Argument Derivations

1. Apr 1, 2005

whozum

Principles of sound reasoning.

For the following argument, derive the conclusion:

$$B$$ & $$C$$
$$(A \leftrightarrow B) v (C\rightarrow D)$$
$$(A v E) \rightarrow H$$
$$(D$$ & $$C) \rightarrow I$$
$$(IvJ) \rightarrow H$$

Conclusion is H.

I know im supposed to make some provisional assumptions, but I really dont know the guidelines to picking the right ones. I have afew more if someoen can help.

2. Apr 1, 2005

phoenixthoth

B&C implies B. Case 1. A <-> B. Therefore, A. Then as (AvE)->H, and we have A, we then have AvE; therefore H. Case 2.C ->D. Well B&C implies C. Therefore D. So we have D&C; therefore I. As we have I, we have IvJ and since (IvJ)->H, we have H. In both cases (which are all the cases), we have H.

3. Apr 1, 2005

whozum

My problem here isnt deriving the argument, its choosing the correct provisional assumptions. Im pretty confident I can derive the argument, but theres some guidelines that your supposed to follow to choose the most useful ones, as opposed to guess and check.

4. Apr 1, 2005

whozum

I dont see the step here. Is 'therefore D' an assumption or a derivation?

5. Apr 1, 2005

phoenixthoth

In case two of the disjunction (line 2 of your tex), C ->D.

Either A<->B (case 1) or C->D (case 2). That is an assumption as it cannot be derived from B&C, which is also an assumption.

B&C implies C. This is a derivation not an assumption.

Therefore D. Where does this come from? We know that C->D is true by assumption. Then since C, we have D.

Now I'm suspecting the way you're supposed to use disjunction PvQ is to convert it tautologically to ~P->Q (or ~Q->P) and go from there. That would eliminate the cases business.

6. Apr 1, 2005

whozum

Sorry to be picky but you're just solving this and I dont know how to interpret what youve done in the general case, its kind of confusing trying to dismember all you've written.
All Im asking is what provisional assumptions should I make and why.

7. Apr 1, 2005

phoenixthoth

Assume things like
1. B can be derived from B&C
2. C can be derived from B&C
3. modus ponens: B can be derived from A & (A->B)
4. ~A can be derived from ~C & (A->C)
5. ~P->Q can be derived from PvQ.

Then, in addition, assume the first couple of lines in the problem.

8. Apr 1, 2005

whozum

Arent #1 and #2 not assumptions but derivations?
3. I think she wants as to assume a certain clause, then work from there. Then when you have p->q in the conclusion then usually assuming P is a good idea. do you have any tips on stuff like that?

9. Apr 1, 2005

phoenixthoth

You assume that certain derivations are valid.

I don't think she does because you don't need to assume any more than the given assumptions plus the automatic assumptions one makes.

If you want to prove Q, then you have to either have proved P->Q AND P
or
assumed P->Q and proved P
or
assumed P->Q and assumed P
etc.

Most likely, you will be PROVING P and assuming P->Q. That will derive Q.

10. Apr 1, 2005

whozum

Ok heres a proof we went through in class, which illustrates the point im trying to make:

Argument:
I will name these the premises, incase theres a vocab conflict here:

$$(B$$ & $$J ) \leftrightarrow (A \rightarrow C)$$
$$J \leftrightarrow (A v E)$$

This is the conclusion:
Prove (A&B) -> (CvD)

The proof is as goes:

1) $$(B$$ & $$J ) \leftrightarrow (A \rightarrow C)$$ || Premise
2) $$J \leftrightarrow (A v E)$$ || Premise
3) B & J || Provisional Assumption
4) B || 3, & E
5) J || 3, & E
6) AvE || 2,5 $$\leftrightarrow$$ E
7) A || 6 vE
8) C || 1,7 $$\rightarrow$$ E
9) CvD || 8,v I
10) A & B || 7,4 &I
11) A & B $$\rightarrow$$ (C$$\rightarrow$$D) || 10,9 $$\rightarrow$$ I
QED

Without the provisional assumption B&J, we couldnt have solved this, ofcourse there might be another assumption I could have made, but thats not my point. My point is how do I know to choose B&J?