Convert statements into first order logic

[Marclan]

Given are the following predicate symbols:
Member(x) : x is a member of the bicycle club
Chairman(x) : x is the chairman of the bicycle club
Bicycle(x) : x is a bicycle
Brand(x, y) : the brand of x is y
Owns(x, y) : x is the owner of y

a. Statement: every member of the bicycle club has the same brand of bicycle.

Note that this statement is ambiguous, the meaning could be:
i: each member of the Club has bycicles of the same brand (monobrand member),
ii: all the members of the Club have bycicles of the same brand (monobrand Club),
iii: all the members of the Club have at least one bicycle of the same brand, but (s)he may
have other bicycles of other brands.

Express these interpretations of the given statement in first order logic, using the predicate
symbols above.

I'm trying for example the first one but I'm having some problems:
i. ∀x ∀y (member(x) ∧ bicycle(y) ∧ owns(x,y) ∧ ∃ z brand(y,z) → ¬∃ z2 brand(y,z2))
Or maybe it's better this:
i. ∀x ∀y (member(x) ∧ bicycle(y) ∧ owns(x,y) → ¬∃ z,z2 (brand(y,z2) ∧ brand(y,z)))

There is a correct answer or there could be an easier way to do this? I'll keep update this post if i will solve it. Thanks

 

[stevendaryl]

I'm trying for example the first one but I'm having some problems:
i. ∀x ∀y (member(x) ∧ bicycle(y) ∧ owns(x,y) ∧ ∃ z brand(y,z) → ¬∃ z2 brand(y,z2))
Or maybe it's better this:
i. ∀x ∀y (member(x) ∧ bicycle(y) ∧ owns(x,y) → ¬∃ z,z2 (brand(y,z2) ∧ brand(y,z)))

Both of those are sort of close, but not exactly. Look at the first one: (I've rewritten it in LaTex):
$\forall x \forall y (member(x) \wedge bicycle(y) \wedge owns(x,y) \wedge \exists z\ brand(y,z) \rightarrow \neg \exists z_2\ brand(y,z_2))$

The clause--$\neg \exists z_2\ brand(y,z_2)$--literally says that "y does not have a brand". What you need to say, instead is that "y does not have a brand that is different from z".

 

[Marclan]

Both of those are sort of close, but not exactly. Look at the first one: (I've rewritten it in LaTex):
$\forall x \forall y (member(x) \wedge bicycle(y) \wedge owns(x,y) \wedge \exists z\ brand(y,z) \rightarrow \neg \exists z_2\ brand(y,z_2))$

The clause--$\neg \exists z_2\ brand(y,z_2)$--literally says that "y does not have a brand". What you need to say, instead is that "y does not have a brand that is different from z".

Thank you for the answer, I'm trying to translate that "different from z" but I'm not finding the right way to do that in FOL, the following statement is wrong too?
∀x ∀y (member(x) ∧ bicycle(y) ∧ owns(x,y) ∧ ∃ z brand(y,z) → ¬∃ z2 (brand(y,z2) ∧ brand(y,z)))

It's probably wrong because it's saying that for each bicycle not exist more than one brand associated with it...

 

[Mark44]

@Marclan, in future posts, please don't delete the three parts of the homework template -- they are required here.

 

[Marclan]

Still trying to understand how to convert the first sentence, can anyone give an advice to help me?
The problem is translating this correct clue "y does not have a brand that is different from z".

 

[Marclan]

In this moment I'm thinking that it's not possible to say that in the first order logic. I'm still on my solution of the third post.

 

[Marclan]

Can anyone check if this can be a solution for the second statement? Thanks ;)
∀x ∀y ∀z ∀x2 ∀y2 (member(x) ∧ bicycle(y) ∧ owns(x,y) ∧ brand(y,z) → member(x2) ∧ bicycle(y2) ∧ owns(x2,y2) ∧ brand(y2,z))

I don't know if it's more correct use ∃z or ∀z.

