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Let A, B and C be propositions. Define ARG(A, B, C) as the following argument:

- A.
- B.
- Therefore, C.

My goal is to create a formula whose truth value **is equivalent** to **"ARG(A, B, C) is valid"**. In other words, I am looking for a formula that yields **true** if and only if the argument is **valid**.

# Attempt #1

ARG(A, B, C) is valid if and only if (A ∧ B) → C

Unfortunately, this attempt does not work. Proof: the following argument is **invalid** (source: third example on wikipedia here), but (A ∧ B) → C yields **true** (since A is false):

- All men are immortal.
- Socrates is a man.
- Therefore, Socrates is mortal.

# Attempt #2

ARG(A, B, C) is valid if and only if A ∧ B ∧ ((A ∧ B) → C)

Unfortuntately, this attempt also does not work, because although it works for the example above, this attempt yields **false** for any argument that has a false premise (and it is known that there are valid arguments with false premises).

I tried to create other attempts but I'm stuck.

**Is this impossible? If yes, why (can you give a proof)? If it's possible, what would be the formula?**

1Are you not confusing validity with truth here? – Mozibur Ullah – 2018-08-30T01:28:41.310

@MoziburUllah Honestly, no. Or so I think, at least. Why did you think that? – Pedro A – 2018-08-30T01:29:36.107

@MoziburUllah I've added a clarification in the beginning of the question (not sure if related or not to your thoughts though). – Pedro A – 2018-08-30T01:32:54.513

There are many types of logic. What is valid in Aristotelian logic is not the same as valid in the Mathematical logic. It looks like you are more into Mathematical logic. If you are trying to find a way to prove validity in every logic system with your program that would be interesting indeed. I also think you are confusing TRUTH with the concept of VALIDITY. Truth be truth table does not always yield TRUE in the real world. – Logikal – 2018-08-30T02:10:59.377

Because of attempt one: which basically has the form 'p is true iff q is true'. – Mozibur Ullah – 2018-08-30T02:11:20.667

Which is only of value when propositions p & q are essentially the same proposition but written in different ways. – Mozibur Ullah – 2018-08-30T02:12:10.617

Ditto for your second attempt. – Mozibur Ullah – 2018-08-30T02:14:08.230

1Validity (and structure of formula) does not depend on truth of premises or conclusions, Since there is no dependence, for any formula we can find a counterexample. – rus9384 – 2018-08-30T02:35:45.357

Obviously, you have to work in the meta-language:

Valid[ARG(A,B,C)]iffTaut[(A ∧ B) → C]. – Mauro ALLEGRANZA – 2018-08-30T06:05:02.557@rus9384 Interesting. That feels right, but I'm not entirely convinced yet. Could you expand that into an answer, giving some more details for this "proof" of yours? That would be great, thanks :) – Pedro A – 2018-08-30T11:16:46.727

1@MauroALLEGRANZA Thanks!! This kind of thing is what I'm looking for. But, I'm not sure I follow: if

Valis the set of all valuations, surely for any given proposition P, I can manually define one valuation that yields 0 for P and then P would not be a tautology. Then I'd conclude that there are no tautologies, implying that all arguments are invalid. What am I missing? – Pedro A – 2018-08-30T11:21:16.853@MauroALLEGRANZA Ah, so not all propositions, just atomic ones. Indeed you've said that before, sorry. But I don't know what is an atomic proposition. Can you put all your comments together in an answer and explain what are atomic propositions? Thanks :) – Pedro A – 2018-08-30T12:52:19.863

I have the feeling this is a technical question that I don't understand. But if you're trying to assess the truth of an argument with logic you're into a looser. An argument can be an agregious fallacy and yet still true. My father tells me that the scar on my lip was caused by a specific incident. Is this true? Well it's argument from authority.. and yet.. true it is. – Richard – 2018-09-01T01:59:56.447