Abstract
The purpose of this paper was to investigate the structure of semi-Heyting chains and the variety \({{\mathcal{CSH}}}\) generated by them. We determine the number of non-isomorphic n-element semi-Heyting chains. As a contribution to the study of the lattice of subvarieties of \({{\mathcal{CSH}}},\) we investigate the inclusion relation between semi-Heyting chains. Finally, we provide equational bases for \({{\mathcal{CSH}}}\) and for the subvarieties of \({{\mathcal{CSH}}}\) introduced in [5].
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1 Introduction and preliminaries
In [5], Sankappanavar introduced a new equational class of algebras, which he called “semi-Heyting Algebras”, as an abstraction of Heyting algebras. This variety includes Heyting algebras and shares with them some rather strong properties. For example, the variety of semi-Heyting is arithmetical; semi-Heyting algebras are pseudocomplemented distributive lattices, and their congruences are determined by filters.
The purpose of this paper was to investigate the properties of semi-Heyting chains and the structure of the variety \({{\mathcal{CSH}}}\) generated by all semi-Heyting chains. In [5], Sankappanavar posed the following problem: for \({{\mathcal{V}}}\) a subvariety of semi-Heyting algebras and n a natural number, if \(f({{\mathcal{V}}},n)\) denotes the number of non-isomorphic n-element semi-Heyting chains in \({{\mathcal{V}}},\) find a formula for \(f({{\mathcal{V}}},n)\). In Sect. 2, we prove some results on semi-Heyting chains and find \(f({{\mathcal{SH}}},n),\) i.e., we determine the number of non-isomorphic structures of semi-Heyting algebra that can be defined over an n-element chain. Section 3 is devoted to investigate the behavior of the subalgebras of a given semi-Heyting chain. Finally, we find equational bases for many subvarieties of \({{\mathcal{CSH}}}\) in Sect. 4, solving the problems 14.6–14.10 posed in [5].
We start by recalling some definitions and results. For basic notation and results, the reader is referred to [1–4].
Definition 1.1
An algebra \({\bf L} = \langle L, \vee, \wedge, \rightarrow, 0, 1\rangle \) is a semi-Heyting algebra if the following conditions hold:
-
(SH1) \(\langle L, \vee, \wedge, 0, 1\rangle \) is a lattice with 0 and 1;
-
(SH2) \(x \wedge (x \rightarrow y) \approx x \wedge y;\)
-
(SH3) \(x \wedge (y \rightarrow z) \approx x \wedge [(x \wedge y) \rightarrow (x \wedge z)];\)
-
(SH4) \(x \rightarrow x \approx 1.\)
We will denote by \({{\mathcal{SH}}}\) the variety of semi-Heyting algebras.
Semi-Heyting algebras are pseudocomplemented distributive lattices, with the pseudocomplement given by \(x^{*} = x \rightarrow 0\) (see [5]).
Lemma 1.2
[5] Let \({{\mathbf{L}}} \in {{\mathcal{SH}}}\) and \(a, b \in L.\)
-
(a)
If \(a \rightarrow b =1\) then \(a \, \leq \, b.\)
-
(b)
If \(a \, \leq \, b\) then \(a \, \leq \, a \rightarrow b.\)
-
(c)
\(a=b\) if and only if \(a \to b = b \to a = 1.\)
-
(d)
\(1 \to a = a.\)
Proof
From \(a \to b = 1\) and (SH3), we get \(a \wedge 1 = a \wedge b,\) i.e., \(a = a \wedge b,\) and we have (a). For (b), by (SH3) and \(a \, \leq \, b\) it follows that \(a =a \wedge (a \to b) \, \leq \, a \to b.\) Property (c) is clear. To prove (d), observe that \(a = 1 \wedge a = 1 \wedge (1 \to a) = 1 \to a. \) \(\square\)
Congruences on semi-Heyting algebras are determined by filters. Besides, we have the following characterization of subdirectly irreducible algebras in \({{\mathcal{SH}}}\) (see [5, Theorem 7.5])
Theorem 1.3
Let\({{\mathbf{L}}} \in {{\mathcal{SH}}}\)with\(|{{\mathbf{L}}}| \, \geq \, 2.\)The following are equivalent:
-
(a)
Lis subdirectly irreducible.