 

[PeroK]

Still trying to understand how to convert the first sentence, can anyone give an advice to help me?
The problem is translating this correct clue "y does not have a brand that is different from z".

This isn't my area, but it terms of normal mathematical logic, for (i) I would start:

For all $x$ there exists $z$ $\dots$

where $x$ is a member and $z$ (depending on $x$) is the brand of bicycle that $x$ has exclusively.

 

[Marclan]

I wrote the iii) and i think it's correct:
∀x ∀y ∀z (member(x) ∧ bicycle(y) ∧ owns(x,y) ∧ brand(y,z) → ∀x2 ∃y2 (member(x2) ∧ bicycle(y2) ∧ owns(x2,y2) ∧ brand(y2,z)))

Always here trying to resolve i) and ii) and learn something new. Thanks!

 

[PeroK]

I wrote the iii) and i think it's correct:
∀x ∀y ∀z (member(x) ∧ bicycle(y) ∧ owns(x,y) ∧ brand(y,z) → ∀x2 ∃y2 (member(x2) ∧ bicycle(y2) ∧ owns(x2,y2) ∧ brand(y2,z)))

Always here trying to resolve i) and ii) and learn something new. Thanks!

That doesn't look right at all. (iii) ought to start with: there exists a brand $z$ ...

 

[PeroK]

@Marclan have you done any (mathematical) logic before? My approach would be to translate the English into logic, then (what I can't do) translate the logic into the first-order mumbo-jumbo. But, if you can't do "normal" logic, it's going to be hard to go straight to the first-order expressions.

For example the phrase I've underlined here is logically superfluous:

iii: all the members of the Club have at least one bicycle of the same brand, but (s)he may
have other bicycles of other brands.

 

[Marclan]

That doesn't look right at all. (iii) ought to start with: there exists a brand $z$ ...

It's not so simple in the first order logic. Can you explain me why my last statement is wrong? Translating that in words sounds like: "For each member who has a bicycle of a brand z, there is for each other member (x2) at least one bicycle Y2 of the same brand z. It's not what iii) means?

 

[Marclan]

iii: all the members of the Club have at least one bicycle of the same brand, but (s)he may
have other bicycles of other brands.

Yes i saw that too, I'm just ignoring it.

 

[PeroK]

It's not so simple in the first order logic. Can you explain me why my last statement is wrong? Translating that in words sounds like: "For each member who has a bicycle of a brand z, there is for each other member (x2) at least one bicycle Y2 of the same brand z. It's not what iii) means?

That's not what (iii) means. (iii) means there is some brand (call it the club's default brand) and all members have at least one bicycle of that brand.

 

[PeroK]

It's not so simple in the first order logic. Can you explain me why my last statement is wrong? Translating that in words sounds like: "For each member who has a bicycle of a brand z, there is for each other member (x2) at least one bicycle Y2 of the same brand z. It's not what iii) means?

It's clearly wrong because (iii) says nothing about "all brands" that are owned by at least one club member, which is the gist of what you have.

I'd start again and try at least to get a clear logical statement of what (i)-(iii) mean. Without that you will just go round in circles.

 

[Marclan]

It's clearly wrong because (iii) says nothing about "all brands" that are owned by at least one club member, which is the gist of what you have.

I'd start again and try at least to get a clear logical statement of what (i)-(iii) mean. Without that you will just go round in circles.

I'm sure about a thing, in the text there isn't a specific brand for the Club (instead of how it could be normal in the reality) so I'm thinking that a right way to solve it is using a general brand. And i don't see the problem "all the brands" because in a normal instantiation of the bicycles, each one will have only one brand so I'm just only saying for each member, each bicycle of him/her and for each brand associated (one) etc...
For example, i can change that in the following statement (exist instead of each for z) but i don't see the advantage:

∀x ∀y ∃z (member(x) ∧ bicycle(y) ∧ owns(x,y) ∧ brand(y,z) → ∀x2 ∃y2 (member(x2) ∧ bicycle(y2) ∧ owns(x2,y2) ∧ brand(y2,z)))

Thank you for the support!

