# Logic Deduction

1. Oct 3, 2007

### moo5003

1. The problem statement, all variables and given/known data
Sigma = { ~S V R, R -> P, S }

Give a deduction with the last component P.

3. The attempt at a solution

I came up with what I think is an answer, but I was a little unsure on what can go in a deduction so I would really like someone to re-check and tell me if i'm right and if not what rules I broke.

<R->P, S, (~SVR)->R, ~SVR, R, P>

R->P is in Sigma
S is in Sigma
(~SVR)->R is a tautology since R always implies R and S is listed before making it always True.
~SVR is in sigma
R by modus ponus.
P by modus ponus.

Is this correct?

Another version I have after doing subsequent problems is this:

First: I showed that Sigma Implies P.

Thus the following is a tautology
(~SVR)->(R->P)->S->P = B

then the deduction looks like:
<B,~SVR,(R->P)->S->P,R->P,S->P,S,P>

1st Term is a tautology
2nd Term in sigma
3rd Term modus ponus
4th Term in sigma
5th Term modus ponus
6th Term in sigma
7th Term modus ponus

This seems alot more straightforward since I was a little unsure about the (~SVR)->R since its not really a tautology unless you assume ~S is always false ie: S is always true.

The 1st term B isnt that hard to prove as a tautology since if Sigma implies P then the wff's in conjuction form a tautology when if wff's then P and then you can use exportation on it to distribute the conjuctions to if/thens for each term.

Last edited: Oct 3, 2007
2. Oct 3, 2007

### AsianSensationK

I like it. It should work.

Last edited: Oct 3, 2007