-
(b)
Lhas a unique coatom.
Observe that as a consequence of this theorem, if L is subdirectly irreducible, then \(1 \in L\) is \(\vee\)-irreducible.
2 Semi-Heyting chains
We say that \({{\mathbf{L}}} \in {{\mathcal{SH}}}\) is a semi-Heyting chain if the lattice reduct of L is totally ordered. In this section, we prove some results on semi-Heyting chains, and we determine a formula for the number of semi-Heyting structures that can be defined on an n-element chain. We generalize the results of Sankappanavar in [5] for chains with 2, 3, and 4 elements.
The following lemma states that part of the operation table of \(\to\) in a semi-Heyting chain is uniquely determined.
Lemma 2.1
[5] Let L be a semi-Heyting chain, \(a,b \in L.\) If \(a < b\) then \(b \rightarrow a = a.\)
Proof
From \(b \wedge a = b \wedge (b \to a)\) and \(a < b,\) we obtain \(a = b \wedge (b \to a).\) As L is a chain, \(a = b\) or \(a = b \to a,\) so \(a = b \to a.\) \(\square\)
The next two lemmas are useful when we have to verify if a given chain is a semi-Heyting algebra.
Lemma 2.2
Let\(\langle L, \vee, \wedge, \rightarrow, 0, 1\rangle \)be a totally ordered bounded lattice L with a binary operation → such that if\(a, b \in L\)with\(a< b\)then\(b \rightarrow a = a.\)The following conditions are equivalent:
-
(1)
-
(a)
\(x \rightarrow x \approx 1;\)
-
(b)
\(x \wedge (x \rightarrow y) \approx x \wedge y;\)
-
(a)
-
(2)
-
(a)
\(x \rightarrow x \approx 1;\)
-
(b)
If \(x < y\) then \(x \wedge (x \rightarrow y) = x \wedge y.\)
-
(a)
Proof
We only have to prove that \((2) \Rightarrow (1).\) We want to prove that \(a \wedge (a \to b) = a \wedge b\) for every \(a, b \in L.\) If \(a < b,\) by (2) (b), \(a \wedge (a \to b) = a \wedge b.\) If \(a = b\) by (2)(a), \(a \wedge (a \to a) = a \wedge 1 = a = a \wedge a.\) Finally, if \(a > b,\) by hypothesis \(a \rightarrow b = b\) and so, \(a \wedge (a \to b) = a \wedge b.\) \(\square\)
Lemma 2.3
Let\(\langle L, \vee, \wedge, \rightarrow, 0, 1\rangle \)be a totally ordered bounded lattice L with a binary operation → such that if\(a, b \in L\)with\(a< b\)then\(b \rightarrow a = a.\)The following conditions are equivalent
-
(1)
\(\langle L, \vee, \wedge, \rightarrow, 0, 1\rangle \in {{\mathcal{SH}}};\)
-
(2)
-
(a)
\(x \rightarrow x \approx 1;\)
-
(b)
\(x \wedge (x \rightarrow y) \approx x \wedge y;\)
-
(c)
If \(y < x < z,\) then \(x \wedge (y \rightarrow x) = x \wedge (y \rightarrow z).\)
-
(a)
Proof
-
(1)\(\Rightarrow\) (2). As \({{\mathbf{L}}} = \langle L, \vee, \wedge, \rightarrow, 0, 1\rangle \in {{\mathcal{SH}}}, {{\mathbf{L}}}\) satisfies (2)(a) and (2)(b). Let \(a, b, c \in L\) such that \(b < a < c.\) Then \(a \wedge (b \to c) = a \wedge [(a \wedge b) \to (a \wedge c)] = a \wedge (b \to a).\)
-
(2)\(\Rightarrow\) (1). It is enough to prove (SH3). Let \(a,b,c \in L,\) and suppose that \(c < b.\)
If \(a < c,\) then \(a \wedge b = a \wedge c = a.\) So \(a \wedge (b \to c) = a \wedge c = a\) and \(a \wedge [(a \wedge b) \to (a \wedge c)] = a \wedge a = a.\) If either \(c < a < b\) or \(c < b < a,\) then \(a \wedge c < a \wedge b.\) Thus, \(a \to [(a \wedge b) \to (a \wedge c)] = a \wedge a = a.\)
