# What exactly is a contradiction and how does it differ from falsity?

I apologize in advance for my lack of knowledge about the terminology of formal logic. I am only interested in informal logic to the extent that a practicing mathematician needs it to proceed. Despite years of experience in mathematics, I am finding myself confused about what a contradiction means. According to this site,

A contradiction is a conjunction of the form "A and not-A"... So, a contradiction is a compound claim, where you’re simultaneously asserting that a proposition is both true and false.

I doubt that this is mathematical definition though, as Wikipedia's article on contradiction defines that

a proposition is a contradiction if false can be derived from it, using the rules of the logic. It is a proposition that is unconditionally false

Two questions:

1. Main question: I'm confused as to the difference between a contradiction and a false statement. If I say that $$x\in S\wedge x\not\in S$$ then is this a contradiction or a false statement? There seems to be two ideas at play, one being a statement that is simply false like "The sky is red" versus something like $$P\wedge \neg P$$ where the $$P$$ can be any statement with a true/false value like a proposition or quantified predicate but regardless of whether $$P$$ is $$0$$ or $$1,$$ the value of $$P\wedge\neg P$$ is $$0$$ (false). In the former case, there is no varying in the underlying components whereas in the latter we compute a truth table to find that we always get $$0.$$ I am running into the issue of distinguishing between the two because this article on proof by contradiction uses the $$\bot$$ symbol and I don't know whether it is refering to a false statement or a logical contradiction, where by a false statement I mean something like "The sky is red" and by a contradiction I mean a statement like $$P\wedge\neg P$$ whose truth table has all $$0$$'s in the final column (I don't know if these are the right definitions for the terms).
2. Side question: Are all contradictions, that is those statements that evaluate to a truth table of all $$0$$'s in the final column, logically equivalent to a statement of the form $$P\wedge \neg P$$? A counterexample or proof would be appreciated.
• Prove that $P \wedge \neg P$ implies $\bot$. The two are equivalent. Is it clear now? Oct 6 '20 at 17:28
• @QiaochuYuan I'm afraid not. Is that in reference to the second question? What exactly is meant by $\bot$? Since $P\wedge \neg P$ implies anything by the principle of explosion, I guess it can prove anything but I don't really know what $\bot$ means. Oct 6 '20 at 17:33
• Favst Some statements can be false sometimes, but not always. $p\to q$ is true alot, but not always. $p\land q$ is false a lot, but not always. A contradiction is always false. The truth table of a contradiction ends with the rightmost column evaluating to FALSE in every row. Oct 6 '20 at 17:35
• But the truth table of a statement $p$ and it's negation $\lnot p$, when arrived at in a proof, define $p\land \lnot p$ to be a contradiction. Oct 6 '20 at 17:42
• Yes, that works. Because you're testing the conjunction $P\land \lnot P$... under each and every truth value assignment you list, the proposition is false. Oct 6 '20 at 17:52

Your understanding is correct. Put simply, a contradiction is a sentence that is always false. More precisely,

A statement is a contradiction iff it is false in all interpretations.

In propositional logic, interpretations are valuation functions which assign propositional variables a truth value, so a contradiction comes down to having 0's as the final column in all rows (= valuations) of the truth table.
In predicate logic, interpretations are structures consisting of a domain of discourse and an interpretation function defining a mapping from symbols to objects, functions and relations on it, so a contradiction is a statement which evaluates to false no matter the choice of objects and interpretation of the non-logical symbols.

Take the expression $$\exists x (x < 0)$$, for instance: This sentence is false in the structure of the natural numbers, but true when we evaluate it in the integers, or under some none-standard interpretation of the natural numbers where e.g. the symbol $$<$$ ist taken to mean "greater than". The statement is not valid (= true in all structures), but it is not contradictory (= false in all structures), either: While it may be coincidentally false in some particular structure/the situation we're currently interested in, it is logically possible for it to become true.
On the other hand, $$\exists x (x < 0) \land \neg \exists x (x < 0)$$ is true in neither of the above three structures structures; in fact, it fails to be true in any structure whatsoever: No matter which domain of objects we take and which interpretation we assign to the symbols $$<$$ and $$0$$, the form of the statement $$A \land \neg A$$ makes it inherently impossible to ever become true.

To pick up your example, "The sky is red" is only coincidentally false in the actual world because our earthly sky just so happens to be blue, but it is possible to imagine a universe in which the atmosphere is constituted differently and the sky is indeed red: The sentence false in the real world, but it is not contradictory. In symbols, the sentence can be formalized as $$p$$, and will have a truth table with both a true and a falsy column.
On the other hand, $$x \in S \land x \not \in S$$ is another statement of the form $$A \land \neg A$$, and thus a contradiction: It is false in all structures, and thus also in our real-world conception of sets in standard ZF set theory. Its truth table has only 0's, no matter which value the component statements take.

The symbol $$\bot$$ is used to refer to a contradiction. And indeed, any contradictory statement is logically equivalent to (and can be transformed into, using rules of inference) both $$A \land \neg A$$ and $$\bot$$: All contradictory statements have the same truth table with only 0's in the last column.

