
ElliottElliott
424
edited Jan 17 at 20:48
Bram28
63.1k44793
asked Jan 16 at 6:33
ElliottElliott
424
edited Jan 17 at 20:48
Bram28
63.1k44793
edited Jan 17 at 20:48
Bram28
63.1k44793
edited Jan 17 at 20:48
Bram28
63.1k44793
63.1k44793
asked Jan 16 at 6:33
ElliottElliott
424
asked Jan 16 at 6:33
ElliottElliott
424
asked Jan 16 at 6:33
ElliottElliott
424
424
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
As coffeemath indicated, the negation should of course end up being that every student either ends up chatting with no other students, or with two or more.
However, the logic expression you got at the end does not translate as such. And that's because you made a small mistake. Note that your $C(x,z)$ became $C(x,y)$, and that should not have happened. Indeed, you final expression should have been:
$$forall x forall y (x =y lor exists z (z not = x land [C(x,z) leftrightarrow z not =y]))$$
Ok, but how does that mean that every student ends up chatting with no other students, or with two or more? That is still not immediately obvious, so let me explain.
First, I would recommend rewriting it as:
$forall x forall y (x not = y rightarrow exists z (z not = x land [C(x,z) leftrightarrow z not = y])) tag{1}$
This is still hard to parse in English, but at least it starts with two different students $x$ and $y$ and now of course we want to make sure that either they didn't chat at all, or that they did chat but there is a third student $z$ that $x$ chatted with as well. In other words, we want to show that if $x$ and $y$ did chat, then there must be some third student $z$ that $x$ chatted with as well.
Ok, so let's suppose $x$ and $y$ chatted. Given $(1)$ and given that $x$ and $y$ are different, we know that there is some $z$ that is different from $x$ such that $C(x,z) leftrightarrow z not =y$.
Now, if $z=y$ then the right side is false, and hence the left side is false as well, and hence $x$ did not chat with $z=y$ .. but we assumed $x$ did chat with $y$, so we have a problem. So, $z$ is some other than $y$, and therefore $x$ did chat with $z$.
And so there you go: as soon as $x$ chats with some other student $y$, then there has to be some third student $z$ that $x$ chats with as well.
answered Jan 17 at 2:13
Bram28Bram28
63.1k44793
$endgroup$
add a comment |
$begingroup$
The negation would be: For all students x, either x has chatted with no other students or x has chatted with more than one other student.
[This is just ruling out x chatting with exactly one other student.]
answered Jan 16 at 7:05
coffeemathcoffeemath
2,8471415
$endgroup$
1
$begingroup$
How does $exists z([z neq x] land [C(x,y) leftrightarrow (z neq y)])$ mean more than one?
$endgroup$
– Elliott
Jan 17 at 0:45
$begingroup$
@Elliott I was negating the statement in yellow of OP's posted question. The sentence of your comment is one the OP obtained, and I didn't think it was right.
$endgroup$
– coffeemath
Jan 17 at 4:10
add a comment |
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2 Answers
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2 Answers
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$begingroup$
As coffeemath indicated, the negation should of course end up being that every student either ends up chatting with no other students, or with two or more.
However, the logic expression you got at the end does not translate as such. And that's because you made a small mistake. Note that your $C(x,z)$ became $C(x,y)$, and that should not have happened. Indeed, you final expression should have been:
$$forall x forall y (x =y lor exists z (z not = x land [C(x,z) leftrightarrow z not =y]))$$
Ok, but how does that mean that every student ends up chatting with no other students, or with two or more? That is still not immediately obvious, so let me explain.
First, I would recommend rewriting it as:
$forall x forall y (x not = y rightarrow exists z (z not = x land [C(x,z) leftrightarrow z not = y])) tag{1}$
This is still hard to parse in English, but at least it starts with two different students $x$ and $y$ and now of course we want to make sure that either they didn't chat at all, or that they did chat but there is a third student $z$ that $x$ chatted with as well. In other words, we want to show that if $x$ and $y$ did chat, then there must be some third student $z$ that $x$ chatted with as well.
Ok, so let's suppose $x$ and $y$ chatted. Given $(1)$ and given that $x$ and $y$ are different, we know that there is some $z$ that is different from $x$ such that $C(x,z) leftrightarrow z not =y$.
Now, if $z=y$ then the right side is false, and hence the left side is false as well, and hence $x$ did not chat with $z=y$ .. but we assumed $x$ did chat with $y$, so we have a problem. So, $z$ is some other than $y$, and therefore $x$ did chat with $z$.