The other cases are similar. \(\square\)
Lemma 2.4
LetLbe a semi-Heyting chain. The following conditions are equivalent:
-
(a)
\(0 \to a = 0\) for some \(a \in L\) with \(a \not= 0.\)
-
(b)
\(0 \to b = 0\) for every \(b \in L\) with \(b \not= 0.\)
Proof
Suppose that \(0 \to a = 0\) for some \(a \in L,\) \(a \neq 0.\) \(0 = a \wedge 0 = a \wedge (0 \to a) = a \wedge ((0 \wedge a) \to (1 \wedge a)) = a \wedge (0 \to 1).\) As \(a \not= 0\) and L is a chain, \(0 \to 1 = 0.\) Let \(b \in L\) such that \(b \not= 0.\) \(0= b \wedge 0 = b \wedge (0 \to 1)\buildrel{(SH3)} \over {=} b \wedge (0 \to b).\) As \(b \not= 0\) and L is a chain, \(0 \to b = 0.\) \(\square\)
In particular, if \(0 \to 1 = 0,\) then \(0 \to b = 0\) for every \(b \in L,\) \(b \not= 0.\)
In order to determine the number of semi-Heyting algebras that can be defined on an n-element chain, let us consider first the following example, since it contains the main ideas underlying in the procedure.
Let L be a 4-element lattice, \(L = \{ 0 , a_1, a_2, 1\},\) with \(0 = a_0 < a_1 < a_2 < a_3 = 1.\) Consider a binary operation \(\to: {\bf L} \times {\bf L} \to {\bf L}\) in such a way that \(\langle L, \wedge, \vee, \to, 0, 1 \rangle\) is a semi-Heyting chain. From Lemma 2.1, we know that for \(x,y \in L,\) if \(x < y\) then \(y \to x = x.\) Consequently, the lower half under the main diagonal, including that diagonal, of the operation table of \(\to\) of L is uniquely determined. So, it remains to fill in the upper half of the table, i.e., the elements \(x \to y\) for \(x < y.\)
Observe that if \(x \, \leq \, y\) then \(x \, \leq \, x \to y\) (Lemma 1.2).
If \(0 \to 1 = 0,\) then by Lemma 2.4, \(0 \to a_1 = 0 \to a_2 = 0.\) So, the first row in the table is
If \(0 \to 1 = a_1,\) then \(0 \to a_2 = a_2\) and \(0 \to a_1 \in [a_1),\) where \([x) = \{y \in L: x \, \leq \, y\}.\)
Indeed, \(a_1 = a_2 \wedge a_1 = a_2 \wedge (0 \to 1) = a_2 \wedge (0 \to a_2),\) so \(0 \to a_2 = a_1;\) and \(a_1 = a_1 \wedge (0 \to 1) \buildrel{(SH3)} \over {=} a_1 \wedge (0 \to a_1),\) i.e., \(0 \to a_1 \, \geq \, a_1.\)
So in this case, the first row of the table is
In a similar way, it can be seen that if \(0 \to 1 = a_2\) or \(0 \to 1 = 1,\) then \(0 \to a_2 \in [a_2)\) and \(0 \to a_1 \in [a_1).\) So we have
Then we can write for the first row
i.e., \((0 \to 1, 0 \to a_2, 0 \to a_1) \in \left\{\{0\} \times \{0\} \times \{0\} \right\} \cup \left\{ \{a_1\} \times \{a_1\} \times [a_1)\right\} \cup \left\{ \{a_2\} \times [a_2) \times [a_1)\right\} \cup \left\{ \{1\} \times [a_2) \times [a_1)\right\}.\)
For the second and third rows, we would have the following possibilities, determined by \(a_1 \to 1\) and \(a_2 \to 1,\) respectively:
Observe that the behavior of each row is independent of the others, and that, for \(x < y,\) the element \(x \to y\) depends on the element \(x \to 1.\)
The following lemma shows that in every semi-Heyting chain, the operation table of \(\to\) behaves as in the previous example.
