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VideCorSpoon

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Mon 4 May, 2009 08:04 am

@ali ant207,

In step 1, we have the proof set up with the first "inference." Because the proof is so difficult that regular inference and replacement rules will not really help us out here, we have to use the indirect proof to get the first step. The indirect proof goes by many names, such as the apagogical argument and the more commonly known reductio ad absurdum. What this particular proof enables us to do is derive a conclusion indirectly by deriving a contradiction from the negation or denial of the conclusion we have to reach which we can then infer the conclusion from. In so many words, we want to show that if there is a fault in the argument where we say the sky is blue and the sky is black, the whole argument is wrong and we can posit whatever we want. To execute an indirect proof, we have to follow these steps :

goapy

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Sun 24 May, 2009 11:47 pm

@VideCorSpoon,

VideCorSpoon wrote:

You mentioned in a previous post that using DeMorgan's would allow a shorter proof, if not for the right conjunct (~Q) being negated. I think DeMorgan's would still work used like this:

1. P-> Q

2. ~P v Q......Implication 1

3. ~P v ~~Q..Double Negation 2

4. ~(P & ~Q)..DeMorgan's 3

VideCorSpoon

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Tue 26 May, 2009 01:31 pm

@goapy,

I agree with your proof as far as line #2, utilizing the replacement rule However, this does not mean that you cannot attack the problem the way you suggest. You could suppose line 3 as another presumed assumption and indent again. However, you would have to derive a conditional or indirect proof before moving on in order to utilize it.

Another thing to note is that DeMorgans cannot derive from line 3 because we are still left with two negation symbols. Double negation is essentially the same thing as you said before.

goapy

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Tue 26 May, 2009 03:04 pm

@VideCorSpoon,

VideCorSpoon wrote:

... The problem stems from the fact that ... negation ... cannot be inferred as a monadic operator to a specific part of a line. So, the rules basically say that you can use the double negation, you just have to use that specific replacement rule in the context of the whole statement.

When you refer to 'the rules', to which rule set are you referring?

What you say may be the case for the rule set you're using. However, I'm familiar with the Copi, Hurley, Layman, and Bergmann rule sets - and those rule sets allow the rules of replacement, including double negation, to be applied to a whole line or any part of a line.

Code:

[I]Hurley, A Concise Introduction to Logic, 9th Ed, pp360;[/I]

The rules of replacement, on the other hand, are not rules of implication but rules of logical equivalence. Since, by the axiom of replacement, logically equivalent statement forms can always replace one another in a proof sequence, the rules of replacement can be applied either to a whole line or any part of a line.

[I]Bergmann, The Logic Book, 5th Ed, pp231;[/I]

Double Negation can be applied to any of the sentential components of a sentence, For instance, from:

G v (H & K)

Double Negation permits us to derive

G v ~~(H & K)

VideCorSpoon wrote:

Another thing to note is that DeMorgans cannot derive from line 3 because we are still left with two negation symbols. Double negation is essentially the same thing as you said before.It is not the case that I am not playing socceris the same as sayingI am playing soccer. In essence, line 3 is not really needed because it is logically redundant.

The Double Negation step was done precisely in order to setup for the use of DeMorgans, since DeMorgans requires both disjuncts be negated when applied in the manner as I did above.

DeMorgan (DeM) [from Bergmann]

~ (

~ P v ~ ~Q is a form of

~

Where the metalinguistic expression

VideCorSpoon

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Tue 26 May, 2009 06:29 pm

@goapy,

The rule set that I use is from Herrick's goapy wrote:

~ P v ~ ~Q is a form of

~Pv ~Q

Where the metalinguistic expressionQ(bold Q) stands for the wff '~Q'.

I understand where you are going with your point, but I just want to put down my point before I address it. For my example, I'm just using a base prefix "it is not the case that.." and "not." Is it the case that;

equivalent (a prime requirement in a replacement rule) to;

goapy

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Tue 26 May, 2009 11:28 pm

@VideCorSpoon,

Herrick's most concise definition of the DM rule appears in this box (Of those four formulations, Hurley/Begrmann allow only the first and second. So, Herrick's DM rule is significantly different from the vast majority of logic texts which allow only the first two formulations above.

