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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.

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    I don't understand the downvote - this seems like a reasonable question. – 2017-02-11
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    You should think of $\to$ as merely a symbol which connects two statements. Abandon your intuition about $\to$, specifically the part which makes you think of $p\to q$ as $p$ causes $q$. The symbol $\to$ is a connective that works for any two statements, specifically $\varphi \to \neg \varphi$ is fine to write. It's not a contradiction due to the reasons you mention. Check out [this](http://math.stackexchange.com/questions/489562/how-to-make-an-introductory-class-in-set-theory-and-logic-exciting#comment1053583_489562) comment. – 2017-02-11
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    Thank you, though I didn't think p -> q meant p causes q. I thought p -> q meant that if p is true, then q will be true. – 2017-02-11
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    @ChrisYoung That's correct - but that means that if $p$ is false, then $p\implies q$ is immediately true. So if it's possible for $p$ to be false, it's possible for $p\implies q$ to be true, regardless of what $q$ is. – 2017-02-11

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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. $C\wedge\sim C$).

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    A key thing worth mentioning here is the principle of explosion that implies that *anything* follows from a falsehood. This includes even immediately contradictory statements. Less formally, once you've assumed something that's not true, there's no reason to expect that you won't be able to prove a contradiction. – 2017-02-11
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    Thank you. You're right about "valid" - I meant does it make sense? It still seems wrong that C => ~C is the same as ~C. Because C => ~C is stating a relation between C and ~C (and one that actually seems false) whereas ~C is only making an assertion about C which may or not be true. But I guess it only doesn't make sense to me. So if C = It is dry, then C => ~C means "it is dry if it is not dry", correct? Which can possibly be true? It just doesn't seem to make any sense when you supply actual sentences for the letters. But I guess that's what we lose when we formalize logic. – 2017-02-11
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    Maybe I should move this to philosophy SE since it is less about the rules of logic (I can see by the truth table a conditional is by definition contingent) than whether those rules make sense in this case. – 2017-02-11
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    @ChrisYoung Other way around; it means "if it is dry, then it is not dry." Basically, it's a way to indicate that "it is dry" can't be true, since if it were true, then it would be false. More to the point, remember that the [material implication](https://en.wikipedia.org/wiki/Material_implication_(rule_of_inference)) isn't expressing causality, or anything deep like that; it's just stating a relationship between truth values. If $C$ is no more true than $\sim C$ is, then $\sim C$ must be true. – 2017-02-11
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    @Noah Ok, I guess. Still "if it is dry, then it is not dry." seems to be demonstrably false too -- not because it's "deep" but because the fact that is dry means it is *not* *not* dry. But your explanation kinda makes sense I guess. Does that mean (C > ~C) v (~C > C) makes sense as a tautology? (If it is dry then it is not dry) OR (If it is not dry then it is dry) which both seem false, together make a sentence that is always true? – 2017-02-11
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    @ChrisYoung Yup, that's a tautology. It is equivalent to $\sim C\vee C$. But when you write "which both seem false", while they may *intuitively* each seem false that is not correct: they are each false, or true, *under certain conditions*, and the point is that the truth conditions of each clause "overlap" to cover all possibilities. This is also a context where real-world statements add some complexity: thinking of purely mathematical examples may be easier. One common kind of proof by contradiction consists of showing that $\sim P$ implies $P$, so we can conclude $P$. – 2017-02-11
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    @ChrisYoung that statement would translate to "if it is dry, it is not dry." If we know it isn't dry, that statement is vacuously true. If it is dry, the statement is false (it contradicts itself). The statements ~C and C => ~C are true exactly when the other is -- thus they are equivalent. Hope that helps! – 2017-02-11
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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/

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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 ($\ref{2}$).

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