Solving Propositional Logic and Quantifier Expressions

[needhelp83]

Not sure where to post this subject, so if it is in the wrong location please forgive.

1. Restore the parentheses to these abbreviated propositional forms?

Q \wedge \backsim S \vee \backsim ( \backsim P \wedge Q )

I got this, but am not sure if it is correct.

[Q \wedge (\backsim S)] \vee ( P \wedge \backsim Q )

2. Translate the following with quantifiers.
a. Some isosceles triangle is a right triangle. (all triangles)

This is what I believe it is. (\exists x)(x \ is \ isosceles \ triangle \ \wedge \ x \ is \ a \ right \ triangle) Is there anyway to reduce this translation?

b. Between any reall number and any larger real number, there is a rational number (Real numbers)

I have no idea how to write this. Any ideas?

3. Which of the following are true for the universe of all real numbers
(a) (\forall x)(\exists y)(x \leq y)

I said true, because for all x's, there exists a y to fit this equation

(b) (\forall x)(\exists!y)(x=y^2)

what does (\exists!y) even mean?

Lastly,
4. Let the function a(x) be an open sentence with variable x.
Prove that (\exists!x)A(x) \Rightarrow (\exists x)A(x)

How to do this?

Any help would be greatly appreciated. Thanks!

 

[konthelion]

1. Use De Morgan's Law.
2. Let A= "all triangles" Let P(x) = "x is an isoceles triangle" , Q(x) = "x is a right triangle"
Then, there exists an x which is an element of A such that? Continue from here.
3-4. \exists! xP(x) is a unique existential quantifier, meaning P(x) is true for one and only x.

 

[needhelp83]

1. Looking at Demorgan's law I determined it would be [Q \wedge (\backsim S)] \vee ( P \vee \backsim Q ). I looked at my solutions manual and I saw the answer to be [Q \wedge (\backsim S)] \vee \backsim (P \wedge Q ) I am not sure how the solutions manual derives to this answer. Could this be a typo?

2. There exists an x triangle which has to be an isosceles and right triangles. Still unsure how to write other.

3a.
(\forall x)(\exists y)(x \leq y)

Looking at this, it would be true. For all real number x exists a real number y where x is greater than or equal to y.

3b. (\forall x)(\exists!y)(x=y^2)

Looking at this one, it would be false. For all reall number x there is a unique y to satisfy x = y^2. There are at least two different cases: 1 and 0.

4. Prove (\exists!x)A(x) \Rightarrow (\exists x)A(x)

If there exists a unique x such that A(x) implies that there exists an x for A(x)

Prove: If (\exists!x)A(x) is true, then truth set for A(x) only has one element. This implies that (\exists x)A(x) is not empty and true<br /> <br /> I think I am now a little more clear on the subject. Does this seem better?

 

[konthelion]

I'm a little sleepy, so I'll look at only 1 and 2.

1. Ok, good. You've used DeMorgan's. Most likely, it's not a typo and there's some trick so that you can further simplify it into that form. Try to use the associative laws and distributive laws. I haven't tried doing the problem, but mess around and see what you get.

2. A= "all triangles" Let P(x) = "x is an isosceles triangle" , Q(x) = "x is a right triangle"
Ok, you can say this like so.
There exists an x in the universe of A, such that x is an isosceles triangle and x is a right triangle. Basically, you wrote the same thing except you need to express it in form of a predicate i.e. P(x). Simply, convert this into logical notation.

 

[konthelion]

3a. That's correct. Archimedes principle/property can be applied here as well.
3b. I don't understand your cases. I believe you only need to provide a counterexample to prove that it is false. Suppose x=4. Then x=y^2 has more than one value for y, what are those values.

4. Since !x implies that there exists one and only unique x. Then you can say that for some y, x \neq y
So, \exists(!x)P(x) \equiv \exists x(Px) \wedge ~\exists x,y(blank)

What is "blank"?