# Negated varieties and their relatively free algebras

During the past days, I asked some questions in order to gain a clear understanding of the notion of "free algebras". I suppose that the question below is the most clear image of the concept I have in my mind:

Let $\mathcal{L}$ be an algebraic language. A negated identity in $\mathcal{L}$ is a formula of the form $$\forall x_1 \ldots \forall x_n: p(x_1, \ldots, x_n)\neq q(x_1, \ldots, x_n),$$ where $p$ and $q$ are terms in $\mathcal{L}$. For an algebra $A$ of type $\mathcal{L}$ assume that $id^+(A)$ and $id^-(A)$ are the set of all identities and negated identities valied in $A$, respectively. As we know $$Var(A)=Mod(id^+(A))$$ is the variety generated by $A$ and during the questions

we tried to understand the relative free algebras in $Var(A)$. Now, suppose $$Var^-(A)=Mod(id^-(A)).$$ It is easy to see that $Var^-(A)$ is $\{ S, P\}$-closed and so it is a pre-variety. Hence, for any set $X$, there exists a free algebra $F_{Var^-(A)}(X)$ in this pre-variety. Now, we ask:

1. How we can determine the structure of $F_{Var^-(A)}(X)$? I mean an answer like polynomial functions given by Anton Klaychko or the extendblity criterion given by Benjamin Steinberg in my previous question.

2. Can we characterize $Var^-(A)$ using class operators? I mean some thing like HSP.

P.S. To answer the question $F_{Var^-(A)}(X)=$?, one should determine the set of all identities with variables from $X$ which are logical consequence of the set $id^-(A)$. More precisely, suppose $$R(X)=\{ (p, q):\ id^-(A)\vDash p=q\}.$$ Then we have $F_{Var^-(A)}(X)=T_{\mathcal{L}}(X)/R(X)$, where $T_{\mathcal{L}}(X)$ is the term algebra. So, to answer the first question, one should say that: which identities are logical consequences of given negated identities. For example, suppose $\mathcal{L}=(0,1, +, -, \times)$. Is it possible to determine all non-trivial identities which can be deduce from the negated identity $$\forall x \forall y: x^2+y^2\neq-1?$$ Honestly, I have no even one example of non-trivial identity deducible from the above negated identity.

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It looks like a negated identity is a version of a Horn sentence, which would mean the class of models is a quasivariety. –  The Masked Avenger Jan 10 at 17:19
Yes, that is Horn sentence but just a special kind, so $Var^-(A)$ is not the Horn class generated by $A$. –  M. Shahryari Jan 10 at 17:25

This refers to the PS in the question (I cannot make comments): the term algebra $T_L(X)$ seems to satisfy the sentence $\forall x\forall y: x^2+y^2\ne -1$ but $T_L(X)$ does not satisfy any non-trivial identity. It follows that no non-trivial identity is a consequence of that sentence.

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This is a complete answer to question 1. I summarize it below as an independent answer. –  M. Shahryari Jan 11 at 11:07

No group can satisfy a negated identity (as defined abvoe) since you may substitute every variable with 1 giving 1 on both sides. Hence $id^{-}(A)=\emptyset$ for every group $A$. If you replace universal quantifiers by existential quatifiers the resulting class is no longer closed under $S$.

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Yes, in the language of groups, there is no negated identity, but if we add constants to the language of groups, we obtain negated identities. –  M. Shahryari Jan 10 at 17:30
In the language $\mathcal{L}=(0,1, +, -, \times)$, we have negated identity $\forall x: x^2\neq -1$ which is belong to $id^-(\mathbb{R})$. –  M. Shahryari Jan 10 at 17:31

This is a complete answer to the question 1, based on the idea of user45359:

We have $F_{Var^-(A)}(X)=T_{\mathcal{L}}(X)$.

Proof. Let $id^-(A)\vDash p\approx q$. By the unique readability of terms, we have $T_{\mathcal{L}}(X)\vDash id^-(A)$ and so we must have $T_{\mathcal{L}}(X)\vDash p\approx q$ and again the unique readability implies $p=q$, therefore $$R(X)=trivial\ congruence.$$ Hence we have $$F_{Var^-(A)}(X)=\frac{T_{\mathcal{L}}(X)}{R(X)}=T_{\mathcal{L}}(X).$$

P.S. So we discovered a general rule of school mathematics: No non-trivial identity can be deduced from a set of negated identities!!!!!

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