Rules of Inference (discrete mathematics)

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I don't understand how to deal with this question:

Show that the argument form with premises p1, p2,...,pn and conclusion q → r is valid if the argument form with premises p1, p2,...,pn, q, and conclusion r is valid.

Assuming r is valid. where will i put the r in the solution?

I tried:

1. p (premise)
2. q (premise)
3. p ^ q (simplification, conclusion of 1,2)
4. p ^ q -> r 
5. r = T

since r is valid, q must be valid based on line 4. So q -> r is T?

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Instead of a formal proof, you can think about this question purely in terms of the definitions of the concepts involved.

We know that the argument form with premises $p_1,...p_n, q$ and conclusion $r$ is valid. This means (by definition of validity) that it is impossible for all of $p_1,...p_n, q$ to be true and $r$ to be false all at the same time. So, if we assume that all of $p_1,...p_n$ are true, then it is impossible to have $q$ true and $r$ false at the same time as well. But by the truth-table of the $\rightarrow$, that means that it is impossible for $q \rightarrow r$ to be false, still under the assumption that all of $p_1,...p_n$ are true. Hence, by definition of validity, any argument form with premises $p_1,...p_n$ and conclusion $q \rightarrow r$ is valid.

If you insist on a formal proof, first of all please know that there are many different formal proof systems with many different rules sets. Also, we can only really sketch such a formal proof, since we are talking about some unknown number of premises. Anyway, here is a pretty generic formal proof.

OK, so we know that the argument form with premises $p_1,...p_n, q$ and conclusion $r$ is valid. That is, we can infer $r$ once we have $p_1,...p_n, q$, i.e. we have a formal proof to the effect of:

$1. p_1$ premise

$2. p_2$ premise

...

$n. p_n$ premise

$n+1. q$ premise

...

$n+m. r$ conclusion

Now, what we need to show is that we can get $q \rightarrow r$ once we have $p_1,...p_n$. OK, so:

$1. p_1$ premise

$2. p_2$ premise

...

$n. p_n$ premise

$n+1. q$ assumption

... (insert above proof of deriving $r$ from $p_1,...p_n, q$ here)

$n+m. r$

$n+m+1. \ q \rightarrow r \ \rightarrow $ Intro