Kenneth Rosen Edition 6th Exercise 1.3 Question 60,61(Page No. 50)

Question

I can solve these ques. with logical reasoning(apti.) but I need to know how to solve questions 60-d and/or 61-e. 

60 a.   ∀ x [P(x) $\rightarrow$ Q(x) ]

b.       ∃ x [R(x) ⋀ $not$ Q(x)]

c        ∃ x [R(x) ⋀ $not$ P(x)]

61 a.   ∀ x [P(x) $\rightarrow$  $not$ Q(x) ] 

  b       ∀ x [R(x) $\rightarrow$  $not$ S(x) ] 

c         ∀ x [  $not$Q(x) $\rightarrow$  S(x) ] 

d         ∀ x [P(x) $\rightarrow$  $not$ R(x) ] 

Answer 1

60

a.  $\forall x [P(x) \implies Q(x)]$

b.  $\exists x [R(x) \wedge \neg Q(x)]$

The first statement says if $P(x)$ is true then $Q(x)$ is true or if $Q(x)$ is not true then $P(x)$ is not true.
Second statement says for some $x$, $R(x)$ is true and $Q(x)$ is not true.
So, combining both we can say, for some $x$, $R(x)$ is true and $P(x)$ is not true.
c. $\exists x [R(x) \wedge \neg P(x)]$

61

a. $\forall x [P(x) \implies \neg Q(x)]$

b. $ \forall x [R(x) \implies  \neg S(x)]$

c. $\forall x [ \neg Q(x) \implies S(x)]$

Form a and c, we can say $\forall x [P(x) \implies S(x)]$. Now using b, we get

d  $\forall  x [P(x) \implies \neg R(x) ]$

Comments

Sir for 60 b .Some excuses are not satisfactory.

Can we write  ∃x ( P(x) $\Rightarrow$ ~Q(x) ).  ??

If not then why?

Answer 2

FOR 60 D.

P(A)->Q(A)[UNIVERSAL SPEC.] ----------1

R(A)⋀ NOT Q(A)=>R(A) IS TRUE----------2

                                 NOT Q(A) IS TRUE---------------3

FROM 1 AND 3 BY MODUS PONENS RULE WE CAN SAY THAT  

                                  notP(A) WILL BE TRUE---------4

NOW FROM 2 AND 4 WE CAN SAY THAT       R(A)⋀notP(A) WILL BE TRUE.

AND FROM EXISTENTIAL GENERALISATION WE CAN SAY ∃ x [R(x) ⋀ notnot P(x)]

61.

YES THE CONCLUSION ALSO FOLLOWS.....SEE ALL ARE UNIVERSAL SPECIFICATION SO NO PROBLEM WITH QUANTIFIERS..

THE PROPOSITION CAN BE SOLVED BY CP RULE IN WHICH( P(x) →→  notnot R(x) )

P(X)IS TAKEN TO BE TRUE THEN WE PROVE THAT notR(X) IS TRUE.

 

1.SO NOW P(X) IS TRUE.

     AND     P(x) →→  notnot Q(x)       IS TRUE (stmt a)

      =>notQ(X) IS TRUE----------

2. NOW notQ(X) IS TRUE.

AND    notnotQ(x) →→  S(x)    IS TRUE    (stmt c)

    =>S(X) IS TRUE

3. NOW S(X) IS TRUE.

   AND     R(x) →→  notnot S(x)  IS TRUE  (STMT B)

=>     notR(x) is true.

which we want 

hence the stmt follows