Universal Quantified Statement being equivalent when variables are swapped
$begingroup$
I was given this statement and asked to express this with universal quantifiers.
Likes(x,y) is a Binary Relation that means that person x likes person y.
The statement given was:
"Everybody is liked by somebody*". Easy Enough, or so I thought? I put:
∀x∃y, (Likes(y,x)). Apparently this is wrong and the correct solution (that a TA displayed as the answer) is ∀y∃x (Likes(x,y)).
For clarity my question is generally :
is ∀x∃y, (P(y,x)) ≡ ∀y∃x, (P(x,y)) ?
I understand the variables are swapped, but i believe that it is logically equivalent, as i just made the "everbody" labled as x and the "somebody" labeled as "y".
Thank you!
discrete-mathematics logic quantifiers logic-translation
edited Dec 11 '18 at 20:48
Bram28
63.9k44793
asked Dec 11 '18 at 5:33
Stephen IsaacStephen Isaac
11
$endgroup$
add a comment |
$begingroup$
I was given this statement and asked to express this with universal quantifiers.
Likes(x,y) is a Binary Relation that means that person x likes person y.
The statement given was:
"Everybody is liked by somebody*". Easy Enough, or so I thought? I put:
∀x∃y, (Likes(y,x)). Apparently this is wrong and the correct solution (that a TA displayed as the answer) is ∀y∃x (Likes(x,y)).
For clarity my question is generally :
is ∀x∃y, (P(y,x)) ≡ ∀y∃x, (P(x,y)) ?
I understand the variables are swapped, but i believe that it is logically equivalent, as i just made the "everbody" labled as x and the "somebody" labeled as "y".
Thank you!
discrete-mathematics logic quantifiers logic-translation
edited Dec 11 '18 at 20:48
Bram28
63.9k44793
asked Dec 11 '18 at 5:33
Stephen IsaacStephen Isaac
11
$endgroup$
$begingroup$
Those statements are logically equivalent, but surely the correct answer to the original problem is $forall xexists y text{Likes}(x,y)$.
$endgroup$
– Lord Shark the Unknown
Dec 11 '18 at 5:39
$begingroup$
Ah yes. My bad. I meant the statement to be the following: "Everybody is liked by somebody"
$endgroup$
– Stephen Isaac
Dec 11 '18 at 5:42
add a comment |
$begingroup$
I was given this statement and asked to express this with universal quantifiers.
Likes(x,y) is a Binary Relation that means that person x likes person y.
The statement given was:
"Everybody is liked by somebody*". Easy Enough, or so I thought? I put:
∀x∃y, (Likes(y,x)). Apparently this is wrong and the correct solution (that a TA displayed as the answer) is ∀y∃x (Likes(x,y)).
For clarity my question is generally :
is ∀x∃y, (P(y,x)) ≡ ∀y∃x, (P(x,y)) ?
I understand the variables are swapped, but i believe that it is logically equivalent, as i just made the "everbody" labled as x and the "somebody" labeled as "y".
Thank you!
discrete-mathematics logic quantifiers logic-translation
edited Dec 11 '18 at 20:48
Bram28
63.9k44793
asked Dec 11 '18 at 5:33
Stephen IsaacStephen Isaac
11
$endgroup$
I was given this statement and asked to express this with universal quantifiers.
Likes(x,y) is a Binary Relation that means that person x likes person y.
The statement given was:
"Everybody is liked by somebody*". Easy Enough, or so I thought? I put:
∀x∃y, (Likes(y,x)). Apparently this is wrong and the correct solution (that a TA displayed as the answer) is ∀y∃x (Likes(x,y)).
For clarity my question is generally :
is ∀x∃y, (P(y,x)) ≡ ∀y∃x, (P(x,y)) ?
I understand the variables are swapped, but i believe that it is logically equivalent, as i just made the "everbody" labled as x and the "somebody" labeled as "y".
Thank you!
discrete-mathematics logic quantifiers logic-translation
discrete-mathematics logic quantifiers logic-translation
edited Dec 11 '18 at 20:48
Bram28
63.9k44793
asked Dec 11 '18 at 5:33
Stephen IsaacStephen Isaac
11
edited Dec 11 '18 at 20:48
Bram28
63.9k44793
asked Dec 11 '18 at 5:33
Stephen IsaacStephen Isaac
11
edited Dec 11 '18 at 20:48
Bram28
63.9k44793
edited Dec 11 '18 at 20:48
Bram28
63.9k44793
edited Dec 11 '18 at 20:48
Bram28
63.9k44793
63.9k44793
asked Dec 11 '18 at 5:33
Stephen IsaacStephen Isaac
11
asked Dec 11 '18 at 5:33
Stephen IsaacStephen Isaac
11
asked Dec 11 '18 at 5:33
Stephen IsaacStephen Isaac
11
11
$begingroup$
Those statements are logically equivalent, but surely the correct answer to the original problem is $forall xexists y text{Likes}(x,y)$.
$endgroup$
– Lord Shark the Unknown
Dec 11 '18 at 5:39
$begingroup$
Ah yes. My bad. I meant the statement to be the following: "Everybody is liked by somebody"
$endgroup$
– Stephen Isaac
Dec 11 '18 at 5:42
add a comment |
$begingroup$
Those statements are logically equivalent, but surely the correct answer to the original problem is $forall xexists y text{Likes}(x,y)$.