 

[PeroK]

I'm sure about a thing, in the text there isn't a specific brand for the Club (instead of how it could be normal in the reality) so I'm thinking that a right way to solve it is using a general brand. And i don't see the problem "all the brands" because in a normal instantiation of the bicycles, each one will have only one brand so I'm just only saying for each member, each bicycle of him/her and for each brand associated (one) etc...
For example, i can change that in the following statement (exist instead of each for z) but i don't see the advantage:

∀x ∀y ∃z (member(x) ∧ bicycle(y) ∧ owns(x,y) ∧ brand(y,z) → ∀x2 ∃y2 (member(x2) ∧ bicycle(y2) ∧ owns(x2,y2) ∧ brand(y2,z)))

Thank you for the support!

I still think you're going wrong on both the logical interpretation of the statement and then the rendering of that statement into first-order logic.

For example, the $\exists \ z$ is superfluous and what you have means something like:

If a club member owns a bicycle of brand z, then all members of the club own a bicycle of brand z.

 

[Marclan]

For example, the $\exists \ z$ is superfluous and what you have means something like:

If a club member owns a bicycle of brand z, then all members of the club own a bicycle of brand z.

You are right, it's the correct meaning. My statement doesn't say that "at least one". In this moment i don't really know how to solve all of these. Thanks anyway.
Always opened for new ideas ;)

 

[PeroK]

You are right, it's the correct meaning. My statement doesn't say that "at least one". In this moment i don't really know how to solve all of these. Thanks anyway.

I'm going offline now anyway. But, my advice is you have to develop your logical thinking. Here's how I look at statements (i)-(iii). I do three things: try to figure out what it means (obviously!); think of an example of when it is true; figure out what it doesn't mean; think of an example of when it is false.

(i) each member of the Club has bicycles of the same brand (monobrand member)

What does it mean? It means every club member has his/her own preferred brand and all his/her bikes are of that brand.

Example: Three members: A has one bike (brand X); B has 6 bikes (all brand Y); C has two bikes (all brand X)

What does it not mean? It doesn't mean there is any relationship between the brands of bicycles preferred by different members.

What else does it mean? It means no one can have two bikes of different brands.

Example: A has two bikes, one brand X, one brand Y. This is not allowed by statement (i).

That's the way I would analyse this. And, at this point I believe I truly understand statement (i).

 

[Marclan]

I'm going offline now anyway. But, my advice is you have to develop your logical thinking. Here's how I look at statements (i)-(iii). I do three things: try to figure out what it means (obviously!); think of an example of when it is true; figure out what it doesn't mean; think of an example of when it is false.

(i) each member of the Club has bicycles of the same brand (monobrand member)

What does it mean? It means every club member has his/her own preferred brand and all his/her bikes are of that brand.

Example: Three members: A has one bike (brand X); B has 6 bikes (all brand Y); C has two bikes (all brand X)

What does it not mean? It doesn't mean there is any relationship between the brands of bicycles preferred by different members.

What else does it mean? It means no one can have two bikes of different brands.

Example: A has two bikes, one brand X, one brand Y. This is not allowed by statement (i).

That's the way I would analyse this. And, at this point I believe I truly understand statement (i).

A good help but I'm still on my past solution:

∀x ∀y (member(x) ∧ bicycle(y) ∧ owns(x,y) ∧ ∃ z brand(y,z) → ¬∃ z2 (brand(y,z2) ∧ brand(y,z)))

And i have just wrote that it's probably wrong because it's only saying that for each bicycle not exist more than one brand associated with it. If only i could know how to write this, i will solve all the exercize.

 

[Stephen Tashi]

If only i could know how to write this, i will solve all the exercize.

Are you required to put all the quantifiers on the left side of the statement?