• I appreciate the answer. Regarding the last point, what would a proof of the equiavlence look like? amWhy gave an example in the comments, but I don't see how it generalizes. If the proof is too elaborate, then a reference to a text would work too. Oct 6 '20 at 18:20
• It depends on the proof system you wnat to use. With the standard equivalence laws as well as in natural deduction, the equivalence between $A \land \neg A$ and $\bot$ is one of the basic laws (often, $\bot$ is defined as shorthand for $A \land \neg A$). As another example, $\neg (A \lor \neg A)$ can be shown to be contradictory using the equivalence laws of De Morgan, double negation and commutativity: $\neg (A \lor \neg A) \equiv \neg A \land \neg \neg A \equiv \neg A \land A \equiv A \land \neg A \equiv \bot$. Oct 6 '20 at 18:23
• A general proof that all contradictory statements are logically equivalent to $A \land \neg A$ is trivial: A contradiction has all 0's in its truth tables by its very definition; $A \land \neg A$ has all 0's in its truth table, as is easily verified; and two statements are logically equivalent iff they have the same truth table (again, by definition). Oct 6 '20 at 19:30
• The sky was red in California recently Oct 7 '20 at 1:55
• @nick012000 Yes, just write the truth tables out. Both will have 0 in all rows and are therefore equivalent. Replacing a subexpression by a logically equivalent one will leave the end result with the same truth values; but replacing a subexpression by an inequivalent one (such as not (A and B) vs (A xor B)) doesn't necessarily change the truth values of the whole expression either. Two statements may have the same final column even though their compound statements differ. Oct 7 '20 at 4:12

The following is less concrete than lemontree's answer and amWhy's comments, which I think are more on-point. However, I do think the following is worth saying, so I'm putting it here.

The snappy version, as you suspect, is:

A contradiction is never true in any situation. A statement is called false if it fails in the particular situation (or class of situations) we care about - but a false statement may nonetheless hold in a different situation (whereas a contradiction cannot).

Below I'll describe the two main ways of making this precise.

## Semantic version

The "semantic" view of logic is that a logical system $$\mathcal{L}$$ is used to describe objects (or structures): basically, such an $$\mathcal{L}$$ consists of a class of sentences $$Sent_\mathcal{L}$$, a class of applicable structures $$Struc_\mathcal{L}$$, and a relation $$\models_\mathcal{L}$$ between applicable structures and sentences with $$\mathfrak{A}\models_\mathcal{L}\varphi$$ being interpreted as "the sentence $$\varphi$$ is true in the structure $$\mathfrak{A}$$."

A contradiction in the sense of $$\mathcal{L}$$, then, is a sentence which is not true in any structure: a $$\psi$$ such that for every $$\mathfrak{A}$$ we have $$\mathfrak{A}\not\models_\mathcal{L}\psi$$. By contrast, when we decide to focus on a particular structure $$\mathfrak{S}$$, we say that $$\varphi$$ is false iff $$\mathfrak{S}\not\models_\mathcal{L}\varphi$$.

## Syntactic version

We can also refrain from talking about structures entirely. The "syntactic" view of logic is that a logical system is used to manipulate sentences (without necessarily assigning them particular meanings). Basically, such an $$\mathcal{L}$$ consists of a class of sentences $$Sent_\mathcal{L}$$ and a relation $$\vdash_\mathcal{L}$$ between sets of sentences and individual sentences with $$\Gamma\vdash_\mathcal{L}\varphi$$ being interpreted as "the sentence $$\varphi$$ is deducible from the set of sentences $$\Gamma$$."

A contradiction in this framework is then a sentence from which we can deduce anything: $$\varphi$$ is a contradiction in the sense of $$\mathcal{L}$$ iff for all $$\psi$$ we have $$\{\varphi\}\vdash_\mathcal{L}\psi$$. By contrast, when we say that a sentence $$\varphi$$ is false, what we mean is that we have in mind some particular "background set of sentences" $$\Gamma$$ and $$\Gamma\cup\{\varphi\}$$ would let us deduce anything (think of this $$\Gamma$$ as our set of axioms).

## Connecting the two

It's worth noting that every semantic logic induces a syntactic logic: given a semantic logic $$\mathcal{L}=(Sent_\mathcal{L}, Struc_\mathcal{L},\models_\mathcal{L})$$ we get a syntactic logic $$\mathcal{L}'=(Sent_{\mathcal{L}'}, \vdash_{\mathcal{L}'})$$ defined as follows:

• $$Sent_{\mathcal{L}'}=Sent_\mathcal{L}$$, that is, we use the same sentences for both logics.

• We set $$\Gamma\vdash_{\mathcal{L}'}\varphi$$ iff whenever $$\mathfrak{A}\in Struc_\mathcal{L}$$ with $$\mathfrak{A}\models_\mathcal{L}\psi$$ for each $$\psi\in\Gamma$$, we have $$\mathfrak{A}\models_\mathcal{L}\varphi$$.

Note that this makes the two notions of "contradiction" line up: if $$\varphi$$ fails in every structure, then vacuously we have $$\{\varphi\}\vdash_{\mathcal{L}'}\psi$$ for every $$\psi$$.

There is also a way to go "syntax-to-semantics" which again makes the two notions of "contradiction" line up, but it's a bit less natural (basically we interpret "structure" as "set of sentences which doesn't deduce everything and is maximal with that property").

## A caveat

Actually, the above isn't entirely accurate: there are logical systems where sentences of the form "$$P\wedge\neg P$$" do not let you deduce everything (these are called "paraconsistent logics;" another relevant (hehe) term is "relevance logics"). This leads to a more nuanced notion of "contradiction" and its relatives. But that's a more advanced topic which I wouldn't approach before first understanding the classical picture.