- #1

Marclan

- 19

- 0

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.

Or maybe it's better this:

i.

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

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

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