PREDICATE LOGIC

Question

Recall that a predicate logic statement is contingent if its truth value depends on the choice of the universe and on the interpretations of the predicate symbol $S$ and the constant symbol $b$ involved.

Consider the following predicate logic statements in which $b$, $x$ and $y$ are elements of the universe $U$.

1. $\forall x\left(S\left(x, b\right)\rightarrow \exists y S \left(x, y\right)\right)$
2. $\forall x \exists y S \left(x, y\right)\rightarrow \exists y \forall x S \left(x, y\right)$
3. $\forall x \left(\neg S \left(x, x\right) \leftrightarrow S \left(b, x\right)\right)$

Which of the following best describes the predicate logic statements?

Statement I - Statement II - Statement III

1. Always true - Always false - Contingent.
2. Always true - Contingent - Always false.
3. Always true - Contingent - Contingent.
4. Contingent - Always true - Always false.

Answer 1

First of all...Excellent Question.

Answer : Option B.

Now, Let's solve it.
Note that in Propositional logic, Domain is, by default, considered Non-empty. So, Do not consider Empty domain since it is not mentioned.

Starting with Statement $II$..

                                                                       Statement $II$

Statement $II$ : $\forall x \exists y S(x,y) \rightarrow \exists y \forall x S(x,y)$

Interpretation of $\forall x \exists y S(x,y)$ : Have you ever heard Raabta Song which goes like this---- "Kehte Hain Khuda Ne Iss Jahan Mein Sabhi Ke Liye Kisi Na Kisi Ko Hai Banaya Har Kisi Ke Liye" ... Yeah..It's the exact interpretation of $\forall x \exists y S(x,y)$.. For every element $x$, there is some element $y$($y$ could be different for different $x$ but for each $x$ there must someone), such that Property(predicate) $S(x,y)$ is True. 

Interpretation of $\exists y \forall x S(x,y)$ : Have you watched "Mahabharata" ? ..If you have, then remember Draupdi...One wife for All five pandavas.. Yes, It's the exact interpretation of $\exists y \forall x S(x,y)$.. One for All. 

Now, Coming to Our Statement $II$.. $\forall x \exists y S(x,y) \rightarrow \exists y \forall x S(x,y)$.. You can now see that It is Neither always true, Nor always false. It will be True when you take the case of Draupdi and it will be false when take the case of Raabta.

Hence, Statement $II$ is Contingent.

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                                                       Statement $I$

  Statement $I$ : $∀x(S(x, b) → ∃yS(x, y))$

This is Always True. Why?

This statement will be True iff for all the elements in the domain, $(S(x, b) → ∃yS(x, y))$ is True. And $(S(x, b) → ∃yS(x, y))$ is an implication sentence, hence, it is always True except when $S(x, b)$ is True and $∃yS(x, y)$ is False. So, If even for at least One element $x1$ in the Domain, $(S(x1, b) → ∃yS(x1, y))$ becomes False  i.e. $S(x1, b)$ becomes True and $∃yS(x1, y)$ becomes False then Statement $I$ will become False. But it's not possible. Because, for $x1$, if $S(x1, b)$ becomes True then $∃yS(x1, y)$ can never be false because there will exist $y = b$ for which $∃yS(x1, y)$ is True.

So, Statement $I$ is Always True.

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                                                                  Statement $III$

 Statement $III$ : ∀x(¬S(x, x) ↔ S(b, x))..

Just for simplicity, Let me change the given statement into its equivalent statement i.e.

$∀x(S(x, x) \oplus S(b, x))$

This expression is Always False. Because when $x$ will be $b$ then for $b$ this statement will always be False. So, For at least One element ($b$) in the domain, $(S(x, x) \oplus S(b, x))$  will always be False. 

Hence, Statement $III$ is Always False.

Hence, Option B is Correct answer.

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NOTE: If Empty domain is also allowed then Statement $I$ will be Always True and the statements $II$ and $III$ will become Contingent.  

Answer 2

B?????????

Comments

How to do this sir??

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For 3rd,

Suppose U={b}

Then ~S(b,b) <- > S(b,b) -------- This is certainly going to be false.

Now Suppose U={a,b}

and a is such that

(~S(a,a) <-> S(b,a)) ^ (~S(b,b) <-> S(b,b)) ------- This is going to be false.

This is true for all universe because b is always going to be element of U(given in the question).

So iii^rd should be always false.

Am I correct?

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@Sunit, Yes, you are correct. I miss read it.

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how second one is contingent