It seems to me the natural way to express things would be to nest them.$\forall m ( member(m) \rightarrow ( \exists c ( bicycle(c) \land owns(m,c) \land \exists b ( brand(c,b) \land \forall y ( bicycle(y) \land owns(x,y) \rightarrow ...$ etc.

 

[PeroK]

A good help but I'm still on my past solution:

∀x ∀y (member(x) ∧ bicycle(y) ∧ owns(x,y) ∧ ∃ z brand(y,z) → ¬∃ z2 (brand(y,z2) ∧ brand(y,z)))

And i have just wrote that it's probably wrong because it's only saying that for each bicycle not exist more than one brand associated with it. If only i could know how to write this, i will solve all the exercize.

There are a number of problems. The main problem is that there is nothing logical to say that $z2 \ne z$. Simply giving a variable a different name doesn't mean it can't have the same value as another. Second, with the last clause, you are actually trying to say that no bike has two brands.

For (i), one of the advantages of my analysis was that I can see an alternative way to write this statement as the contrapositive:

No member has two bikes of different brands.

Logically that becomes:

There does not exist member $x$ who owns bikes $y_1$ and $y_2$ with Brand of $y_1$ not equal to Brand of $y_2$.

That is an alternative approach you may wish to think about.

The equivalent positive statement for (i) is:

Every member has bikes of only a single brand. (this is the logical structure I'll use)

For all members $x$, if $x$ owns bikes $y_1$ and $y_2$ then Brand of $y_1$ = Brand of $y_2$.

You can see how I'm building this up and can check at each stage that my statements agree with the previous step.

One more step for you:

For all members $x$, if $x$ owns bikes $y_1$ and $y_2$ and Brand($y_1, z_1)$ and Brand($y_2, z_2$), then $z_1 = z_2$

The final step is to expand this into the full first-order. But, as I keep saying, your problem is not really with first-order logic, your problem is that you haven't yet learned to think logically. It's a bit like trying to run a hurdles race before you can run. You need to learn to run first (think logically) then run and jump the hurdles (first-order logic).

 

[Marclan]

There are a number of problems. The main problem is that there is nothing logical to say that $z2 \ne z$. Simply giving a variable a different name doesn't mean it can't have the same value as another. Second, with the last clause, you are actually trying to say that no bike has two brands.

For (i), one of the advantages of my analysis was that I can see an alternative way to write this statement as the contrapositive:

No member has two bikes of different brands.

Logically that becomes:

There does not exist member $x$ who owns bikes $y_1$ and $y_2$ with Brand of $y_1$ not equal to Brand of $y_2$.

That is an alternative approach you may wish to think about.

The equivalent positive statement for (i) is:

Every member has bikes of only a single brand. (this is the logical structure I'll use)

For all members $x$, if $x$ owns bikes $y_1$ and $y_2$ then Brand of $y_1$ = Brand of $y_2$.

You can see how I'm building this up and can check at each stage that my statements agree with the previous step.

One more step for you:

For all members $x$, if $x$ owns bikes $y_1$ and $y_2$ and Brand($y_1, z_1)$ and Brand($y_2, z_2$), then $z_1 = z_2$

The final step is to expand this into the full first-order. But, as I keep saying, your problem is not really with first-order logic, your problem is that you haven't yet learned to think logically. It's a bit like trying to run a hurdles race before you can run. You need to learn to run first (think logically) then run and jump the hurdles (first-order logic).

The big problem is that simple z1=z2 is not possible to write in FOL (as far as I'm concerned). Yesterday i thought about a new way to say the i), and this is it:

∀x ∀y ∃z (member(x) ∧ bicycle(y) ∧ owns(x,y) ∧ brand(y,z) → ∀y2 bicycle(y2) ∧ owns(x,y2) ∧ brand(y2,z))

It's seems correct to me, but I'm waiting your opinion ;)

 

[Marclan]

Are you required to put all the quantifiers on the left side of the statement?

