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# Seriously, I need help with this proof

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Aristophanes Posted: Thu, Jun 28 2012 11:05 AM

A -> C

/ (A^B) -> C

I would think that it has to start with a conditional exchange, but have no idea where to start.  There is a lot of reference to logic on the mises forums, so how about someone on the forums show me that they understand it.

I need help with this one too...

A -> ~A

/ ~A

I thought I was getting good at proofing, but not if there is only one premise.

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Smiling Dave replied on Thu, Jun 28 2012 11:35 AM

This may help a bit:

1. A and B implies A. A implies C. therefore A and B implies C.

2.  A imples not A [given].

not A implies not A [tautology]

Therefore A or not A implies not A. [Some law about or]

But A or not A is always the case [Law of excluded middle]

Therefore always the case that not A

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Aristophanes replied on Thu, Jun 28 2012 11:43 AM

Thanks for responding.

1. A and B implies A. A implies C. therefore A and B implies C.

What allows me to infer from conditionals?  I am not allowed to use conditional proofsod indirect proofs for these two.

i got two from this.  thanks!

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Balrogo replied on Thu, Jun 28 2012 11:57 AM

If "A and B" is true then "A" is true and "B" is true (or whatever jargon you want to use) is called simplification or conjunctive elimination

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Smiling Dave replied on Thu, Jun 28 2012 12:13 PM

A implies B

B implies C

Therefore A implies C.

You don't have that one in your repetoire?

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Aristophanes replied on Thu, Jun 28 2012 12:18 PM

If "A and B" is true then "A" is true and "B" is true (or whatever jargon you want to use) is called simplification or conjunctive elimination

We do not know that "A and B" is true.  We have A implying C and that is it.  If (A^B) was a premise I wouldn't have needed help.

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Aristophanes replied on Thu, Jun 28 2012 12:21 PM

A implies B

B implies C

Therefore A implies C.

You don't have that one in your repetoire?

Well, you are trying to use Hypothetical Syllogism with two premises.  My problem has only one premise and A implies B is not one of them.  We do not know that B implies C.  We are trying to prove that B doesn't interfere with A's implication of C.

The point of the exercises is to manipulate the inference rules and replacement rules.

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Smiling Dave replied on Thu, Jun 28 2012 12:29 PM

You have A and B implies A, because that's a tautology.

You have A implies C

Thus A and B implies C. [by Hypothetical Syllogism with two premises.]

Of course, I don't know what the rules of the game are in your case, so it may not be what you are looking for.

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Aristophanes replied on Thu, Jun 28 2012 12:32 PM

You have A and B implies A, because that's a tautology.

It is not a premise.  We do not have "A and B implies A" becase we do not have "A and B."

Of course, I don't know what the rules of the game are in your case, so it may not be what you are looking for.

I am allowed to use any inference rule or any replacement rule.  Excluded are ONLY CP and IP (it is the next part of the HW).

Have you ever done a proof?

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Smiling Dave replied on Thu, Jun 28 2012 12:36 PM

Have you ever done a proof?

It's been a while. I should think tautologies can be snuck in there.

Anyway, that's all I've got. Am interested in the answer, when you find it.

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Friedmanite replied on Thu, Jun 28 2012 12:54 PM

Unfortunately, the precise way of proving logical propositions like these aren't standardized.  For your first, you can infer A from "A and B" from a boolean elimination rule called "conjuction elimination".

For your second one, I suppose you can do a proof by contradiction.  Suppose A, then since A implies not A, we have A and not A, a contradiction.  QED

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Aristophanes replied on Thu, Jun 28 2012 1:10 PM

Unfortunately, the precise way of proving logical propositions like these aren't standardized.

I think it is a good thing that there are many ways to do them.  I am just not confident in the moves I make all the time.

For your first, you can infer A from "A and B" from a boolean elimination rule called "conjuction elimination".

Can I infer things from the conculsion?  I thought I could merely rearrange them so that I can arrive at the conclusion on different terms.

For your second one, I suppose you can do a proof by contradiction.  Suppose A, then since A implies not A, we have A and not A, a contradiction.  QED

I cannot use Conditional or Indirect proofs (assumptions) on these particular problems.

For the second one I got, simply enough...

1.) A -> ~A   Pr

/ ~A

2.) ~A v ~A     (Conditional Exchange; line 1)

3.) ~A    (Duplication; line 2)

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Friedmanite replied on Thu, Jun 28 2012 1:14 PM

I think you should consult your instructor on these.

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Friedmanite replied on Thu, Jun 28 2012 1:14 PM

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Aristophanes replied on Thu, Jun 28 2012 1:17 PM

My instructor sucks.  He is in his fifth (!!) year of the graduate program...it is only a three year graduate program in philosophy at my University...

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Balrogo replied on Thu, Jun 28 2012 1:29 PM

I mean its true

A->C

(A^B)/A

(A^B)->C

But without inferring the logical equivalent of the statement of (A^B), which is what that conjunction rule does, idk how you'd even get to the form of the conclusion. You got me man

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Friedmanite replied on Thu, Jun 28 2012 1:33 PM

I don't know what to tell you since the terminology you're using is specific to your class.   And your last proof is not logically coherent at all.  Look at it carefully.

Here's a proof by cases.

Case 1: A

By the premise, A implies not A.  Therefore, not A.

Case 2: not A

Then clearly, not A.

QED

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Aristophanes replied on Thu, Jun 28 2012 1:40 PM

I know it is true, you cannot proof invalid arguments. Here is a proof from later in the HW.  It is supposed to be harder than the first one...

1.) (A -> B) -> ~(C -> D)   Pr

2.) ~(A v F)   Pr

/ ~(D v F)

3.) ~A ^ ~F   (DeMorgan's Law; line 2)

4.) ~F   (Simplification; line 3) - Here half of my conclusion has been proven.

5.) (A -> B) -> (~C ^ ~D)   (Conditional Exchange; line 1)

6.) (~A v B) -> (~C ^ ~D)   (Conditional Exchange; line 5)

7.) ~A   (Simplification; line 3)

8.) (~C ^ ~D)   (Dysjunctive Syllogism; lines 6 & 7)

9.) ~D   (Simplification; line 8)

10.) ~D ^ ~F   (Conjunction; lines 9 & 4)

11.) ~(D v F)   (DeMorgan's Law; line 10) -  Done.

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Aristophanes replied on Thu, Jun 28 2012 1:46 PM

And your last proof is not logically coherent at all.  Look at it carefully.

It makes perfect sense. I added links so you can see why it makes sense.

If I am trying to prove ~A, then all I have to do is replace the conditional with a disjunct to present a choice of ~A or ~A.  The implication is replaced by a new choice in the dysjunct.

1.) A -> ~A   Pr

/ ~A

2.) ~A v ~A     (Conditional Exchange; line 1)

3.) ~A    (Duplication; line 2)

Smiling Dave, Duplication is tautology. So, you were right about it fitting in there.

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shackleford replied on Thu, Jun 28 2012 4:15 PM

For the second one, your conditional statement is true. Make a truth table.

It only makes sense when A is False and ~A is True. Otherwise, you have contradictions and a case when the conditional is False.

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