Convert statements into first order logic

[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?

 

[PeroK]

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))