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...
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.
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"...
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).
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...
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...
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 ;)
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...
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?
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!
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.