Every dog and cat who is well trained is a good pet.

(F: a is a dog; G: a is a cat; H: a is well trained; I: a is a good pet.)

Here are my options:

a) ∀x((Fx∨Gx)∧Ix→Hx)

b) ∀x((Fx∨Gx)∧Hx→Ix)

c) ∀x((Fx∧Gx)∧Hx→Ix)

d) ∀x∀y(Fx∧Gy∧Hx→Ix)

For a) I interpret it as: for all dogs or cats and if the dog or cat is well trained, then the dog or cat is a good pet. Makes sense...

for b) I interpret it as: for all dogs or cats and if it's a good pet, then it is well trained. I don't think this one is right since it means that being a good pet implies well trained but the statement says: well trained implies good pet

for c) for all things that are dogs and cats and is well trained it implies good pet. (i don't think it's this one because you cant be a dog and a cat)

for d) I'm not sure about this one? I read that the ordering of the quantifiers matter but I'm not too sure. Also the it only applies for either a dog or a cat since it is Hx -> lx and not Hy -> Ly. can someone help me with this one?

i think it's a)... need guidance thank you.


Option b looks correct to me. Your descriptions for a and b seem to be the wrong way round.

Option c says that anything that is both a dog and a cat and well-trained is a good pet.

Option d is weird - it says that if anything is a cat then well-trained dogs are good pets.

  • oops Thank you, those were the other way around. But for d) shouldn't it be: if everything that is a cat and a well trained dog then it is a good pet? and would it change the meaning if the quantifiers switched positions? so if it was ∀y∀x instead of ∀x∀y – user125535 Nov 4 '15 at 17:55
  • The order of the quantifiers doesn't matter here. There is no Iy in the consequent, so it is not saying that cats are good pets. – Bumble Nov 4 '15 at 19:05

A_ | B | A and B ___ __ __ ______ A_| B | A or B

T_ | T |___ T__________________ T_ | T |__T

T_ | F | ___F__________________ T_ | F |__T

F_ | T | ___F__________________ F_ | T |__T

F_ | F | ___F__________________ F_ | F |__F

As you can see, when conjuncting A, B(A and B) is only true if both prepositions are true. When disjuncting A, B(A or B) is true if both are true or if one of them is true. So what can you interpret from the question? Every dog and cat who's well trained is a good pet. Do you think it needs to be both cat and dog for the sentence to become true or just one of them to sentence be true?

The way I read it:

a) For all xx, dog(true) or cat(true), good pets imply well trained.

The answers needs that both dog and cat are true, since there is or means that both
don't need to be true at the same time.

b) For all xx, dog(true) or cat(true), is well trained implies they are good pets.

Again it says dog or cat not both, it needs to be both.

c) For all xx, dog(true) and cat(true), are well trained so good pets.

I believe more in c) than in a) or b) or d) because says that both dog and 
cat are well trained and therefore good pets and for all xx means 'every' so
= Every dog and cat who is well trained is a good pet.

d) For all xx and yy, dog(x-true) and cat(y-true), dog well trained means good pet.

That means:every dog and every cat, and dog who's well trained means good
pet, because the implication is true only for all xx and cat is not x is y.
So cat is not included in the implication (→)

I'm pretty sure is answer c).

b) does not groups cats and dogs. It says that either a cat or a dog is well trained so is a good pet. The way I read it is: Every dog or cat who is well trained is a good pet.

  • c would make something simultaneously a dog and a cat. While that's actually a possible reading of the English, it doesn't make very much sense... – virmaior Nov 5 '15 at 3:37
  • Maybe like that would be better: (∀xFx∧ ∀xGx)∧Hx→Ix – Inês Barata Feio Borges Nov 5 '15 at 3:43
  • That would actually be gibberish or at the least very confusing because the universal quantifier would not apply to Hx or Ix making them instantiations using x while in ∀xFx∧ ∀xGx there are two different universal instantiations of x that make it so that you're saying "If (For any x, x is a cat and For any x, x is a dog") and y (not same as previous universal but oddly being called "x") is well-trained, then that particular y is a good pet. – virmaior Nov 5 '15 at 3:45
  • (∀xFx∧∀xGx)∧∀x(Hx→Ix). if you do multiplication by all caracteres insede the brakets it's going to give the same thing – Inês Barata Feio Borges Nov 5 '15 at 3:50

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