Mathematical Logic: Inconsistent models

I would like to find out if there are any examples of an inconsistent model in first order logic. I understand that to be an inconsistent model, a formula is consistent in that model and the negation of the formula is also in that model. So then, the question will be looking for a particular structure, an assignment function and a formula that gives rise to this. But I am not exactly sure how to go about doing this..

Likewise, I would also like to know if there are any models which a formula and the negation of the formula is not satisfied by the model?

Many thanks!

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Isn't consistency about theories or formulas, instead of models? – leo Feb 21 '13 at 2:38
There is no such thing as an ‘inconsistent model’. One can only speak of ‘inconsistent theories’. Models are mathematically consistent because they are actual objects of the set-theoretic universe, which we assume to be contradiction-free. – Haskell Curry Feb 21 '13 at 3:32

For any specific $L$-structure $M$, and with the usual definition of truth, we have:

(1) Every sentence of $L$ is either true in $M$ or false in $M$.

(2) No sentence of $L$ is both true in $M$ and false in $M$.

With formulas that have free occurrences of variable symbols, the situation is different, but uninteresting. Sometimes truth of a formula in $M$ is defined, for technical reasons. But there is no need for the notion.

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A footnote. Let's say that a logical system $S$ allows truth-value gaps if for some $S$-sentences (closed wffs without free variables) and some $S$-valuations, a sentence comes out neither true nor false. And a logical system $S$ allows truth-value gluts if for some $S$-sentences and some $S$-valuations, a sentence comes out both true and false.

As other answers have noted, the text-book classical version of first-order logic doesn't allow truth-value gluts. Nor does it allow truth-value gaps. But there are other first-order logics.

It is worth saying, then, that there are perfectly sensible first-order logics (logics for regimenting first-order quantificational inference) which allow gaps. For suppose we drop the requirement that every term denotes (after all, we may use names in ordinary language which, by misadventure, fail to denote: and in maths we do use partial functions so that $\varphi_e(n)$ can fail to denote). Then it is natural to say that an atomic sentence $Ft$ lacks a truth-value when $t$ fails to denote. And we can develop free versions of first-order logic (versions free of the assumption that all terms denote) based on this natural idea. See, for an accessible review, http://plato.stanford.edu/entries/logic-free/

You can also -- perhaps more surprisingly -- develop systems that allow truth-value gluts. Obviously if these are to be non-trivial, we'll have to drop the explosion rule that from $A$ and $\neg A$ we can infer any $C$. But we might want to drop explosion anyway. Why should we allow gluts? Well, it is one way to go with the paradoxes. Add a truth-predicate $T$ to a theory of arithmetic, lay down some plausible rules governing it, and (oh dear!) you find that -- because you can do some Gödelian self-reference-via-coding -- you can construct a liar sentence L where $T(L)$ comes out both true and false. What to do? Some have gone for biting the bullet and said, yep, some special sentences are both true and false, and we can assert this without the contagion of apparent paradox spreading everywhere. And (as I said, perhaps surprisingly) this idea can be developed into a coherent logical system. I'm not recommending it, but it can be done! For an accessible review see http://plato.stanford.edu/entries/dialetheism/ and http://plato.stanford.edu/entries/logic-paraconsistent/

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If you allow the formula to have free variables, it's easy. Take any language with one constant symbol $c$, and any structure with more than two elements will satisfy $\varphi(x):= x=c$ for some assignments and will satisfy its negation for other assignments.

If, however, you want to limit yourself to sentences (i.e. formulas without free variables) then the answer is no. The reason is how we define the truth value of a sentence in a structure. In particular we define $M\models\lnot\varphi$ if and only if $M\not\models\varphi$.

This means that it is impossible to have a sentence which is satisfied and its negation is satisfied.

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Ok understood. What about for a model that does not satisfy a formula and the negation of a formula? Do such models exist? – Confused Feb 21 '13 at 2:53
I don't understand what do you mean by this. Satisfy in what sense? Some assignment does? All assignments do? What formula? A formula? All formulas? – Asaf Karagila Feb 21 '13 at 3:01
@Confused it can't, because the model itself is used to define satisfaction. – leo Feb 21 '13 at 3:33

Things goes like this: you have a first order language $L$, an $L-$structure $\cal{M}$ is an object where does it makes sence to interpret the constants, functions, and relations symbols of the language. For example consider the language $$L=\{e,\ast\}$$ where $e$ is a constant and $\ast$ is a binary function. Then an $L-$structure is $${\cal M}=\langle \Bbb R,+,0 \rangle,$$ where $+$ is the usual sum in $\Bbb R$, as well as $${\cal N}=\langle \Bbb Z_{\geq 0},+,0 \rangle,$$ where $+$ is the usual sum of integers.

You can make formulas with the language.

As André says in his answer, the structure says wheter a given formula is true or false. Usually we say that a closed formula or a theory (set of closed formulas) is consistent if there is at least one model where the formula/theory is true.

For example, with language $L$ given above, consider the formula $$\varphi=\forall x\exists y(y+x=e).$$ Notice that $\cal M$ satisfies $\varphi$ but $\cal N$ does not.

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If someone can improve or add something, no doubt in do so. – leo Feb 21 '13 at 3:31
Just a matter of typography: \cal is applied until the next font modifier or the next }; \mathcal is applied on the next character or the next {...} group. This caused the $0$ to appear strange. – Asaf Karagila Feb 21 '13 at 3:33
@AsafKaragila I see. Thanks, didn't know that. – leo Feb 21 '13 at 3:36
The same holds for \scr,\bf,\sf,\frak,\it,\tt and all fonts except \bb which is not a command, for that you can use \Bbb but its scope is similar to the scope of its \mathbb counterpart. Interesting! – Asaf Karagila Feb 21 '13 at 3:38