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2

$begingroup$

Is ∀x∀y∀z[φ(x,y)∧p(y,z)->p(x,z)] equivalent to ∀x∀y∀z[φ(x,y)∧p(x,z)->p(y,z)] ?

The only thing I can think of is that this question can be answered if we show that p->q is equals (↔) το q->p, but that's not true because p->q ↔ ¬p->¬q, hence p->q is not equals to q->p. However, I do not know if my logic is correct and if it can be accepted as an answer.

I also think that the (first) ∀x∀y∀z[φ(x,y)∧ part is irrelevant and that i have to focus only to the last part.

logic discrete-mathematics propositional-logic

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edited 9 hours ago

George Z.

asked 9 hours ago

George Z.George Z.

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add a comment | 

2

$begingroup$

Is ∀x∀y∀z[φ(x,y)∧p(y,z)->p(x,z)] equivalent to ∀x∀y∀z[φ(x,y)∧p(x,z)->p(y,z)] ?

The only thing I can think of is that this question can be answered if we show that p->q is equals (↔) το q->p, but that's not true because p->q ↔ ¬p->¬q, hence p->q is not equals to q->p. However, I do not know if my logic is correct and if it can be accepted as an answer.

I also think that the (first) ∀x∀y∀z[φ(x,y)∧ part is irrelevant and that i have to focus only to the last part.

logic discrete-mathematics propositional-logic

share|cite|improve this question

edited 9 hours ago

George Z.

asked 9 hours ago

George Z.George Z.

11355 bronze badges

New contributor

George Z. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

Is ∀x∀y∀z[φ(x,y)∧p(y,z)->p(x,z)] equivalent to ∀x∀y∀z[φ(x,y)∧p(x,z)->p(y,z)] ?

The only thing I can think of is that this question can be answered if we show that p->q is equals (↔) το q->p, but that's not true because p->q ↔ ¬p->¬q, hence p->q is not equals to q->p. However, I do not know if my logic is correct and if it can be accepted as an answer.

I also think that the (first) ∀x∀y∀z[φ(x,y)∧ part is irrelevant and that i have to focus only to the last part.

logic discrete-mathematics propositional-logic

share|cite|improve this question

edited 9 hours ago

George Z.

asked 9 hours ago

George Z.George Z.

11355 bronze badges

New contributor

George Z. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 9 hours ago

George Z.

asked 9 hours ago

George Z.George Z.

11355 bronze badges

New contributor

George Z. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 9 hours ago

George Z.

asked 9 hours ago

George Z.George Z.

11355 bronze badges

New contributor

George Z. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 9 hours ago

George Z.George Z.

11355 bronze badges

New contributor

George Z. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 9 hours ago

George Z.George Z.

11355 bronze badges

1 Answer
1

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2

$begingroup$

It does hold that
$$
forall x forall y forall z , [p(y,z) to p(x,z)] quad longleftrightarrow quad
forall x forall y forall z , [p(x,z) to p(y,z)],
$$
since $forall x forall y$ is the same thing as $forall y forall x$. So your logic is incorrect.

Similarly, if $varphi$ is promised to be symmetric then the equivalence holds, for similar reasons.

A simple example where the equivalence doesn't hold is $varphi(x,y) = x$, $p(x,y) = x$.

In this case the first statement is
$$
forall x forall y forall z , [x land y to x],
$$
whereas the second statement is
$$
forall x forall y forall z , [x land x to y].
$$
(This assumes that the statements are interpreted as $(varphi land p) to p$ rather than $varphi land (p to p)$.)

share|cite|improve this answer

answered 9 hours ago

Yuval FilmusYuval Filmus

207k1515 gold badges200200 silver badges368368 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is there any explanation why ∀x∀y∀z[x∧y→x] is not equivalent to ∀x∀y∀z[x∧x→y]? .... I mean, how would I prove it?
  $endgroup$
  – George Z.
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm afraid you'll have to work it out on your own.
  $endgroup$
  – Yuval Filmus
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see... I tried few things but still not something clear. Anyway thanks for your help.
  $endgroup$
  – George Z.
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Use the definitions.
  $endgroup$
  – Yuval Filmus
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  [x∧y→x] => ¬(x∧y)∨x =>¬x∨¬y∨x which is always true. But x∧x→y => x->y which can be false (if y is false). What about this? :)
  $endgroup$
  – George Z.
  8 hours ago
  
  

  
  
  
  

 | 
show 1 more comment

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2

$begingroup$

It does hold that
$$
forall x forall y forall z , [p(y,z) to p(x,z)] quad longleftrightarrow quad
forall x forall y forall z , [p(x,z) to p(y,z)],
$$
since $forall x forall y$ is the same thing as $forall y forall x$. So your logic is incorrect.

