# Predicate logic translations with identity

While translating/transcribing the English sentences into predicate logic, straightforward result is not easily obtained in many times. Although I solved several examples, I am not quite sure if I got the translation correct. Also many translations require unveil the hidden structures. The strategy I use for solving them is the one I acquired from one of Russell's that is used when he explains how to refer "the" in predicate logic. Below, I will ask some questions and will provide the progress I did to solve them. It would be super nice if you comment on their correctness in fruitful way. Thanks in advance.

Let $D(x) = x$ is a dog, $H(x) = x$ is happy, $P(x) = x$ is a person, $L({x,y}) = x$ likes $y$, $S({x,y}) = x$ is stronger than $y$. Given these:

1) If Hector likes Achilles, then he likes a happy dog.

Here we have if, then statement which indicates an implication. Let h denotes Hector and a denotes achilles. So antecedent part of the conditional is $L(h,a)$. However I am not sure about the consequent. It simply states that Hector likes a happy dog, another we cannot b sure if the happy dog is referring him liking achilles or any other dog. Therefore I thought he likes some other dog, which maybe is Achilles that we are unsure of.
$L(h,a) \rightarrow \exists x \Big((D(x)\land H(x))\land L(h,x)\Big) \tag 1$
2)A happy dog likes nothing.

any happy dog that is taken, any thing that is taken, the dog will not like it. So two universals will do.
$\forall x \Big( (D(x) \land H(x)) \rightarrow \lnot \forall y L(x,y)\Big) \tag 2$
3)Only Hector and Achilles are happy dogs.

We can break these sentence into the following sentences.
*Hector is a happy dog.
*Achilles is a happy dog.
*If something is a happy dog, it is either Hector and Achilles. However I am not sure about stating Hector and Achilles as happy dogs in the beginning. Some argues that we can only understand that they are dogs at first sight. So there are two possible answers.
$((D(s) \land H(s)) \land (D(a) \land H(a)))\land \forall x \Big( (D(x) \land H(x)) \rightarrow (x=s \lor x=a)\Big)\tag{3.1}$
$(D(s) \land (D(a))\land \forall x \Big( (D(x) \land H(x)) \rightarrow (x=s \lor x=a)\Big)\tag{3.2}$
6)Someone likes the happy dog.

I am confused about the happy dog part. Does this mean that for all things that are dogs, there is only one dog that is happy and this sentence is referring this happy dog? I thought this is the case and therefor I translated this sentence as there is exactly one happy dog who is read by at least one person.
$\exists x \Big( (D(x) \land H(x) ) \land \forall y ((D(y) \land H(y) ) \rightarrow x=y) \land \exists z(P(z) \land L(z,x))\Big)\tag 6$
7)There is exactly one happy dog.

This is simply the main part of previous question.
$\exists x \Big( (D(x) \land H(x) ) \land \forall y ((D(y) \land H(y) ) \rightarrow x= y) \Big)\tag 7$
10)The happy dog is stronger than Hector and Achilles.

We have a single thing who is happy and a dog. Also, this thing is stronger than both Hector and Achilles. However, we do not know that Hector and Achilles are dogs.
$\exists x \Big( (D(x) \land H(x) ) \land \forall y ((D(y) \land H(y) ) \rightarrow x=y) \land (S(x,h) \land S(x,a))\Big)\tag{10}$

## 2 Answers

Let $D(x) = x$ is a dog, $H(x) = x$ is happy, $P(x) = x$ is a person, $L({x,y}) = x$ likes $y$, $S({x,y}) = x$ is stronger than $y$.

1) If Hector likes Achilles, then he likes a happy dog.

we cannot be sure if the happy dog is referring him liking achilles or any other dog.

Interestingly, in English that sentence is often used to imply "Achilles is a happy dog"! Yours is more like a logician's interpretation, so okay, but you need to specify what "$h$" and "$a$" mean.

2)A happy dog likes nothing.

$\forall x \Big( (D(x) \land H(x)) \rightarrow \lnot \forall y L(x,y)\Big)$

Wrong, as pointed out already. "he likes nothing" is the same as "he does not like $x$, for every $x$" or "there does not exist any $x$ that he likes".

3)Only Hector and Achilles are happy dogs.

However I am not sure about stating Hector and Achilles as happy dogs in the beginning. Some argues that we can only understand that they are dogs at first sight.

No; in English the sentence clearly says not only that they are dogs but that they are happy, because "happy" modifies "dogs". If the sentence was "Only Hector and Achilles can be happy dogs.", then it will not even imply that they are dogs.

6)Someone likes the happy dog.

I am confused about the happy dog part. Does this mean that for all things that are dogs, there is only one dog that is happy and this sentence is referring this happy dog? I thought this is the case and therefor I translated this sentence as there is exactly one happy dog who is read by at least one person.

Your confusion stems from the typical sloppiness in logic textbooks that do not clearly explain how definite references in natural language work. When we say "the book on the table", we do not ever mean that there is only one table in the whole world. Rather, it is merely in the current context that there is a unique table. Furthermore, that table may have more than one book, but the phrase is valid if we are already talking about one particular book.

This is the same issue we face in resolving pronoun references. In all cases we must resolve references before translating to logic. Your answer is hence acceptable but actually the real answer must involve the context. The easiest way to do so is to simply create a constant-symbol "$hd$" to refer to the happy dog in the current context, and then the sentence is translated to simply "$\exists x\ ( Person(x) \land Likes(x,hd) )$".

7)There is exactly one happy dog.

Correct.

10)The happy dog is stronger than Hector and Achilles.

However, we do not know that Hector and Achilles are dogs.

Yup, but no need to mention this.

1. Correct!

2. The $\neg \forall y L(x,y)$ should be $\forall y \neg L(x,y)$ or $\neg \exists y L(x,y)$

3. Use $h$ instead of $s$ throughout.

3.1 is correct

3.2 is not. 3.2 does not force Hector and Achilles to be happy.

For 3, you can do: $\forall x ((D(x) \land H(x)) \leftrightarrow (x= h \lor x=a))$

1. Correct!

2. Correct!

3. Correct!