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So, the definition of a valid argument form is that the truth of the conclusion is guaranteed via the truth of the premises. Soundness is often said to be a valid argument where the premises are true. But note that people say it is a sound argument, rather than an argument form. There is a distinction between

  1. If it is raining then it is cloudy. It is raining therefore it is cloudy.
  2. (p --> q), p , therefore q

The soundness of the first is going to be pretty much dependent on whether or not if it is actually cloudy and raining for it to be a sound argument. But what about the second? It is an argument form rather than a concrete argument. It seems odd to think that for an argument form to be sound it requires its premise set to be tautological. I'd be inclined to think in the case of an argument form it is sound if it is both valid and the premise set consistent. Or is an argument form simply not something that can be sound in the first place?

  • I haven't heard of this notion of soundness of an argument. It seems like it wouldn't be very mathematically interesting, for the reason that you said. – Trevor Wilson Mar 18 '13 at 07:37
  • Indeed, these concepts are only encountered in more philosophically-oriented texts (both the concept of "argument form" and the concept of an argument being "sound" if is valid and its hypotheses and conclusion are true). For example this SEP article has a detailed discussion of validity and soundness in this sense: http://plato.stanford.edu/entries/logic-informal/ . – Carl Mummert Mar 18 '13 at 12:20

2 Answers2

1

I am pretty sure that, as an argument form, an argument form cannot be sound. When we consider only the form, we do not have recourse to the actual truth or falsity of the premises but only to the logical structure of the argument. Perhaps this is a little pedantic, but even an argument form such as

p or not p

therefore p or not p

is not sound when considered as a form. As an actual argument, however, it will always be sound.

It's like someone asks you to give them a real number and you say "any real number will do". It's true that any real number will do but "any real number will do" is not itself a number, and you have not yet specified a value and haven't satisfied the request. In the same way, we must determine whether the atomic propositions (p, q, etc) are true of false (specify a value) and only then can we say that the premises are actually true (or false), even if we know they will always be true ("any truth value will do").

As for your example (1), it is of course valid. I apologise for being a pedant, again, but I'd like to point out that its soundness would require both that it is raining and that is is cloudy.

jim
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Yes, an argument form can qualify as sound, given that we have some way of defining all the connectives of the formulas and we can have tautological formulas. Soundness will depend on what the logical theory looks like. In classical logic, you can take any axiomization {a, ..., n} of propositional logic, and restate theorems {t, ..., z} using their proofs basically as an argument form as follows:

a, ..., j, therefore t

a, ..., k, therefore u

.

.

.

a, ..., l, therefore z

This might at first seem trivial. However, if you start to look at multi-valued logics, if it holds that formulas a, ..., n also hold in a given multivalued logic K, and K and classical logic has the same rules of inference used in a given classical logic proof, then K also has the same formulas as could get derived from formulas a, ..., n in classical logic, because of the soundness of argument forms. In other words, if we have a derviation in classical logic of theorem T using axioms A and B of classical logic and rules C and D, and A and B hold also hold in a multivalued logic M, as well as C and D ending up as valid rules, then via the soundness of the argument form A, B |- T, it follows that T holds in multivalued logic M.

As an example, we know that we can axiomitize classical propositional logic as follows

1: (p->(q->p))

2: (((p->(q->r))->((p->q)-(p->r)))

3: (($\lnot$q->$\lnot$p)->(p->q))

You can find proofs that show (p->p) a theorem which only use axioms 1 and 2. Thus, (p->(q->p)), (((p->(q->r))->((p->q)-(p->r))) therefore (p->p) is a sound argument form. Consequently, if we have an axiom system M which is not classical, but has 1 and 2 as axioms, (p->p) will also appear in M.