Similarly, if $varphi$ is promised to be symmetric then the equivalence holds, for similar reasons.

A simple example where the equivalence doesn't hold is $varphi(x,y) = x$, $p(x,y) = x$.

In this case the first statement is
$$
forall x forall y forall z , [x land y to x],
$$
whereas the second statement is
$$
forall x forall y forall z , [x land x to y].
$$
(This assumes that the statements are interpreted as $(varphi land p) to p$ rather than $varphi land (p to p)$.)

share|cite|improve this answer

answered 9 hours ago

Yuval FilmusYuval Filmus

207k1515 gold badges200200 silver badges368368 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is there any explanation why ∀x∀y∀z[x∧y→x] is not equivalent to ∀x∀y∀z[x∧x→y]? .... I mean, how would I prove it?
  $endgroup$
  – George Z.
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm afraid you'll have to work it out on your own.
  $endgroup$
  – Yuval Filmus
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see... I tried few things but still not something clear. Anyway thanks for your help.
  $endgroup$
  – George Z.
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Use the definitions.
  $endgroup$
  – Yuval Filmus
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  [x∧y→x] => ¬(x∧y)∨x =>¬x∨¬y∨x which is always true. But x∧x→y => x->y which can be false (if y is false). What about this? :)
  $endgroup$
  – George Z.
  8 hours ago
  
  

  
  
  
  

 | 
show 1 more comment

2

$begingroup$

It does hold that
$$
forall x forall y forall z , [p(y,z) to p(x,z)] quad longleftrightarrow quad
forall x forall y forall z , [p(x,z) to p(y,z)],
$$
since $forall x forall y$ is the same thing as $forall y forall x$. So your logic is incorrect.

Similarly, if $varphi$ is promised to be symmetric then the equivalence holds, for similar reasons.

A simple example where the equivalence doesn't hold is $varphi(x,y) = x$, $p(x,y) = x$.

In this case the first statement is
$$
forall x forall y forall z , [x land y to x],
$$
whereas the second statement is
$$
forall x forall y forall z , [x land x to y].
$$
(This assumes that the statements are interpreted as $(varphi land p) to p$ rather than $varphi land (p to p)$.)

share|cite|improve this answer

answered 9 hours ago

Yuval FilmusYuval Filmus

207k1515 gold badges200200 silver badges368368 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is there any explanation why ∀x∀y∀z[x∧y→x] is not equivalent to ∀x∀y∀z[x∧x→y]? .... I mean, how would I prove it?
  $endgroup$
  – George Z.
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm afraid you'll have to work it out on your own.
  $endgroup$
  – Yuval Filmus
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see... I tried few things but still not something clear. Anyway thanks for your help.
  $endgroup$
  – George Z.
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Use the definitions.
  $endgroup$
  – Yuval Filmus
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  [x∧y→x] => ¬(x∧y)∨x =>¬x∨¬y∨x which is always true. But x∧x→y => x->y which can be false (if y is false). What about this? :)
  $endgroup$
  – George Z.
  8 hours ago
  
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

It does hold that
$$
forall x forall y forall z , [p(y,z) to p(x,z)] quad longleftrightarrow quad
forall x forall y forall z , [p(x,z) to p(y,z)],
$$
since $forall x forall y$ is the same thing as $forall y forall x$. So your logic is incorrect.

Similarly, if $varphi$ is promised to be symmetric then the equivalence holds, for similar reasons.

A simple example where the equivalence doesn't hold is $varphi(x,y) = x$, $p(x,y) = x$.

In this case the first statement is
$$
forall x forall y forall z , [x land y to x],
$$
whereas the second statement is
$$
forall x forall y forall z , [x land x to y].
$$
(This assumes that the statements are interpreted as $(varphi land p) to p$ rather than $varphi land (p to p)$.)

share|cite|improve this answer

answered 9 hours ago

Yuval FilmusYuval Filmus

207k1515 gold badges200200 silver badges368368 bronze badges

$endgroup$

share|cite|improve this answer

answered 9 hours ago

Yuval FilmusYuval Filmus

207k1515 gold badges200200 silver badges368368 bronze badges

answered 9 hours ago

Yuval FilmusYuval Filmus

207k1515 gold badges200200 silver badges368368 bronze badges

answered 9 hours ago

Yuval FilmusYuval Filmus

207k1515 gold badges200200 silver badges368368 bronze badges