Lemma 2.5
Let L be a semi-Heyting chain and let \(a,b,c \in L,\) \(a \not= 1.\) If \(a \rightarrow 1 = b\) then for \(c > a\)
Proof
Suppose that \(b<c.\) From \(a \to 1 = b\) we get \(c \wedge (a \to 1) = c \wedge b = b,\) i.e., \(c \wedge [(c \wedge a) \to (c \wedge 1)] = b.\) Thus, \(c \wedge (a \to c) = b.\) As L is a chain and \(b < c,\) then \(b = a \to c.\)
The case \(b \, \geq \, c\) is similar. \( \square\)
Let \({{\mathbf{L}}}_{n} = \langle L_n, \wedge, \vee, 0, 1 \rangle\) be an n-element totally ordered lattice:
Let \(\to: {\bf L}_{n} \times {\bf L}_{n} \to {\bf L}_{n}\) be a function such that \(\langle L_{n}, \wedge, \vee, \to, 0 , 1 \rangle \in {{\mathcal{SH}}}.\)
Motivated by the previous example, we are going to introduce the following sets:
For \(a_{j} \in L_{n},\) \(0 \, \leq \, j \, \leq \, n-1,\) and for \(1 \, \leq \, k \, \leq \, n-1,\)
and for \(a_i \, \leq \, a_j,\)
Finally, let \(F_{i}\) be the set
where \(F_i\) would represent the row in the table that corresponds to the elements of the form \(a_{i} \to y\) with \(a_{i} < y.\)
In the previous example (n = 4),
We have \((0 \to 1, 0 \to a_2, 0 \to a_1) \in F_0,\) \((a_1 \to 1, a_1 \to a_2) \in F_1\) and \(a_2 \to 1 \in F_2\) and we will write \(((0 \to 1, 0 \to a_2, 0 \to a_1),(a_1 \to 1, a_1 \to a_2), a_2 \to 1) \in F_{0} \times F_{1} \times F_{2}.\)
This motivates the introduction of the following set F that would represent the upper half above the main diagonal of the table:
Lemma 2.6
\((a_{i} \to 1;a_{i} \to a_{n-2}, \ldots, a_{i} \to a_{i+1}) \; \in \; F_{i}.\)
Proof
As \(a_{i} \, \leq \, 1,\) then \(a_{i} \to 1 \, \geq \, a_{i}.\) Suppose that \(a_{i} \to 1 = a_{k}\) with \(i \, \leq \, k \, \leq \, n-1.\) We will see that \((a_{i} \to 1;a_{i} \to a_{n-2}, \ldots, a_{i} \to a_{i+1})\; \in E_{a_{k}}^{a_{i}}.\) By Lemma 2.5,
We will see that \(a_{i} \to a_{j} \in E_{n-j}^{k}\) with \(i+1 \, \leq \, j \, \leq \, n-1.\) If \(n-j < n-k,\) then \(j > k.\) Hence, by Eq. 1, \(a_{i} \to a_{j} = a_{k}\) and, consequently, \(a_{i} \to a_{j} \in E_{n-j}^{k}.\) If \(n-j \, \geq \, n-k,\) then \(j \, \leq \, k.\) Thus, by Eq. 1, \(a_{i} \to a_{j} \, \geq \, a_{j},\) and then \(a_{i} \to a_{j} \in [a_{n-(n-j)}).\) So, \(a_{i} \to a_{j} \in E_{n-j}^{k}\) and \((a_{i} \to 1;a_{i} \to a_{n-2}, \ldots, a_{i} \to a_{i+1}) \in E_{a_{k}}^{a_{i}}.\) Therefore, \((a_{i} \to 1;a_{i} \to a_{n-2}, \ldots, a_{i} \to a_{i+1}) \; \in \; F_{i}. \square\)
Notation 2.7