It seems to me the natural way to express things would be to nest them.$\forall m ( member(m) \rightarrow ( \exists c ( bicycle(c) \land owns(m,c) \land \exists b ( brand(c,b) \land \forall y ( bicycle(y) \land owns(x,y) \rightarrow ...$ etc.

No but using more quantifiers you can on the left side of the statement helps to reduce the starting set to consider and make more simple the writing of the formula (instead of writing → followed by one other → etc).

 

[PeroK]

The big problem is that simple z1=z2 is not possible to write in FOL (as far as I'm concerned). Yesterday i thought about a new way to say the i), and this is it:

∀x ∀y ∃z (member(x) ∧ bicycle(y) ∧ owns(x,y) ∧ brand(y,z) → ∀y2 bicycle(y2) ∧ owns(x,y2) ∧ brand(y,z))

It's seems correct to me, but I'm waiting your opinion ;)

I think that says something like "if x owns a bicycle of brand z, then all bicycles are owned by x and are of brand z)"

Given the way z1 = z2 translates to FOL, perhaps a better preparatory statement would be:

For all members x, there exists (a brand) z, such that all bikes owned by x are of brand z.

 

[Marclan]

I think that says something like "if x owns a bicycle of brand z, then all bicycles are owned by x and are of brand z)"

Given the way z1 = z2 translates to FOL, perhaps a better preparatory statement would be:

For all members x, there exists (a brand) z, such that all bikes owned by x are of brand z.

I'm trying to write your last statement but the last part (where you noted the problem) is where I'm blocked.
I have to say that "such that all bikes owned by x are of brand z" but I'm only able to write this:
∀y2 bicycle(y2) ∧ owns(x,y2) ∧ brand(y2,z)).
How to say "for each bike owned by x" instead of what I'm wrongly writing "all the bikes are owned by x"?
∀y2 bicycle(y2) ∧ owns(x,y2) → brand(y2,z)) : for example this is not a solution.

I have to find a way to separate "all the bikes" and that "own"

 

[PeroK]

How to say "for each bike owned by x" instead of what I'm wrongly writing "all the bikes are owned by x"?
∀y2 bicycle(y2) ∧ owns(x,y2) → brand(y,z)) : for example this is not a solution.

That looks a lot better, although you need $y2$ throughout.

 

[Marclan]

That looks a lot better.

But the problem you saw before remain, right? (i wrongly wrote y instead of y2, edited).

I have to put that "own" before "all the bikes" but i don't know how because it's a predicate associated with a bike.

 

[PeroK]

But the problem you saw before remain, right?

You want to say:

For each member (x) there exists a brand (z) such that: (all bikes owned by x are of brand z).

And you now have the component in FOL expressing "all bikes owned by x are of brand z".

You should be nearly there for (i).

PS I'm going to change my analogy. I think initially you were trying to jump all the hurdles at the same time. You need to approach these problems by breaking them down into their logical components.

 

[Marclan]

You want to say:

For each member (x) there exists a brand (z) such that: (all bikes owned by x are of brand z).

And you now have the component in FOL expressing "all bikes owned by x are of brand z".

You should be nearly there for (i).

PS I'm going to change my analogy. I think initially you were trying to jump all the hurdles at the same time. You need to approach these problems by breaking them down into their logical components.

As you surely saw, the predicate brand have two arguments so i can't say only (For each member (x)) there exists a brand (z) but i have to use also the bikes when i write it (and it implies the using of the predicate owns too). So i have to come back on my solution, where i have only changed the last ∧ with a → .
∀x ∀y ∃z (member(x) ∧ bicycle(y) ∧ owns(x,y) ∧ brand(y,z) → ∀y2 bicycle(y2) ∧ owns(x,y2) → brand(y2,z))

Before you said : I think that says something like "if x owns a bicycle of brand z, then all bicycles are owned by x and are of brand z)".
So, the problem of saying with → ∀y2 bicycle(y2) ∧ owns(x,y2) → that all the bikes of the club are owned by x is still there right? (instead of what i want to say "all the bikes owned by x").