And so there you go: as soon as $x$ chats with some other student $y$, then there has to be some third student $z$ that $x$ chats with as well.
answered Jan 17 at 2:13
Bram28Bram28
63.1k44793
$endgroup$
add a comment |
$begingroup$
As coffeemath indicated, the negation should of course end up being that every student either ends up chatting with no other students, or with two or more.
However, the logic expression you got at the end does not translate as such. And that's because you made a small mistake. Note that your $C(x,z)$ became $C(x,y)$, and that should not have happened. Indeed, you final expression should have been:
$$forall x forall y (x =y lor exists z (z not = x land [C(x,z) leftrightarrow z not =y]))$$
Ok, but how does that mean that every student ends up chatting with no other students, or with two or more? That is still not immediately obvious, so let me explain.
First, I would recommend rewriting it as:
$forall x forall y (x not = y rightarrow exists z (z not = x land [C(x,z) leftrightarrow z not = y])) tag{1}$
This is still hard to parse in English, but at least it starts with two different students $x$ and $y$ and now of course we want to make sure that either they didn't chat at all, or that they did chat but there is a third student $z$ that $x$ chatted with as well. In other words, we want to show that if $x$ and $y$ did chat, then there must be some third student $z$ that $x$ chatted with as well.
Ok, so let's suppose $x$ and $y$ chatted. Given $(1)$ and given that $x$ and $y$ are different, we know that there is some $z$ that is different from $x$ such that $C(x,z) leftrightarrow z not =y$.
Now, if $z=y$ then the right side is false, and hence the left side is false as well, and hence $x$ did not chat with $z=y$ .. but we assumed $x$ did chat with $y$, so we have a problem. So, $z$ is some other than $y$, and therefore $x$ did chat with $z$.
And so there you go: as soon as $x$ chats with some other student $y$, then there has to be some third student $z$ that $x$ chats with as well.
answered Jan 17 at 2:13
Bram28Bram28
63.1k44793
$endgroup$
add a comment |
$begingroup$
As coffeemath indicated, the negation should of course end up being that every student either ends up chatting with no other students, or with two or more.
However, the logic expression you got at the end does not translate as such. And that's because you made a small mistake. Note that your $C(x,z)$ became $C(x,y)$, and that should not have happened. Indeed, you final expression should have been:
$$forall x forall y (x =y lor exists z (z not = x land [C(x,z) leftrightarrow z not =y]))$$
Ok, but how does that mean that every student ends up chatting with no other students, or with two or more? That is still not immediately obvious, so let me explain.
First, I would recommend rewriting it as:
$forall x forall y (x not = y rightarrow exists z (z not = x land [C(x,z) leftrightarrow z not = y])) tag{1}$
This is still hard to parse in English, but at least it starts with two different students $x$ and $y$ and now of course we want to make sure that either they didn't chat at all, or that they did chat but there is a third student $z$ that $x$ chatted with as well. In other words, we want to show that if $x$ and $y$ did chat, then there must be some third student $z$ that $x$ chatted with as well.
Ok, so let's suppose $x$ and $y$ chatted. Given $(1)$ and given that $x$ and $y$ are different, we know that there is some $z$ that is different from $x$ such that $C(x,z) leftrightarrow z not =y$.
Now, if $z=y$ then the right side is false, and hence the left side is false as well, and hence $x$ did not chat with $z=y$ .. but we assumed $x$ did chat with $y$, so we have a problem. So, $z$ is some other than $y$, and therefore $x$ did chat with $z$.
And so there you go: as soon as $x$ chats with some other student $y$, then there has to be some third student $z$ that $x$ chats with as well.
answered Jan 17 at 2:13
Bram28Bram28
63.1k44793
$endgroup$
As coffeemath indicated, the negation should of course end up being that every student either ends up chatting with no other students, or with two or more.
However, the logic expression you got at the end does not translate as such. And that's because you made a small mistake. Note that your $C(x,z)$ became $C(x,y)$, and that should not have happened. Indeed, you final expression should have been:
$$forall x forall y (x =y lor exists z (z not = x land [C(x,z) leftrightarrow z not =y]))$$
Ok, but how does that mean that every student ends up chatting with no other students, or with two or more? That is still not immediately obvious, so let me explain.
First, I would recommend rewriting it as:
$forall x forall y (x not = y rightarrow exists z (z not = x land [C(x,z) leftrightarrow z not = y])) tag{1}$
This is still hard to parse in English, but at least it starts with two different students $x$ and $y$ and now of course we want to make sure that either they didn't chat at all, or that they did chat but there is a third student $z$ that $x$ chatted with as well. In other words, we want to show that if $x$ and $y$ did chat, then there must be some third student $z$ that $x$ chatted with as well.
