# Do we have false statement in inconsistent systems?

This is a rather soft question.

My understanding:

Suppose we have $$x \in ℝ$$ and $$x^2 = -1$$ [in the normal interpretation].

Then the statement "there exists $$r \in ℝ$$ such that $$r^2 = -1$$" is true.

This is because $$x \in ℝ$$ and $$x^2 = -1$$ form a contradiction, and under contradictory settings any statement follows. That is, in an inconsistent system, any statement is true. [UPDATE: this should be "any statement can be proved" as pointed out in the following answers/comments]

My question:

So, is that still valid to say that the negation of a true statement in an inconsistent system is false? If yes, then we would have any statement in an inconsistent system is simultaneously true and false. [UPDATE: this implication is actually wrong and has been corrected in the following answers/comments]

Or do we rather leave false to be undefined in inconsistent systems? (since I think the definition of false is to some extent redundant in such systems)

Motivation:

I am thinking about what does it actually mean when we say some statement is true.

In a vacuous implication, we say that the premise is false. However, for example, when we are using proof by contradiction to test if a statement is false, we actually treat the statement as if it is a true statement until we hit a contradiction, and then conclude that the statement is false, under the given settings. In other words, a statement is not necessary to be false if we don’t expect consistent system in the first place.

• I am not sure if this is true or not, but here are some of my thoughts for some dialogue. If you start off with a false statement, anything conclusions drawn from that will be false as well relative to a consistent system. However, the negation of a true statement in an inconsistent system would be true w.r.t the inconsistent system, but false w.r.t a consistent system. In your case, it is like accepting two truths at once, e.g. when proving $\sqrt{n}$ irrational by contradiction if $n$ is not a perfect square, we get that $n$ is both an integer and not an integer. Jul 24 '20 at 6:10
• A system must be consistent to have any merit at all, otherwise it can derive every statement (and of course also the negation of every statement). Thus we must assume consistency although we cannot guarantee it. "True" and "false" ony become meaningful if we have an interpretation. A statement can be true with respect to an interpretation and false with respect to another one. In this case, it can neither be proven nor disproven. Jul 24 '20 at 6:26
• In a formal system (in the sense that you have axioms and rules for deriving new statements from the axioms) there are, strictly speaking, no true or false statements, only provable and unprovable ones. In a structure (say in $\mathbb R$, supposing it "physically exists") every statement is either true or false. One question is whether a formal system admits a model, i.e. a structure where provable statements are true, and if the formal system (theory) is inconsistent then it doesn't. Jul 24 '20 at 6:32
• Statements being true in very intepretation are provable and a system that can disprove it, is inconsistent and cannot be used. Analogue with statements false in every interpretation. They can be disproved and a system that can prove it is inconsistent. Jul 24 '20 at 6:33
• @user8268 Could you please recommend me some books on the topics included in your comment? I haven't studied them, but I’m quite interested in them. Thanks Jul 24 '20 at 6:50

There are different (equivalent) definitions of consistency.

Basically, an inconsistent system is a system that proves a sentence $$\varphi$$ and its negation $$¬ \varphi$$.

If so, due to the fact that the negation of a True sentence is False, and vice versa, an inconsistent system is a system that proves True sentences as well as False ones.

Thus,

YES, we have false statement in inconsistent systems.

Regarding your example, we assume that we know facts about real numbers (i.e. mathematical objects whose collection is named with $$\mathbb R$$), where for simplicity I'll equate a "mathematical fact" with the content expressed by a mathematical theorem.

It is a theorem that, for every real number $$r : r^2 \ge 0$$.

This means that if we can prove that, for some real $$x$$, we have $$x^2=-1$$, this fact contradicts the above theorem.

This amounts to having found an inconsistency in the system we have used to prove it.

Does it mean that every statement in an inconsistent system is simultaneously true and false?

If we agree that there are mathematical objects called (real) numbers and there are objective facts regarding them that we can "discover" through proofs in a suitable system describing them, we accept the "classical" concept of Truth and thus we cannot have statements that are both True and False.

Thus, if we have an inconsistent theory of real numbers, i.e. a system that proves both a statement $$\varphi$$ and its negation $$\lnot \varphi$$, we have to conclude that the system is a wrong description of the reals and we have to fix it (as happened already in the past).

