Help with if-then sentences in Predicate Logic

Question

I need help to translate some English sentences to predicate logic. I am able to translate sentences like "If no philosophers are happy, then some philosopher is wise." but I have no idea how to translate when a sentence is like "1 is more clever than 2" or "1 reads 2"

For example, let $$P(x): =\text{“x is a philosopher”; }$$ $$H(x): =\text{ “x is happy” ;}$$ $$C(a,b): = \text{“a is more clever than b” ;}$$

$$R(a,b):=\text {“a reads b”.} $$

How would these sentence translate into predicate logic in this instance?

1. No happy philosopher reads every wise philosopher.

2. All wise philosophers are read by some philosopher.

3. If Socrates reads every philosopher, then he reads every wise philosopher.

4. Not every wise philosopher fails to read some happy philosopher.

5. If not every happy philosopher is read by some philosopher, then some happy philosophers aren’t read by any philosopher.

   I would like to get explained answers rather than direct ones. Thanks in advance.

Answer 1

To get you started, I've translated directly, the first statement.

"No happy philosopher reads every wise philosopher."

$$\lnot \exists x \Big(P(x)\land H(x) \land \forall y\big((P(y) \land W(y))\rightarrow R(x, y)\big)\Big)\tag 1$$

$$\forall x\Big(\big(P(x)\land W(x)\big)\rightarrow \exists y(P(y) \land R(y, x))\Big)\tag 2$$

You can also define constants that name a specific individual. For the third question, I'll use $s$ to name "Socrates".

"If Socrates reads every philosopher, then he reads every wise philosopher."

$$\forall x, \Big((P(x) \rightarrow R(s,x))\Big) \rightarrow \forall y\Big(\big((W(y) \land P(y))\rightarrow R(s, y)\big)\Big)\tag 3$$

$$\lnot \forall x\Big((P(x) \land W(x))\rightarrow \exists y(P(y) \land H(y) \land \lnot R(x, y))\Big)\tag 4$$

Answer 2

For question 5:$ \lnot \forall x \Big(P(x)\land H(x) \rightarrow \exists y(P(y) \land R(x,y))\big) \rightarrow \Big(\exists x (P(x)\land H(x)) \rightarrow (\forall y P(y) \rightarrow \lnot R(x,y))\Big) \tag 5$

Answer 3

If $P(x)$ means that $x$ is a philosopher, than negating the same predication means $x$ is not a philosopher, i.e. $\lnot P(x)$ means "x is not a philosopher".

If you want to work with one philosopher, you use the existential quantifier. For a relation that holds for an arbitrary case or all cases, you use the universal quantifier. The negation of an existential quantifier is universal, and vice versa. The rule though, is that you distribute the negation across the quantifiers to the statements (when proving theorems using the contrapositive, etc).

For every element of the domain, the predication holds such that:

$\forall x(...x...)$

For every element of the domain, there is one element such that the predication holds:

$\exists x(...x...)$

So when you start symbolozing relations, look for the signs that the property or relation is for any, for all, for one, or for none.

Here is an example: "All philosphers dislike Kant"

The key is to notice how many are doing what, and what the relationship is. The "all" makes the quantifier for philosophers easy, we'll use $P$ as the predication for philosopher and $P(x)$ for "x is a philosopher." Since Kant is a proper noun, we could use a single capital letter, but realize this is a different order logic! We'll add a lower case variable instead. Let $y$ be Kant. The relation between Kant and philosophers is that they all dislike Kant. So the relation is between all philosophers and Kant. The relation to Kant is also a "no," or negative relation so we'll need to negate something. Now there was only one Kant (thankfully, amiright?) so we use the existential for any predications of $y$. We'll set $L$ for "likes."

We can then say:

$ \forall x \exists y \ [(Px \uparrow Ky) \implies \lnot Lxy]$

We can use $K$ to stand for Kant, which simplifies the statement, otherwise you would need to refine the predication of $y$ to be:

$\exists y (Py \uparrow Ky)$

For, "There is a philosopher, and that philosopher is Kant."

The standard "if-then" conditionals hinge on the truth table of the conditional operator. If the consequent is true, the antecedent can be true or false. If the antecedent is true, then the consequent must be true for the conditional operator to be true.

"If you are a philosopher, then you hate Kant. All philosophers hate Kant."

$\forall x \exists y \ [Px \implies (\lnot Lxy \uparrow Ky) ]$

When you move to more advanced predicate calculus you'll want to take care in how often you combine the universal and existential quantifiers for single propositions like theorems. You'll also want to be aware of which variables are existential or universal.

Now we'll moved to ordered relations. A relation that is reflexive uses an ordered pair of variables sharing the same predication. For example, only clever philosophers like Kant. Using the same notation as before, but now introducing $C$ for "clever" and $Cxy$ for "x is more clever than y" we can symbolize this as:

$\forall (x)(z) \exists (y) \ [(Px \uparrow Pz ) \implies \lnot Lxy] \downarrow [(Pz \uparrow Czx) \implies Lzy]$

This symbolizes the statement:

"There are two types of philosophers, those who don't like Kant or those who are more clever and like Kant."

Hope this helps!

Here is a link to a two page pdf about quantifiers hosted at UC-Berkeley.