Is $C \to \neg C$ a "proper statement"?

Question

EDIT: Thanks to whoever corrected "valid" to "proper statement". The original title used "valid" which means something else entirely, so it's appreciated.

First time poster, so apologies if anything amiss or terminology wrong. I searched but didn't seem to find anything addressing this question directly.

By ~C I mean negation of C. By -> I mean the conditional connective

Currently studying logic, and learning about truth tables. It stumped me that the following sentence is contingent because it seems to imply C is the case if C is not the case -- which seems it will always be false. If anything, it seemed a contradiction.

However looking at the definition of conditional I can see it's defined to be true unless a true hypothesis leads to a false conclusion. And that made more sense if we had, for instance, C -> ~A, since I guess if C was false we can't imply anything about ~A being true.

The problem is it seems we can imply something in this case (namely that if C is true then ~C is false) which led we to wonder if it even makes sense for the conclusion to be the negation of the hypothesis in a conditional. Or even more broadly, can the conclusion reference the negation of the hypothesis in any way, since it seems that C -> (~C v B) or something would suffer the same problem. But C -> C seems to make sense (and trivially so) and C -> (C v B) seems to make sense too, and less trivially.

In trying to make sense of it, I concocted the following examples:

C = It is dry C -> ~C

It is dry if it is not dry --> seems obvious contradiction by logic alone, independent of the truth value of C, but by the rules of formal logic it's contingent?

while with

A = It is raining C = It is dry C -> ~A

It makes sense that if it is not dry, we can't say anything about whether or not it's raining.

Answer 1

It is perfectly meaningful: the sentence $C\implies\sim C$ is equivalent to "$\sim C$," that is, it is true iff $C$ is false. Indeed, any sentence built out of propositional atoms and Boolean operations is meaningful as long as it is grammatically correct (something like "$\wedge\wedge C$" is obviously not meaningful).

Incidentally, you shouldn't use the term "valid" here - "valid" has a technical meaning in logic. A valid sentence is one which is true in every possible truth assignment - so e.g. $C\vee \sim C$ is valid since, regardless of whether $C$ is true or false, $C\vee\sim C$ is true. The sentence $C\implies \sim C$ is not valid in this sense: it is only true if $C$ is false. It is, however, a satisfiable sentence (= true in some truth assignment); there are unsatisfiable, yet perfectly meaningful, sentences (e.g. $C\wedge\sim C$).

Answer 2

If the conditional and negation are taken to be classical the formula is indeed satisfiable. But your contrary intuition has an honorable tradition leading back at least to Aristotle. Connexive Logic tries to give it formal expression, so it might be worth having a look at that system:

https://plato.stanford.edu/entries/logic-connexive/

Answer 3

When reading the statement in plain English; you have made a common mistake of adding additional information about the claim. Consider an English sentence: If it is not raining, then it is raining. Often interpreted in common speech to mean ($\ref{1}$):

"It is true that it is not raining; and it is true that if it is not raining, then it is raining."$\label{1}\tag{1}$

In Propositional Logic, what you have done symbolically is equivalent to writing ($\ref{2}$)

$\lnot A\land (\lnot A\rightarrow A)\label{2}\tag{2}$

,where A represents the fact that it is raining.

Which; according to its truth table, is always false. As you noted earlier; a contradiction arises when we AND not A with A ($\ref{2}$).

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To direct you toward something that follows the English language more closely; consider looking up the definition of Modus Ponens. It allows us to infer the truth value of the consequent by affirming the truth value of the conditional and antecedent. https://en.wikipedia.org/wiki/Modus_ponens