Homework help: predicate logic

In summary, the problem is that the student is not comfortable with predicate logic and is having trouble getting started. Specific things the student does not understand include finite vs. infinite quantifiers, the difference between universal and particular, and how to exchange quantifiers. The student is looking for help with de Morgan's laws, specific existential quantifiers, and how to eliminate the y from a particular equation. The student has been successful in getting one problem solved, but is struggling with the others.
  • #1
ashleemorgan65
1
0
I am taking a logic class and we are getting into Predicate Logic and i have no idea how to do it can someone help me?
 
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  • #2
What are some specific things you don't understand? Are you comfortable with everything in propositional (or sentential) logic?
 
  • #3
I'm also studying predicate logic, and I can't say I'm hugely comfortable with propositional logic.

Here's a problem I'm working on right now:

V(universal quantifier)xFx |- VxGx--->Vx(Fx & Gx)

Having trouble getting started. Due in two hours. I know that I should hypothesize...

Here's basically where I got stuck:

1. V(universal quantifier)xFx |- VxGx--->Vx(Fx & Gx) A
2. | VxGx H
3. | Vx(Gx v Fx) 1, 2 vI
4. VxGx ----> Vx(Gx v Fx) 2,3--->I

From there I'm lost. De Morgan's doesn't get me exactly what I need.

Eh...looks like I need to get out of the propositional logic box.
 
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  • #4
Hell, all these are hard. The next ones are:

3. -](makeshift particular quantifier)x]yLxy |- Vx-Lxx

Quantifier exchange...but how do I get rid of the y?

4. |- ]xFx v Vx-Fx

Theorom...
 
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  • #5
lazycritic said:
V(universal quantifier)xFx |- VxGx--->Vx(Fx & Gx)
You can use Universal Instantiation, Conditional Proof, Conjunction Introduction (P, Q l- (P & Q)), and Universal Generalization.
3. -](makeshift particular quantifier)x]yLxy |- Vx-Lxx

Quantifier exchange...but how do I get rid of the y?

4. |- -]xFx v Vx-Fx
Is "-" negation or part of the quantifiers? "~" is negation, A and E are quantifiers.
3) ~(ExEy(Lxy)) l- Ax~(Lxx) ?
4) l- ~(Ex(Fx) v Ax~(Fx)) ?
 
  • #6
According to your definitions, those problems look like:

3. ~Ex EyLxy |-Ax ~Lxx (close to the same thing)

4.|- ExFx v Ax ~Fx (looks like my post had a typo - no negation on the first existential quantifier)

Thanks, btw. Pretty sure I got the first one. I have:

2. Fa AE
3. | Ga
4. | Ga & Fa
5. Ga ---> Ga & Fa
6. Ga ---> Fa & Ga
7. AxGx--->Ax(Fx & Gx)

Don't worry too much about it...need help on these others.
 
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  • #7
Anyhow, thanks for the help. Might've gotten the last one, but I'm heading off to class now. Serves me right for procrastinating. :rolleyes:
 
  • #8
lazycritic said:
According to your definitions, those problems look like:

3. ~Ex EyLxy |-Ax ~Lxx (close to the same thing)
Sorry, I'm not sure how to get rid of the y either.
4.|- ExFx v Ax ~Fx (looks like my post had a typo - no negation on the first existential quantifier)
Double negation. What is ~~Ax(~Fx)? Or ~~Ex(Fx)?

2. Fa AE
3. | Ga
4. | Ga & Fa
5. Ga ---> Ga & Fa
6. Ga ---> Fa & Ga
7. AxGx--->Ax(Fx & Gx)
Looks good.

Edit: Ah, I spent too long trying to find a stupid rule for 3.
 
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  • #9
I did manage to get #4 right before I left for class. It's just:

4.|- ExFx v Ax ~Fx

1. | ~ExFx
2. | Ax ~Fx 1, QE
3. ~ExFx ---> Ax ~Fx 1-2 -->I
4. ~~ExFx v Ax ~Fx 3, MI
5. ExFx v Ax ~Fx

That double negation might've been what you were hinting at. I was having trouble figuring out how Quantum Exchange worked exactly, heh. Oh well. I got another homework assignment...might end up posting it too. :-p
 
  • #10
For #3, can't you use this?

Ay Py
-----
Px

Or some sort of substitution rule?
 
  • #11
Hurkyl said:
For #3, can't you use this?

Ay Py
-----
Px

Or some sort of substitution rule?
3. ~Ex EyLxy |-Ax ~Lxx
If the negation applies only to Ex, I don't know what to do- it doesn't really even make sense to me. So I'll assume it applies to ExEy.
I only know some predicate logic, and I'm least comfortable with quantifier inference rules, but I've read that you can't instantiate a negated quantifier (which makes sense); So you at least know you need to move the negation to Lxy beforehand: AxAy~(Lxy). Unless otherwise noted, x and y aren't necessarily distinct, so AxAy~(Lxy) implies Ax~(Lxx), but I don't know how to derive it. If there isn't a restriction on UI such that
1. AxAy~(Lxy)
2. ~Luu (u is an arbitrary constant) [1, UI]
is invalid, the proof is a cinch. If that is invalid, I don't know what to do.
 

Related to Homework help: predicate logic

1. What is predicate logic?

Predicate logic is a formal system of logic that deals with the relationships between propositions using predicates and quantifiers. It allows for precise and rigorous reasoning about complex statements and arguments.

2. How is predicate logic used in homework help?

Predicate logic is commonly used in mathematics and computer science to solve problems and prove theorems. In homework help, it can be used to analyze and evaluate arguments, identify logical fallacies, and construct logical proofs.

3. What are some common symbols used in predicate logic?

Some common symbols used in predicate logic include quantifiers (∀ for "for all" and ∃ for "there exists"), logical operators (¬ for "not", ∧ for "and", ∨ for "or"), and variables (x, y, z). These symbols help to express complex logical relationships in a concise and precise manner.

4. How can I improve my understanding of predicate logic?

To improve your understanding of predicate logic, it is important to practice solving problems and constructing proofs. You can also read textbooks or watch online lectures to learn about different techniques and strategies for using predicate logic.

5. Are there any common mistakes to avoid in predicate logic?

One common mistake in predicate logic is confusing the order of quantifiers. For example, the statement "for every x, there exists a y" is not the same as "there exists a y for every x." It is important to carefully consider the placement and scope of quantifiers when constructing logical statements.

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