Contradiction: false or meaningless?

[nomadreid]

There seems to be two concepts of what constitutes a "meaningful" proposition in logic: one is that you can build up the proposition via the syntactical rules. However, another approach is to say that a meaningful sentence is that it is assigned a set in a model (even if it's only the empty set). (That is, that there is a model in which it is either true or false.) By the first approach, a contradiction is clearly meaningful. In the second one, however, I see a problem: yes, it can be assigned the empty set, so at first glance one would say that it is meaningful. On the other hand, a theory with a contradiction, i.e., an inconsistent theory, has no model, and therefore there cannot be a model to contain the empty set to which the contradiction should be assigned. Hence, it would seem that the contradiction is meaningless. So, which is it? False under all interpretations, or meaningless?

 

[gopher_p]

Whether or not a formula constitutes a proposition depends on your definition of "proposition". I think most logicians would require that a formula have no free variables to be considered a proposition/sentence/statement (in general), in which case the formula would not define a nontrivial set in any universe of any model of any theory of the language.

A contradiction in a first-order language is a formula of the form $\Phi(x)=\phi(x)\land\lnot\phi(x)$ for some formula $\phi(x)$. If it is a proposition (has no free variables), then it is false in every consistent theory in the language; in other words, $\lnot\Phi$ is a member of every "full" theory. If it is a formula with free variables, then it defines "an" empty set in every model of every consistent theory in that language; i.e. $\{a\in M^n:\Phi(a)\}=\emptyset\subset M^n$ when $\Phi$ has $n$ free variables and $M$ is the universe of some model of a theory in the language.

Note that in the case where you have an inconsistent theory, there is no model in which to interpret meaning. So in a sense every proposition in the theory is meaningless.

Also note that our choice to exclude all formulas of the form $\phi(x)\land\lnot\phi(x)$ from all "meaningful" theories is part of what gives meaning to the symbols $\land$ and $\lnot$.

 

[nomadreid]

gopher_p, thanks for the excellent answer. So in summary my question required more context: a contradiction is false in every consistent theory and meaningless in an inconsistent one.

Although it is not crucial, I am a little puzzled by the double negation (actually triple, but the double I mean is the last two in the following):

" a formula [with] no free variables ...would not define a nontrivial set in any universe ..."

By "trivial set", I presume you mean the empty set.

Getting rid of the underlined double negative, I arrive at
"would define a trivial set"

To show what I do not understand, it sounds to me that if we have
<N,$\in$> $\models$ ∃x (0 < x < 5) (so no free variables), this appears to me to defining {1,2,3,4} in P(N), which is not trivial.

Alternatively, if I am reducing the negations incorrectly, and you mean that no formula (without free variables) would define a trivial set, then I am led astray by ∃x (x≠ x).

Please clarify where my mistake or misunderstanding is. Thanks.

 

[gopher_p]

gopher_p, thanks for the excellent answer. So in summary my question required more context: a contradiction is false in every consistent theory and meaningless in an inconsistent one.

Although it is not crucial, I am a little puzzled by the double negation (actually triple, but the double I mean is the last two in the following):

" a formula [with] no free variables ...would not define a nontrivial set in any universe ..."

By "trivial set", I presume you mean the empty set.

Getting rid of the underlined double negative, I arrive at
"would define a trivial set"

To show what I do not understand, it sounds to me that if we have
<N,$\in$> $\models$ ∃x (0 < x < 5) (so no free variables), this appears to me to defining {1,2,3,4} in P(N), which is not trivial.

Alternatively, if I am reducing the negations incorrectly, and you mean that no formula (without free variables) would define a trivial set, then I am led astray by ∃x (x≠ x).

Please clarify where my mistake or misunderstanding is. Thanks.

I'm not sure what your definition of definability is, but the one that I use is

Given a language $\mathcal{L}$, an $\mathcal{L}$-structure $\mathcal{M}$ with universe $M$, and $\mathcal{L}$-formula $\phi$ with free variables among $x_1,...,x_n$, we say that $\phi$ defines the set $R_\phi:=\{a\in M^n\ |\ \mathcal{M}\models\phi(a)\}$.

So I originally said "would not define a set" and then realized that it could technically define $\emptyset\subset M^0$. In my opinion, it doesn't provide any benefit to consider that formulas without free variables define sets. Sets in $M^0$ are uninteresting.

As far as your example goes (and I'm unfortunately going to need to get a bit pedantic here);

1. $\exists x( 0<x\land x<5)$ is a formula, call it $\phi$, in any language $\mathcal{L}$ containing the binary relation symbol $<$ and the constant symbols $0$ and $5$. It has no free variables, and is therefore a sentence/statement/proposition. In any $\mathcal{L}$-structure $\mathcal{M}$ either $\mathcal{M}\models\phi$ or $\mathcal{M}\models\lnot\phi$. In the structure $\mathcal{N}=<\mathbb{N},\mathcal{L}^\mathbb{N}>$ (i.e. the structure with universe the natural numbers and the symbols $<,0,5$ interpreted as one would expect), this is a true statement, and so $\mathcal{N}\models\phi$. Then the set defined by $\phi$ is $\{a\in \mathbb{N}^0\ |\ \mathcal{N}\models\phi\}=\mathbb{N}^0$.

$0<x\land x<5$ is a formula with one free variable $x$ which, in the structure $\mathcal{N}$, defines the set $\{1,2,3,4\}$.

2. Whatever you mean by $<N,\in>$ or $\models$, $<N,\in>\models\exists x( 0<x\land x<5)$ is not considered a formula in the sense that we're talking about and it does not define any sets in the sense that we're talking about formulas defining sets. It is a string of symbols used as shorthand by logicians to indicate that the structure $<N,\in>$ is a model of the the statement $\exists x( 0<x\land x<5)$, which roughly means that the proposed meaning of the symbols in $\exists x( 0<x\land x<5)$ makes that a "true" statement in the model. Note that $\exists x( 0<x\land x<5)$ is not shorthand for anything in this context; it is a fully expressed statement in the language $\mathcal{L}$ which is, in a certain sense, given semantic meaning by the logical axioms of our system and additional meaning when interpreted in the context of a given structure. I say this because, unlike previous math courses where the symbols were used solely to help communicate the subject, in this case some of the symbols (such as $\models$ and $\mathcal{L}$) are used to facilitate communication while others ($\exists,\forall,\lor,\land,\lnot,...$) are the core of the subject being studied.

Remark: Throughout, I have used "free" to mean "occurs in the formula and is not bound". It's likely that your text has defined what it means for a variable to be "bound" and then defines "free" to simply mean "not bound". There is a difference, and it does lead to some messiness which I prefer to avoid.

 

[nomadreid]

I'm unfortunately going to need to get a bit pedantic here

Actually, I prefer the pedantic approach. It was my sloppiness in the usage of "definable" that led to the questions which you very nicely cleared up, and I thank you for that slap on the wrist.