References:

• So, in the last part, do you mean we don't accept the classical concept of truth (bivalence) in inconsistent systems? [btw thanks for the references :) ] Jul 24 '20 at 8:47
• @J-A-S - No, on the contrary. If we accept it, there are no statement that are both True and False. What we have is that an inconsistent system proves both $s$ and $\lnot s$ and thus proves True statements as well as False ones. Jul 24 '20 at 8:50
• @J-A-S - Yes: if $s$ is True, then $\lnot s$ is False and vice-versa. Thus, if a theory proves both, it proves True as well as False statements. Regarding the second point, compare Intuitionism with "classical" mathematics. Jul 24 '20 at 9:40
• According to Intuitionism "The truth of a mathematical statement can only be conceived via a mental construction that proves it to be true." If so, no proof, no truth-value. This is not the "mainstream" approach (called Platonism) according to which Fermat's Las Th have a definite truth value that we have only recently discovered. Jul 24 '20 at 9:42
• [Sorry please let me reword] Thank you for your reply, may I ask for further clarification as follows. Do you mean that in inconsistent system we can prove both 𝑠 and ¬𝑠, while each of 𝑠 and ¬𝑠 still possesses only one truth value? If so, I am a little bit confused about the subtlety that, when we proved ¬𝑠, then it should mean that ¬𝑠 is true. Then does it imply that 𝑠 is false? If so, we would have 𝑠 now being both True and False, contradicting the bivalence. Jul 24 '20 at 9:44

First off, "There exist $$x \in \mathbb{R}$$ such that $$x^2 = -1$$" is by itself not at all contradictory. It is just not true in the actual world by our usual understanding of the symbols $$\mathbb{R}, -x$$ etc, in which there just so happen to be no negative squares. A contradiction only arises if the theory additionally proves that there are no negative squares, in which case the theory proves both the statement and its negation. This is what I will assume in the following.

Remember that statements aren't just true or false by themselves: Truth is defined relative to interpretations. So what exactly is it that you're asking? In which structures would you like the statements to be false?

Are there any theorems that are false in all models of the inconsistent theory?

In a consistent theory, the answer would be "no, trivially", because the models of a theory are defined as those structures in which all theorems hold, i.e. in which no statement of the theory is false.
But an inconsistent theory has no model: There is no structure in which a contradiction is true. So the answer to this question is: Yes, vacuously, because there are no models to begin with, so in particular there are none in which there are not any statements of the theory that are false in it.

Are there any theorems that are false in any conceivable structure whatsoever?

In classical logic, with the principle of explosion, an inconsistent theory proves everything. This means in particular that it proves $$\phi$$ and $$\neg \phi$$ for any statement $$\phi$$. But although both may be provable, $$\phi$$ and $$\neg \phi$$ can never be simultaneously true under a given interpretation. So in any conceivable structure, for all the infinitely many sentences $$\phi$$, either $$\phi$$ is true but $$\neg \phi$$ false in that structure or vice versa, whereas both of them are theorems. So here the answer is: Yes, there are infinitely many such structures in which infinitely many statements of the theory are false.

In the context of theories, truth is often understood as truth in the standard model with the "intended interpretation" for the non-logical symbols: By saying "$$s(0) + s(0) = s(s(0))$$ is true" we mean that it is true in the structure of the natural numbers with the successor function and addition defined as usual.
But again: Since an inconsistent theory doesn't have any models, it doesn't have a standard model either. So the question

Are there any theorems that are false in the standard model of the inconsistent theory?

Again, for every of the infinitely many provable pairs of statements $$\phi, \neg \phi$$, one of them must be false under each interpretation, such as the real world. So the answer here is again yes: An inconsistent theory proves statements that are false in the real world, namely those whose negation is true in the real world.
This is a crucial point to understand in symbolic logic: Truth exists only relative to interpretations, and the real world/standard model with the intended meaning of the symbols is just one of them. We very well can also have non-standard interpretations in which we, say, take the symbol "$$\_^2$$" to mean "square root", that yield different truth values for the same sentences. When asking about truth, you have to specify which interpretation you are talking about.