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Gödel-Dummett logic is an extension of the intuitionist logic (IPL) by the following axiom: $$ (p → q) ∨ (q → p) $$

A logic has the Scroggs' Property if it isn't characterized by any finite logical model/matrix, although every proper extension of it has a finite characteristic model/matrix.

[By proper extension I mean that the theory of the original logic is a proper-subset of the theory of the extended one]

How can I show that the Gödel-Dummett logic has the Scroggs' Property?

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The simplest finite model of your axiom $\bf L$ is the two-element boolean algebra. More interestingly, there are the models $G_n$ (say) of intuitionistic logic used by Gödel to show that intuitionistic propositional logic has infinitely many truth values. The Heyting algebra $G_n$ has $n$ elements, linearly ordered by logical strength, with $a \lor b = \max(a, b)$ and $a \land b = \min(a, b)$. So the two-element boolean algebra is $G_2$. The existence of the $G_n$ shows that the your axiom schema $\bf L$ has arbitrarily large finite models.

What you call Tautologies($\bf L$) (which I would just call the theory of $\bf L$) is not the same as the theory of $M$ (your Tautologies($M$)) for any finite model $M$: in a model with $n$ elements, a formula $\phi_n$ over propositional variables $p_1, \ldots, p_{n+1}$ asserting that at least two of $p_i$ and $p_j$ are equivalent holds, but $\phi_n$ does not hold in $G_{n+1}$ and hence is not in the theory of $\bf L$ (since $\bf L$ is sound for linearly ordered Heyting algebras).

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    If I understand correctly, Gödel-Dummett doesn't have a model with finite number of truth value (and as you mention, Gödel used the symbol G_infinity). However I'm not quite sure why adding a new formula to the theory of this logic would somehow make it possible to create a finite model for it. – 2017-02-28
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    And what I'm looking for is a model M such that Theory(L)=Theory(M). In some papers this is sometimes called "a weakly sound and weakly complete" model or matrix. And this isn't achievable with G_2 – 2017-02-28
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    Sure: adding an axiom to a theory can never increase its class of models. But that isn't what I am saying. Intuitionistic propositional logic has models with $n$ elements for any finite $n \ge 1$. But for any such model $M$, there are formulas $\phi$ and $\psi$ that have the same truth value no matter how they are interpreted in $M$ but take different truth values in a bigger model. In that sense, intuitionistic propositional **logic** must have infinitely many truth values. – 2017-02-28
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    You wrote: *how can I prove ... (1) ... or better ... (2) ... ". Please forgive me for taking the two parts of your question in order. The first part of my answer about $G_2$ and the $G_n$ provides an answer to (1) and the rest of my answer answers (2). – 2017-02-28
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    You wrote: adding an axiom to a theory can never increase its class of models. I'm not sure about it; there are logics that aren't characterized by any finite logical model/matrix although every proper extension of them has a finite characteristic model/matrix. This is the _Scroggs' Property_ [link](http://www.oxfordreference.com/view/10.1093/acref/9780191816802.001.0001/acref-9780191816802-e-382) My question is how can I show that Gödel-Dummett logic has this property – 2017-02-28
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    Let me say it again: adding an axiom to a theory can never increase its class of models, if $M \not\models \Delta$ then *a fortiori* $M \not\models \Delta \cup \{\phi\}$. Adding an axiom can turn a theory that is not characterized by a finite class of models into one that is, but the argument in my answer shows that adding the G-D axiom to IPL does not result in a theory that is characterized by a finite class of models. – 2017-02-28
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    If your question really means "how do I show that IPL + the CD-axiom has the Scrogg's property", then please edit it to say that. – 2017-02-28