Give a formal proof of the sentencepfrom the single premise ¬¬pusing only Modus Ponens and the standard axiom schemata. Warning: This is surprisingly difficult. Though it takes no more than about ten steps, the proof is non-obvious.

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- November 14th 2010, 07:01 AMnoviceBy Modus Ponens only
Give a formal proof of the sentence

*p*from the single premise ¬¬*p*using only Modus Ponens and the standard axiom schemata. Warning: This is surprisingly difficult. Though it takes no more than about ten steps, the proof is non-obvious. - November 14th 2010, 01:52 PMnovice
The post above is solved.

- November 15th 2010, 05:33 AMemakarov
I am interested in how you solved it. When a derivation involves the third axiom (not B -> not A) -> (A -> B), which is responsible for double-negation elimination and classical logic, I know of only one systematic way of doing this. One can construct a derivation using the Deduction Theorem, which is usually much simpler. Since the proof of the theorem is constructive, one can then emulate it in this concrete instance to get a complete derivation.

- November 15th 2010, 09:23 AMnovice
I got the question from internet—I don't remember where, but it's quite interesting. I was taken by the warning, but it turned out to be a bluff.

As I understood it, only in the execution of a new line, Modus Ponens is required. The negation of an atomic formula is not an issue, particularly when used in the assumptions, so I did it as follows:

Given Argument: ~~P |- P

Proof:

1. P.................................Hypotheses

2. ~ ~ P...........................Premise

3. P.................................Hypotheses

4....... ~ P........................Assumption

5...... ~~P -->~~P............Hypotheses

6..............~~P.................Assumption

7. ......~~P.......................2,5, Modus ponens

Explanations:

Line 6 satisfies the hypotheses on line 5—cross out line 6. Line 5 becomes a premise.

Line 4 and 7 satisfy the hypotheses on line 3—cross out lines 4,5, and 7. Line 3 becomes a premise.

Line 3 satisfies the hypotheses on line 1. Lines 1 and 2 make up a valid argument.

If you don't like the contradiction for an inference, you can try this:

1. P.................................Hypotheses

2. ~ ~ P...........................Premise

3. P-->~~P.......................Hypotheses

4. P..................................Assumption

5...... ~~P -->~~P............Hypotheses

6..............~~P.................Assumption

7. ......~~P.......................2,4, Modus ponens

Line 6 satisfies line 5.

7 satisfies 3

4 satisfies 1 - November 15th 2010, 10:25 AMemakarov
I am now more confused than before. What deductive system are you using: natural deduction, Hilbert system, or something else? In my usage, hypothesis, assumption and premise are all synonyms.

- November 15th 2010, 04:09 PMnovice
Sorry, brother. I don't have the slightest idea.

Will this help now?

1.Show P

..|------------------------------|

2|. ~ ~ P...........................|

3|. Show P........................|

..|.|-------------------------|....|

4|.|~ P.........................|....|

5|.|Show ~~P -->~~P.|....|

..|.|...|--------|...............|....|

6|.|.. |.~~P..|...............|....|

..|.|...|--------|...............|....|

7|.|.~~P......................|....|

..|.|------------------------|....|

..|------------------------------| - November 16th 2010, 09:20 AMemakarov
This looks like Fitch variant of Natural Deduction, but I still can't recognize it. For one, one does not write what one needs to prove (like "Show P") in natural deduction derivations; all formulas are either assumptions or have already been proven.

Bit that's OK if the solution works for you. - November 16th 2010, 03:48 PMnovice
I have learned the method you mentioned, but with only one given it will not take you any where.

This method I have just begun learning is more powerful.

A free book for you:

Philosophy 110 on-line text