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nomadreid

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- #1

nomadreid

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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

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nomadreid

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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

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,[itex]\in[/itex]> [itex]\models[/itex] ∃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.

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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 ....wouldnotdefine anontrivial 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,[itex]\in[/itex]> [itex]\models[/itex] ∃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##

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 gonna 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

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.

- #5

nomadreid

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I'm unfortunately gonna 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.

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