General

Re: Seriously, I need help with this proof

shackleford — Thu, 28 Jun 2012 22:15:34 GMT

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.

Re: Seriously, I need help with this proof

Aristophanes — Thu, 28 Jun 2012 19:46:47 GMT

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.

Re: Seriously, I need help with this proof

Aristophanes — Thu, 28 Jun 2012 19:40:19 GMT

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.

Re: Seriously, I need help with this proof

Friedmanite — Thu, 28 Jun 2012 19:33:41 GMT

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

Re: Seriously, I need help with this proof

Balrogo — Thu, 28 Jun 2012 19:29:06 GMT

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

Re: Seriously, I need help with this proof

Aristophanes — Thu, 28 Jun 2012 19:17:26 GMT

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...

Re: Seriously, I need help with this proof

Friedmanite — Thu, 28 Jun 2012 19:14:57 GMT

It's best to ask your instructor.

Re: Seriously, I need help with this proof

Friedmanite — Thu, 28 Jun 2012 19:14:16 GMT

I think you should consult your instructor on these.

Re: Seriously, I need help with this proof

Aristophanes — Thu, 28 Jun 2012 19:10:14 GMT

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)

Re: Seriously, I need help with this proof

Friedmanite — Thu, 28 Jun 2012 18:54:03 GMT

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

Re: Seriously, I need help with this proof

Smiling Dave — Thu, 28 Jun 2012 18:36:53 GMT

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.

Re: Seriously, I need help with this proof

Aristophanes — Thu, 28 Jun 2012 18:32:13 GMT

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?

Re: Seriously, I need help with this proof

Smiling Dave — Thu, 28 Jun 2012 18:29:59 GMT

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.

Re: Seriously, I need help with this proof

Aristophanes — Thu, 28 Jun 2012 18:21:53 GMT

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.

Re: Seriously, I need help with this proof

Aristophanes — Thu, 28 Jun 2012 18:18:35 GMT

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.