Proving Validity of Argument Form with Contrapositive and Modus Ponens

[illidari]

Homework Statement

a) p
b) ~q -> ~p
c) q -> r
d) (r V t)-> s

proves e) s


Use a demonstration to show the following argument form is valid

Homework Equations

The Attempt at a Solution

1) ~q-> ~p proves p -> q contrapositive

2) p-> q , p proves q modus ponens

3) q->r , q proves r modus ponens


This is where I get iffy and confused, I think this is allowed but want to verify

4) (r V t) -> s proves ~s -> ~(rVt) implication to disjuction

5) ~s -> ~(rVt) proves ~s->~r^~t DeMorgans

6) ~s -> ~r ^~t proves sV~r^~t by implication to disjunction

7) sV~r ^ ~t proves sV~r specialization

8) sV~r, r proves s



does this look legit if anyone has taken a discrete math course that has a sheet with this stuff :(?

 

[Mark44]

It looks to me like all this is given (renumbered so as to lessen any possible confusion with too many letters):
1. p
2. ~q -> ~p
3. q -> r
4. r V t -> s

and you need to show that if the hypotheses (1 through 4) are true, then s is true.

As you noticed, 2 is equivalent to p -> q, so the four hypotheses are the same if we replace the second one with p -> q.

So, we have p being true.
p being true and p -> q being true imply that q is true.
q being true and q -> r being true imply that r is true.
r being true implies that r V t is true. (t is completely irrelevant.)
r V t being true and (r V t) -> s being true imply that s is true.

 

[illidari]

It looks to me like all this is given (renumbered so as to lessen any possible confusion with too many letters):
1. p
2. ~q -> ~p
3. q -> r
4. r V t -> s

and you need to show that if the hypotheses (1 through 4) are true, then s is true.

As you noticed, 2 is equivalent to p -> q, so the four hypotheses are the same if we replace the second one with p -> q.

So, we have p being true.
p being true and p -> q being true imply that q is true.
q being true and q -> r being true imply that r is true.
r being true implies that r V t is true. (t is completely irrelevant.)
r V t being true and (r V t) -> s being true imply that s is true.

That was the part that bothers me. I think i have to somehow get rVt separated before I can get that claim.

rVt -> s and something else would have to prove rVt somehow :/

Then I can use rVt to move on. I'm suppose to write the rule name followed by which step I pulled the p and q from.

 

[Mute]

That was the part that bothers me. I think i have to somehow get rVt separated before I can get that claim.

rVt -> s and something else would have to prove rVt somehow :/

Then I can use rVt to move on. I'm suppose to write the rule name followed by which step I pulled the p and q from.

As Mark44 said, r being true implies r V t is true. It doesn't matter at all what t is. It could even be ~r. The point is that because you already know r is true (it is given), r v anything is automatically true by virtue of the fact that r is true.

The name of this rule is simply "addition". See http://en.wikipedia.org/wiki/Addition_(logic ) (or apparently disjunctive introduction if you want to be fancy).

 

[illidari]

Okay figured out my confusion , it was labeled generalization on my paper :)

I didn't realize p proves pVq meant that I could pick any q regardless if I had it proven or not.

Made this way harder than I should of >.<

Thanks !