Ok, so let's suppose $x$ and $y$ chatted. Given $(1)$ and given that $x$ and $y$ are different, we know that there is some $z$ that is different from $x$ such that $C(x,z) leftrightarrow z not =y$.
Now, if $z=y$ then the right side is false, and hence the left side is false as well, and hence $x$ did not chat with $z=y$ .. but we assumed $x$ did chat with $y$, so we have a problem. So, $z$ is some other than $y$, and therefore $x$ did chat with $z$.
And so there you go: as soon as $x$ chats with some other student $y$, then there has to be some third student $z$ that $x$ chats with as well.
answered Jan 17 at 2:13
Bram28Bram28
63.1k44793
answered Jan 17 at 2:13
Bram28Bram28
63.1k44793
answered Jan 17 at 2:13
Bram28Bram28
63.1k44793
answered Jan 17 at 2:13
Bram28Bram28
63.1k44793
63.1k44793
add a comment |
add a comment |
$begingroup$
The negation would be: For all students x, either x has chatted with no other students or x has chatted with more than one other student.
[This is just ruling out x chatting with exactly one other student.]
answered Jan 16 at 7:05
coffeemathcoffeemath
2,8471415
$endgroup$
1
$begingroup$
How does $exists z([z neq x] land [C(x,y) leftrightarrow (z neq y)])$ mean more than one?
$endgroup$
– Elliott
Jan 17 at 0:45
$begingroup$
@Elliott I was negating the statement in yellow of OP's posted question. The sentence of your comment is one the OP obtained, and I didn't think it was right.
$endgroup$
– coffeemath
Jan 17 at 4:10
add a comment |
$begingroup$
The negation would be: For all students x, either x has chatted with no other students or x has chatted with more than one other student.
[This is just ruling out x chatting with exactly one other student.]
answered Jan 16 at 7:05
coffeemathcoffeemath
2,8471415
$endgroup$
1
$begingroup$
How does $exists z([z neq x] land [C(x,y) leftrightarrow (z neq y)])$ mean more than one?
$endgroup$
– Elliott
Jan 17 at 0:45
$begingroup$
@Elliott I was negating the statement in yellow of OP's posted question. The sentence of your comment is one the OP obtained, and I didn't think it was right.
$endgroup$
– coffeemath
Jan 17 at 4:10
add a comment |
$begingroup$
The negation would be: For all students x, either x has chatted with no other students or x has chatted with more than one other student.
[This is just ruling out x chatting with exactly one other student.]
answered Jan 16 at 7:05
coffeemathcoffeemath
2,8471415
$endgroup$
The negation would be: For all students x, either x has chatted with no other students or x has chatted with more than one other student.
[This is just ruling out x chatting with exactly one other student.]
answered Jan 16 at 7:05
coffeemathcoffeemath
2,8471415
answered Jan 16 at 7:05
coffeemathcoffeemath
2,8471415
answered Jan 16 at 7:05
coffeemathcoffeemath
2,8471415
answered Jan 16 at 7:05
coffeemathcoffeemath
2,8471415
2,8471415
1
$begingroup$
How does $exists z([z neq x] land [C(x,y) leftrightarrow (z neq y)])$ mean more than one?
$endgroup$
– Elliott
Jan 17 at 0:45
$begingroup$
@Elliott I was negating the statement in yellow of OP's posted question. The sentence of your comment is one the OP obtained, and I didn't think it was right.
$endgroup$
– coffeemath
Jan 17 at 4:10
add a comment |
1
$begingroup$
How does $exists z([z neq x] land [C(x,y) leftrightarrow (z neq y)])$ mean more than one?
$endgroup$
– Elliott
Jan 17 at 0:45
$begingroup$
@Elliott I was negating the statement in yellow of OP's posted question. The sentence of your comment is one the OP obtained, and I didn't think it was right.
$endgroup$
– coffeemath
Jan 17 at 4:10
1
1
$begingroup$
How does $exists z([z neq x] land [C(x,y) leftrightarrow (z neq y)])$ mean more than one?
$endgroup$
– Elliott
Jan 17 at 0:45
$begingroup$
How does $exists z([z neq x] land [C(x,y) leftrightarrow (z neq y)])$ mean more than one?
$endgroup$
– Elliott
Jan 17 at 0:45
$begingroup$
@Elliott I was negating the statement in yellow of OP's posted question. The sentence of your comment is one the OP obtained, and I didn't think it was right.
$endgroup$
– coffeemath
Jan 17 at 4:10
$begingroup$
@Elliott I was negating the statement in yellow of OP's posted question. The sentence of your comment is one the OP obtained, and I didn't think it was right.
$endgroup$
– coffeemath
Jan 17 at 4:10
add a comment |
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