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honestrosewater

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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.

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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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...

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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You can use Universal Instantiation, Conditional Proof, Conjunction Introduction (P, Q l- (P & Q)), and Universal Generalization.lazycritic said:V(universal quantifier)xFx |- VxGx--->Vx(Fx & Gx)

Is "-" negation or part of the quantifiers? "~" is negation, A and E are quantifiers.3. -](makeshift particular quantifier)x]yLxy |- Vx-Lxx

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

4. |- -]xFx v Vx-Fx

3) ~(ExEy(Lxy)) l- Ax~(Lxx) ?

4) l- ~(Ex(Fx) v Ax~(Fx)) ?

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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.

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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- #8

honestrosewater

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Sorry, I'm not sure how to get rid of the y either.lazycritic said:According to your definitions, those problems look like:

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

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

Looks good.2. Fa AE

3. | Ga

4. | Ga & Fa

5. Ga ---> Ga & Fa

6. Ga ---> Fa & Ga

7. AxGx--->Ax(Fx & Gx)

Edit: Ah, I spent too long trying to find a stupid rule for 3. :grumpy:

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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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For #3, can't you use this?

Ay Py

-----

Px

Or some sort of substitution rule?

Ay Py

-----

Px

Or some sort of substitution rule?

- #11

honestrosewater

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3. ~Ex EyLxy |-Ax ~LxxHurkyl said:For #3, can't you use this?

Ay Py

-----

Px

Or some sort of substitution rule?

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

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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