We denote \(\alpha_{\to}(i) = (a_{i} \to 1;a_{i} \to a_{n-2}, \ldots, a_{i} \to a_{i+1})\) with \(0 \, \leq \, i \, \leq \, n-2.\)
Corollary 2.8
\((\alpha_{\to}(0), \alpha_{\to}(1), \ldots, \alpha_{\to}(n-2))\; \in \; F.\)
Now we are going to construct an implication from a given element \(x \in F.\)
Consider the set F and \(x \in F.\) Then \(x = (x(0), x(1), \ldots, x(n-2)),\) where \(x(i) \in F_{i}\) for \(0 \, \leq \, i \, \leq \, n-2.\) Now, for each i, there exists \(j_{i}\) with \(i \, \leq \, j_{i} \, \leq \, n-1\) such that \(x(i) \in E_{a_{j_{i}}}^{a_{i}}.\) Hence, \(x(i) = (x(i)_{j_{i}}(1),x(i)_{j_{i}}(2), \ldots,x(i)_{j_{i}}(n-i-1))\) with \(x(i)_{j_{i}}(k) \in E_{k}^{j_{i}},\) \(1 \, \leq \, k \, \leq \, n-i-1.\) We define in \(L_{n}\) an operation \(\Rightarrow\) in the following way:
The following lemma proves that \(\Rightarrow\) is a semi-Heyting implication.
Lemma 2.9
\(\langle L_{n}, \wedge, \vee, \Rightarrow, 0, 1\rangle \; \in\; {{\mathcal{SH}}}.\)
Proof
First observe that if \(r < l,\) then \(1 \, \leq \, n-l \, \leq \, n-r-1.\) So \(\Rightarrow\) is well defined.
Now, we will see that \(\Rightarrow\) it is an implication of semi-Heyting algebras.
By definition of \(\Rightarrow, L \models x \Rightarrow x \approx 1.\)
Let us see that \(L \models x \wedge (x \Rightarrow y) \approx x \wedge y.\) Let \(x,y \in L\) with \(x < y.\) Then \(x = a_{r}\) and \(y = a_{l}\) with \(0 \, \leq \, r,l \, \leq \, n-1\) and \(r<l.\) Thus, \(a_{r} \Rightarrow a_{l} = x(r)_{j_{r}}(n-l)\) that belongs to
As \(r \,\leq \, j_{r}\) and \(r < l,\) \(a_{r} \wedge (a_{r} \Rightarrow a_{l}) = a_{r} = a_{r} \wedge a_{l}.\) Hence, \(x \wedge (x \Rightarrow y) = x \wedge y.\) By Lemma 2.2, \(L \models x \wedge (x \Rightarrow y) \approx x \wedge y.\)
It is long but computational to prove that \(L \models x \wedge (y \Rightarrow z) \approx x \wedge [(x \wedge y) \Rightarrow (x \wedge z)].\)
By Lemma 2.3, \(\Rightarrow\) is a semi-Heyting implication. \(\square\)
From the previous results, we have the following:
Theorem 2.10
Let\(S_{n} = \{\to: L_{n}^{2} \to L_{n} | \langle L_{n}, \wedge, \vee, \to, 0, 1\rangle \in {{\mathcal{SH}}}\}.\)There exists a bijective correspondence between\(S_{n}\) and F.
Proof
We define \(\alpha: S_{n} \to F\) as follows: \(\alpha(\to) = (\alpha_{\to}(0), \alpha_{\to}(1), \ldots, \alpha_{\to}(n-2))\) for each \(\to \in S_{n}.\) By Corollary 2.8, \(\alpha(\to) \in F.\) By Lemma 2.9, \(\alpha\) is onto.
