# Convert statements into first order logic

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
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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.

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
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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.

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.

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PeroK
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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)!)

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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).

(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
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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?

PeroK
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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?

The qualifiers at the beginning were key. Once you get those in the correct order, it's just:

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

Marclan