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# Rules for Conditional Proofs

Mon 2 May, 2011 03:32 am
Certain rules confuse me in certain circumstances. Any clarification would be great.
1. In the DS rule, the formula is ex: PvQ, and if given ~P then Q. Well is it possible if it is ~PvQ, and given P, can that work to get Q as well?

2. Ex: of Simplying rule: (C & (AvB))-->D. Can that be simplified into (C)-->D?

I am not tuned into thinking this way and I find that my source is a horrible resource for information and am struggling so much on this. These questions are me reaching and hoping that it works but I doubt the do.

I'm trying to make a Conditional Proof for this:

1. ((AvB) & C)-->D
2. D-->(Ev~B)
3. B & (EvF) Therefore: C-->E

Thoughts, tips? This is supposed to be before incorporating "Indirect Proof."

Thanks for any help
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Oylok

2
Wed 4 May, 2011 01:24 am
@ericalynn012,
Hi,

It might be easier to translate symbolic logic into English for you if you told us a little about yourself. Not much though, just what your favourite TV show is, for example.

Also, what course of study are you doing? Philosophy? Math? (Please don't give out personal info like where you are going to school.)

I'm fairly new to this stuff, too, but I have a BA in Math...

ericalynn012 wrote:

1. In the DS rule, the formula is ex: PvQ, and if given ~P then Q. Well is it possible if it is ~PvQ, and given P, can that work to get Q as well?

Think of two hoodlums, Paula and Quinn, who are responsible all for the shoplifting in town. Every theft is the fault of at least one of them, although sometimes they work together. That is what "P or Q" means in Math: "P or Q or both." The guilty party in any crime is either Paula or Quinn (or possibly both).

Now suppose a theft occurs at Family Dollar, and we want to know who was responsible. We have that "P v Q" is true; it was Paula or Quinn (or both). Suppose we find out "~P"; in other words, it wasn't Paula. Then logic (specifically, the rule of disjunctive syllogism) tells us "Q"; in other words, it was Quinn.

Now you asked: ~PvQ, and given P, can that work to get Q as well?

What you seem to be asking here is, "if Paula was involved in a shoplifting, does that logically imply that Quinn was involved?"

The answer is no. Knowing Paula was involved allows us to say nothing about whether Quinn was involved. Maybe he was and maybe he wasn't. So in other words "P" does not give us "Q", nor does it give us "~Q".

Quote:
2. Ex: of Simplying rule: (C & (AvB))-->D. Can that be simplified into (C)-->D?

I'm going to say "no".

It's actually a very good question. After all, doesn't Rule #4 at this website on logical inference tell us "C & (whatever)" allows us to conclude "C" ?

Let's give meanings to all these letters, though ...

Suppose you own a dog and yard. Let's say "C" means that a cat is in the yard. "A" means that your dog is awake. "B" means you bother the dog and wake it up. "D" means your dog runs into the yard to chase the cat.

To me, "(C & (AvB))-->D" means "if a cat is in the yard, and if the dog's awake or you bother it, then it will run into the yard after the cat." And "C --> D" means "if a cat is in the yard, then the dog runs into the yard." Clearly you can't conclude the second statement from the first. After all, the dog won't chase unless it's awake or you wake it.

So it looks the answer to your question is "no, you can't." But how do we reconcile that with the old "simplification rule"? My guess would be that "simplification" just doesn't work inside an implies statement. (But a more expert opinion would be welcome here.)

Quote:
I'm trying to make a Conditional Proof for this:

1. ((AvB) & C)-->D
2. D-->(Ev~B)
3. B & (EvF) Therefore: C-->E

(See "rules of inference" at this site: http://www.barnzilla.ca/blog/?p=2152 )

4. C. (Conditional proof assumption.)
5. ((AvB) & C) --> (Ev~B). (Using "hypothetical syllogism" on 1 and 2.)
6. B. (Simplification on 3.)
7. A v B. ("Addition" on 6.)
8. (AvB) & C. (Conjunction on 7 and 4.)
9. (Ev~B). (Modus ponens on 5 and 8.)
10. ~(~B). (Double negation on 6.)
11. E. (Disjunctive syllogism on 9 and 10.)
12. C --> E. (Follows from lines 4-12; conditional proof.)

That's rather long, so you'll have to check if it works.
Oylok

0
Wed 4 May, 2011 01:50 am
@Oylok,
Oylok wrote:

12. C --> E. (Follows from lines 4-12; conditional proof.)

12. C --> E. (Follows from lines 4-11; conditional proof.)

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