But the above definition of DM is not Herrick's definitive all-inclusive rule. As Herrick explains below, his DM algorithm allows the replacements defined in the above box

Let's try his algorithm on the problem from this thread. That problem, again:

P -> Q / ~(P & ~Q)

1. P -> Q

2. ~P v Q [Impl 1]

'~P v Q' is a formula or subformula that has an ampersand or wedge as its main connective, so we may use Herrick's DM algorithm:

Given (line 2): ~P v Q

1. First we change the ampersand to a wedge or the wedge to an ampersand. In this case, we therefore change the wedge to an ampersand:

~P & Q

2. Next, we negate each side of the ampersand or wedge. Here we negate the ~P and also negate the Q:

~~P & ~Q

3. Finally, we negate the formula as a whole:

~(~~P & ~Q)

As Herrick points out at the bottom of page 188 and at the top of page 189, we can use Double Negation to 'cancel out the double negatives' in the result from step three:

After applying Double Negation to our result from step three, the result is

~(P & ~Q)

So, using precisely the same steps as Herrick used in the box above, formula #2 was derived from the formula #1 by applying the DM algorithm:

formula #1: ~P v Q

formula #2: ~(P & ~Q)

Therefore, following Herrick's algorithm and instructions exactly as illustrated in his example, the following proof for our problem meets Herrick's requirements:

P -> Q / ~(P & ~Q)

1. P -> Q

2. ~P v Q [Impl 1]

3. ~(~~P & ~Q) [DM 2]

4. ~(P & ~Q) [DNeg 3]

-----

I'm convinced that Herrick's Double Negation rule is exactly the same as Hurley/Bergmann, and that it can be applied to any of the sentential components of a sentence, just like Hurley/Bergmann. For example, 'P v Q' :. 'P v ~~Q, is a correct application of Double Negation using Herrick's Double Negation rule.

In the box below, Herrick uses his Double Negation rule on the sentential components of a sentence. So, 'P v Q' :. 'P v ~~Q, is a correct application of Double Negation under Herrick:

goapy

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Wed 27 May, 2009 02:23 pm

@VideCorSpoon,

Is there a more recent edition of Herrick's text than the version from which I quoted?There is another thread here:

http://www.philosophyforum.com/forum/philosophy-forums/philosophy-101/2961-needs-help-logic-homework.html

In which the first post is exactly one of Herrick's rules of DM:

~(~P v ~Q) :. (P & Q)

and needs no further proof under Herrick's system, yet nearly everyone disagrees about how DM should be applied.

Later in the thread, it is incorrectly stated that:

~(~P v ~Q), therefore ~P & ~Q

is an application of DM, but this is not even an equivalent form.

---------- Post added at 02:05 PM ---------- Previous post was at 01:23 PM ----------

VideCorSpoon;65268 wrote:

I disagree to some extent with the previous proof and the example proof. Here are the two proofs side by side. I think that there is to some extent a discrepancy and a procedural error here.

P-> Q

2. ~P v Q......Implication 1

3. ~P v ~~Q..Double Negation 2

4. ~(P & ~Q)..DeMorgan's 3

P -> Q / ~(P & ~Q)

1. P -> Q

2. ~P v Q [Impl 1]

3. ~(~~P & ~Q) [DM 2]

4. ~(P & ~Q) [DNeg 3]

Of course, the first proof was not meant to adhere to Herrick's rules, since I was unaware of them. The first proof does follow Hurley/Bergmann/Layman rules. If entered into Layman's online proof checker, it passes:

Power Of Logic Tutor -- not logged in

As for the second proof, I followed Herrick's DM algorithm and Double Negation rule exactly, so you'd have to be more precise with your description of the procedural error.

djronbxs

Reply
Fri 27 Nov, 2009 03:47 am

@VideCorSpoon,

Hi,I would like to have some help on how to know on what criteria an ASSUMPTION must be made in a proof ? ie in a proof that needs an ASSUMPTION to solve how should I tackle it ?

thanks

ron

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