Let us prove that \(\alpha\) is injective. Let \(\to_{1}, \to_{2} \in S_{n}\) such that \(\alpha(\to_{1}) = \alpha(\to_{2}).\) By Lemma 2.1, if \(a_{r} > a_{l}\) then \(a_{r} \to_{1} a_{l} = a_{l} = a_{r} \to_{2} a_{l}.\) If \(a_{r} = a_{l},\) \(a_{r} \to_{1} a_{l} = 1 = a_{r} \to_{2} a_{l}.\) As \(\alpha(\to_{1}) = \alpha(\to_{2}),\) \( (\alpha_{\to_{1}}(0), \alpha_{\to_{1}}(1), \ldots, \alpha_{\to_{1}}(n-2)) = (\alpha_{\to_{2}}(0), \alpha_{\to_{2}}(1), \ldots, \alpha_{\to_{2}}(n-2)).\) Hence, \(\alpha{\to_{1}}(i) = \alpha{\to_{2}}(i)\) for all i with \(0 \, \leq \, i < n-1.\) Then \(a_{i} \to_{1} 1 = a_{i} \to_{2} 1, a_{i} \to_{1} a_{n-2} = a_{i} \to_{2} a_{n-2}, \ldots, a_{i} \to_{1} a_{i+1} = a_{i} \to_{2} a_{i+1}\) for all i with \(0 \, \leq \, i < n-1.\) Hence, \(\to_{1} = \to_{2},\) and \(\alpha\) is injective. \(\square\)
Our next objective is to determine the cardinal of the set F.
Lemma 2.11
\(E_{a_{j}}^{a_{i}} \cap E_{a_{k}}^{a_{i}} = \emptyset \) with \(j\not= k.\)
Proof
Let \(x \in E_{a_{j}}^{a_{i}} \cap E_{a_{k}}^{a_{i}}\) and suppose that \(j < k.\) As \(x \in E_{a_{j}}^{a_{i}}\) then \(x = (x(1), x(2), \ldots, x(n-i-1)),\) with \(x(m) \in E^{j}_{m}, 1 \, \leq \, m \, \leq \, n-i-1.\) As \(x \in E_{a_{k}}^{a_{i}}\) then \(x = (z(1), z(2), \ldots, z(n-i-1))\) with \(z(m) \in E^{k}_{m}, 1 \, \leq \, m \, \leq \, n-i-1.\) Now, if \(1 < n-j,\) \(x(1) = a_{j}\) and if \(1 \, \geq \, n-j,\) then \(x(1) = 1.\) If \(1 < n-k,\) \(z(1) = a_{k}\) and if \(1 \, \geq \, n-k,\) then \(z(1) = 1.\) Moreover, as \(j < k,\) \(n-j> n-k.\) If \(1 < n-k < n-j,\) \(x(1) = a_{j}\) and \(z(1) = a_{k}\) a contradiction since \(x(1) = z(1) .\) The other cases are similar. Consequently, \(E_{a_{j}}^{a_{i}} \cap E_{a_{k}}^{a_{i}} = \emptyset. \square\)
Lemma 2.12
\( \left| F \right| = \prod\limits_{i = 0}^{n - 2} {\left[ {1 + (n - i - 1)!\sum\limits_{j = i + 1}^{n - 1} \frac{1}{(n - j - 1)!}}\right]}\)
Proof
By Lemma 2.11, \(\left| F_i \right| = \sum\limits_{j = i}^{n - 1}\left| E_{a_j }^{a_i} \right|\) with \(\left| {E_{a_j}^{a_i}} \right| = \left| {E_1^j } \right| \cdot \left| {E_2^j } \right| \cdots \left| {E_{n - i - 1}^j } \right|.\)
Hence, \(\left| {E_{a_j }^{a_i}} \right| = \left\{ \begin{array}{lll} (n - j) \cdot (n - j + 1) \cdots (n - i - 1) & \text{if} & j > i \\ 1 & \text{if} & j = i \\ \end{array}\right..\) So,
Thus, \(\left| F \right| = \prod\limits_{i = 0}^{n - 2}\left[ 1 + (n - i - 1)!\sum\limits_{j = i + 1}^{n - 1} \frac{1}{(n - j - 1)!} \right].\) \(\square\)
By Lemma 2.12 and Theorem 2.10, it follows immediately the next corollary:
Corollary 2.13
There are
non-isomorphic n-element semi-Heyting chains, for \(n \, \geq \, 2.\)
For \(n = 2,3,4\) this formula gives 2, 10, and 160, respectively, which coincide with the numbers determined by Sankappanavar in [5]. For \(n = 5,\) there are 10,400 non-isomorphic semi-Heyting chains with five elements.
