Homework help: predicate logic

  • #1
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
honestrosewater
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What are some specific things you don't understand? Are you comfortable with everything in propositional (or sentential) logic?
 
  • #3
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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
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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
honestrosewater
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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
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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
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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
honestrosewater
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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. :grumpy:
 
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  • #9
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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. :tongue:
 
  • #10
Hurkyl
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Science Advisor
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For #3, can't you use this?

Ay Py
-----
Px

Or some sort of substitution rule?
 
  • #11
honestrosewater
Gold Member
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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.
 

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