Deduce P from Sigma { ~S V R, R -> P, S }

[moo5003]

Homework Statement

Sigma = { ~S V R, R -> P, S }

Give a deduction with the last component P.

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 a lot 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 isn't 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.

 

[AsianSensationK]

I like it. It should work.