3 Isomorphic subalgebras of a semi-Heyting chain
When one investigates the lattice of subvarieties of \({{\mathcal{CSH}}},\) it is important to characterize the subalgebras of a finite semi-Heyting chain and to study the inclusion relation between chains in \({{\mathcal{CSH}}}.\)
In this section, we consider the following related problem, which will show the complexity of the general problem. Given an n-element chain L, we wish to know when an \((n+1)\)-element semi-Heyting chain L′ contains a subalgebra isomorphic to L, and the number of semi-Heyting chains L′ satisfying this condition.
Consider the following example. Let \({{\mathbf{L}}} = \langle L, \wedge, \vee, \to , 0,1 \rangle\) be a 4-element semi-Heyting chain, \(L : 0 < a < b < 1.\) We want to know the number of 5-element semi-Heyting chains L′ such that L is (isomorphic to) a subalgebra of L′. Let L′ be a 5-element semi-Heyting chain such that \({{\mathbf{L}}}\in {{\mathbb{IS}}}({{\mathbf{L}}}')\) and assume that \(L' = L \cup \{c\}.\)
Suppose first that \(0 < a < c < b < 1.\) We have just to consider the elements \(x \to c\) for \(x < c\) and the elements \(c \to y\) for \(c < y.\)
If \(0 \to b \, \leq \, a,\) then \(c \wedge (0 \to c) = c \wedge (0 \to b) = 0 \to b.\) Since L is a chain, \(0 {\to}c =0 {\to}b.\)
If \(0 \to b \, \geq \, b,\) then \(c \wedge (0 \to c) = c \wedge (0 \to b) = c.\) So, \(0 \to c \, \geq \, c,\) i.e.,
Similarly, if \(a \to b = a\) then \(a \to c = a \to b,\) and if \(a \to b \, \geq \, b\) then \(a \to c \, \geq \, c,\) i.e.,
Besides, \(c \to 1 \, \geq \, c\) and
So, we conclude that
In a similar way, we should consider the cases \(0 < a < b < c < 1\) and \(0 < c < a < b < 1.\)
In general, let \({\bf L} = \langle L, \wedge, \vee, \to, 0, 1\rangle \) be an n-element semi-Heyting chain, \(n\, \geq \, 3,\) with
For a given \(0 \, \leq \, i \, \leq \, n-2 ,\) let \(\langle L^{i}, \wedge, \vee, 0, 1\rangle \) be an \((n+1)\)-element chain with \(L^i = L \cup \{b_i\},\) \(b_{i} \notin L,\) and
We want to find conditions on an implication \(\to: L^{i} \times L^{i} \to L^{i}\) in order for \({\bf L}\) to be a subalgebra of \({\bf L}^{i} = \langle L^{i}, \wedge, \vee, \to, 0, 1\rangle \) and to determine the number of implication operations that satisfy those conditions.
