Discrete Math implications by rules of inference

[r0bHadz]

Homework Statement

p→(q→r)
¬q →¬p
p
-----------------------
∴r

Homework Equations

The Attempt at a Solution

My book gives the following solution:

(1) p - premise
(2) ¬q→¬p premise
(3) q, (1) and (2) and rule of detachment,
(4) p and q, law of conjuctive addition
.
.
.

Can anyone explain to me why you can use p on step 4?
Since in step 3, you are using step 1 and step 2, and p comes from step 1

Does the p not get "used" up? I don't understand why you're able to use it again.

 

[andrewkirk]

No the p does not get used up. Since it is a given assumption rather than a hypothesis used to open a conditional proof, it is valid throughout the proof. The only case where statements get 'used up' is where they are made in a conditional proof, in which case they are not valid outside the conditional proof. Since there are no conditional proofs used in what you wrote above, that does not happen here.

 

[r0bHadz]

No the p does not get used up. Since it is a given assumption rather than a hypothesis used to open a conditional proof, it is valid throughout the proof. The only case where statements get 'used up' is where they are made in a conditional proof, in which case they are not valid outside the conditional proof. Since there are no conditional proofs used in what you wrote above, that does not happen here.

Hmm great explanation.

Am I right when i say, so because P is the premise, I can use that whenever I want because its never going to not be the premise?

Similarly, because I deducted q in step 3, I can always use q just like I can always use p?

 

[andrewkirk]

Hmm great explanation.

Am I right when i say, so because P is the premise, I can use that whenever I want because its never going to not be the premise?

Similarly, because I deducted q in step 3, I can always use q just like I can always use p?

Yes. Both of those are valid.