$endgroup$
– Lord Shark the Unknown
Dec 11 '18 at 5:39
$begingroup$
Ah yes. My bad. I meant the statement to be the following: "Everybody is liked by somebody"
$endgroup$
– Stephen Isaac
Dec 11 '18 at 5:42
$begingroup$
Those statements are logically equivalent, but surely the correct answer to the original problem is $forall xexists y text{Likes}(x,y)$.
$endgroup$
– Lord Shark the Unknown
Dec 11 '18 at 5:39
$begingroup$
Those statements are logically equivalent, but surely the correct answer to the original problem is $forall xexists y text{Likes}(x,y)$.
$endgroup$
– Lord Shark the Unknown
Dec 11 '18 at 5:39
$begingroup$
Ah yes. My bad. I meant the statement to be the following: "Everybody is liked by somebody"
$endgroup$
– Stephen Isaac
Dec 11 '18 at 5:42
$begingroup$
Ah yes. My bad. I meant the statement to be the following: "Everybody is liked by somebody"
$endgroup$
– Stephen Isaac
Dec 11 '18 at 5:42
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Yes, they are equivalent.
To actually formally show this, use the following two equivalences:
$$forall x varphi(x) Leftrightarrow forall y varphi(y)$$
$$exists x varphi(x) Leftrightarrow exists y varphi(y)$$
where $varphi$ is any formula, and where $varphi(x)$ is the same as $varphi(y)$, but with all free variables $x$ and $y$ swapped.
With these, we can do:
$$forall x exists y P(y,x) Leftrightarrow forall z exists y P(y,z) Leftrightarrow forall z exists x P(x,z) Leftrightarrow forall y exists x P(x,y)$$
answered Dec 11 '18 at 20:48
Bram28Bram28
63.9k44793
$endgroup$
add a comment |
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$begingroup$
Yes, they are equivalent.
To actually formally show this, use the following two equivalences:
$$forall x varphi(x) Leftrightarrow forall y varphi(y)$$
$$exists x varphi(x) Leftrightarrow exists y varphi(y)$$
where $varphi$ is any formula, and where $varphi(x)$ is the same as $varphi(y)$, but with all free variables $x$ and $y$ swapped.
With these, we can do:
$$forall x exists y P(y,x) Leftrightarrow forall z exists y P(y,z) Leftrightarrow forall z exists x P(x,z) Leftrightarrow forall y exists x P(x,y)$$
answered Dec 11 '18 at 20:48
Bram28Bram28
63.9k44793
$endgroup$
add a comment |
$begingroup$
Yes, they are equivalent.
To actually formally show this, use the following two equivalences:
$$forall x varphi(x) Leftrightarrow forall y varphi(y)$$
$$exists x varphi(x) Leftrightarrow exists y varphi(y)$$
where $varphi$ is any formula, and where $varphi(x)$ is the same as $varphi(y)$, but with all free variables $x$ and $y$ swapped.
With these, we can do:
$$forall x exists y P(y,x) Leftrightarrow forall z exists y P(y,z) Leftrightarrow forall z exists x P(x,z) Leftrightarrow forall y exists x P(x,y)$$
answered Dec 11 '18 at 20:48
Bram28Bram28
63.9k44793
$endgroup$
add a comment |
$begingroup$
Yes, they are equivalent.
To actually formally show this, use the following two equivalences:
$$forall x varphi(x) Leftrightarrow forall y varphi(y)$$
$$exists x varphi(x) Leftrightarrow exists y varphi(y)$$
where $varphi$ is any formula, and where $varphi(x)$ is the same as $varphi(y)$, but with all free variables $x$ and $y$ swapped.
With these, we can do:
$$forall x exists y P(y,x) Leftrightarrow forall z exists y P(y,z) Leftrightarrow forall z exists x P(x,z) Leftrightarrow forall y exists x P(x,y)$$
answered Dec 11 '18 at 20:48
Bram28Bram28
63.9k44793
$endgroup$
Yes, they are equivalent.
To actually formally show this, use the following two equivalences:
$$forall x varphi(x) Leftrightarrow forall y varphi(y)$$
$$exists x varphi(x) Leftrightarrow exists y varphi(y)$$
where $varphi$ is any formula, and where $varphi(x)$ is the same as $varphi(y)$, but with all free variables $x$ and $y$ swapped.
With these, we can do:
$$forall x exists y P(y,x) Leftrightarrow forall z exists y P(y,z) Leftrightarrow forall z exists x P(x,z) Leftrightarrow forall y exists x P(x,y)$$
answered Dec 11 '18 at 20:48
Bram28Bram28
63.9k44793
answered Dec 11 '18 at 20:48
Bram28Bram28
63.9k44793
answered Dec 11 '18 at 20:48
Bram28Bram28
63.9k44793
answered Dec 11 '18 at 20:48
Bram28Bram28
63.9k44793
63.9k44793
add a comment |
add a comment |
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$begingroup$
Those statements are logically equivalent, but surely the correct answer to the original problem is $forall xexists y text{Likes}(x,y)$.
$endgroup$
– Lord Shark the Unknown
Dec 11 '18 at 5:39
$begingroup$
Ah yes. My bad. I meant the statement to be the following: "Everybody is liked by somebody"
$endgroup$
– Stephen Isaac
Dec 11 '18 at 5:42