Trying to explain a bit better, ∀y2 bicycle(y2) ∧ owns(x,y2) say which of those?:
-all the bikes in the club are owned by x or
-all the bikes of the club owned by x.

In this moment i agree with your past thought and so the first one.

 

[PeroK]

As you surely saw, the predicate brand have two arguments so i can't say only (For each member (x)) there exists a brand (z) but i have to use also the bikes when i write it (and it implies the using of the predicate owns too). So i have to come back on my solution, where i have only changed the last ∧ with a → .
∀x ∀y ∃z (member(x) ∧ bicycle(y) ∧ owns(x,y) ∧ brand(y,z) → ∀y2 bicycle(y2) ∧ owns(x,y2) → brand(y2,z))

Before you said : I think that says something like "if x owns a bicycle of brand z, then all bicycles are owned by x and are of brand z)". So, the problem of saying with → ∀y2 bicycle(y2) ∧ owns(x,y2) → that all the bikes of the club are owned by x is still there right? (instead of what i want to say "all the bikes owned by x").

The order of the quantifies is important. By putting the $\exists z$ after the $\forall y$ you make the brand depend on the bicycle. You need:

$\forall x \ \exists z \ \forall y \ (member(x) \land bicycle(y) \land owns(x, y) \rightarrow brand(y, z))$

Which says: for all members, there exists a brand (depends on the member) such that all bikes owned by that member are of brand z.

Which is what you want.

That's very different from starting with:

$\forall x \ \forall y \ \exists z \ \dots$

Which means the brand depends on the member and the particular bicycle. That gets you off on the wrong foot immediately. (Although it might be useful for part (ii) or (iii)!)

 

[Marclan]

The order of the quantifies is important. By putting the $\exists z$ after the $\forall y$ you make the brand depend on the bicycle. You need:

$\forall x \ \exists z \ \forall y \ (member(x) \land bicycle(y) \land owns(x, y) \rightarrow brand(y, z))$

Which says: for all members, there exists a brand (depends on the member) such that all bikes owned by that member are of brand z.

Which is what you want.

That's very different from starting with:

$\forall x \ \forall y \ \exists z \ \dots$

Which means the brand depends on the member and the particular bicycle. That gets you off on the wrong foot immediately. (Although it might be useful for part (ii) or (iii)!)

A very useful reply and there is also the answer for the i). Thanks, i'll use your advices to do the ii) and iii).

 

[Marclan]

(Although it might be useful for part (ii) or (iii)!)

Tried the ii and iii:
ii) ∀x ∀y ∃z (member(x) ∧ bicycle(y) ∧ owns(x,y) → brand(y,z))
iii) ∀x ∃y ∃z (member(x) ∧ bicycle(y) ∧ owns(x,y) ∧ brand(y,z) → ∀x2 ∃y2 (member(x2) ∧ bicycle(y2) ∧ owns(x2,y2) ∧ brand(y2,z)))

 

[PeroK]

Tried the ii and iii:
ii) ∀x ∀y ∃z (member(x) ∧ bicycle(y) ∧ owns(x,y) → brand(y,z))
iii) ∀x ∃y ∃z (member(x) ∧ bicycle(y) ∧ owns(x,y) ∧ brand(y,z) → ∀x2 ∃y2 (member(x2) ∧ bicycle(y2) ∧ owns(x2,y2) ∧ brand(y2,z)))

(ii) looks familiar!

For (iii), the existence of the brand must come first. There is a brand that everyone owns (at least one bike each).

 

[Marclan]

(ii) looks familiar!

For (iii), the existence of the brand must come first. There is a brand that everyone owns (at least one bike each).

ii) Haha

iii) ∃z ∀x ∃y (member(x) ∧ bicycle(y) ∧ owns(x,y) ∧ brand(y,z) → ∀x2 ∃y2 (member(x2) ∧ bicycle(y2) ∧ owns(x2,y2) ∧ brand(y2,z)))

So, can i mark this thread as finally solved?