Consider the following sets
-
(a)
\(F_{j}^{k} = \left\{ \begin{array}{lll} \left\{ a_k \right\} & \text{if} & k < j \\ \left[ a_j \right) & \text{if} & k \ge j \\ \end{array} \right.\)
-
(b)
\(F_{k}^{i} = \left\{ a_k \right\} \times \prod\limits_{j = i + 1}^{n - 2} F_j^k \)
-
(c)
\(F_{0}^{i} = \left\{ b_i \right\} \times \prod\limits_{j = i + 1}^{n - 2}\left\{b_i\right\}\)
-
(d)
\(E_{j}^{i} = \left\{ \begin{array}{lll} \left\{ a_j \to a_{i + 1}\right\} & \text{if} & a_j \to a_{i + 1} \le a_i \\ \left[b_i \right) & \text{if} & a_j \to a_{i + 1} > a_i \\ \end{array} \right.\)
-
(e)
\(E^{i} = \prod\limits_{j = 0}^i E_j^i \)
-
(f)
\(G^{i} = E^i \times \left[\left(\bigcup\limits_{k = i + 1}^{n - 1} F_k^i\right) \cup F_0^i\right]\)
Lemma 3.1
Let L and \({\bf L}^{i}\) be as defined earlier. Then
Proof
For \(0 \, \leq \, j \, \leq \, i,\) \(a_{j} < a_{i+1},\) and then by Lemma 1.2, \(a_{j} \to a_{i+1} \, \geq \, a_{j}.\) If \(a_{j} \to a_{i+1} = a_{l}\) with \(j \, \leq \, l \, \leq \, i,\) as \(b_{i} > a_{l},\) by Lemma 2.5 \(a_{j} \to b_{i} = a_{l} = a_{j} \to a_{i+1}.\) If \(a_{j} \to a_{i+1} = a_{l}\) with \(i+1 \, \leq \, l \, \leq \, n-1,\) by Lemma 2.5 \(a_{j} \to b_{i} \, \geq \, b_{i}.\) Then \(a_{j} \to b_{i} \in E_{j}^{i}.\) Consequently, \((a_{0} \to b_{i}, a_{1} \to b_{i}, \ldots, a_{i} \to b_{i}) \in E^{i}.\) Moreover, as \(b_{i} < 1,\) by Lemma 1.2, \(b_{i} \to 1 \, \geq \, b_{i}.\) Hence, \(b_{i} \to 1 = b_{i}\) or \(b_{i} \to 1 = a_{k}\) with \(i+1 \, \leq \, k \, \leq \, n-1.\)
Suppose that \(b_{i} \to 1 = b_{i}.\) Consider \(j\) with \(i+1 \, \leq \, j \, \leq \, n-2.\) By Lemma 2.5, \(b_{i} \to a_{j} = b_{i}.\) Then \(( b_{i} \to 1, b_{i} \to a_{i+1}, \ldots, b_{i} \to a_{n-2}) \in F_{0}^{i}.\)
If \(b_{i} \to 1 = a_{k}\) with \(i+1 \, \leq \, k \, \leq \, n-1,\) consider j with \(i+1 \, \leq \, j \, \leq \, n-2.\) If \(j \, \leq \, k,\) by Lemma 2.5, \(b_{i} \to a_{j} \, \geq \, a_{j}.\) If \(j > k,\) by Lemma 2.5, \(b_{i} \to a_{j} =a_{k}.\) Then \(( b_{i} \to 1, b_{i} \to a_{i+1}, \ldots, b_{i} \to a_{n-2}) \in F_{k}^{i}.\)
Therefore, \(( b_{i} \to 1, b_{i} \to a_{i+1}, \ldots, b_{i} \to a_{n-2}) \in \bigcup_{k=i+1}^{n-1} F_{k}^{i} \cup F_{0}^{i}.\) Thus, \(((a_{0} \to b_{i}, a_{1} \to b_{i}, \ldots, a_{i} \to b_{i}), (b_{i} \to 1, b_{i} \to a_{i+1}, b_{i} \to a_{i+2}, \ldots, b_{i} \to a_{n-2})) \in G^{i}. \square\)
Now we will construct an implication from a given element of the set \(G^{i}.\)
Consider now the set \(G^{i}.\) Let \(\alpha \in G^{i}.\) \(\alpha = (\alpha_{1}, \alpha_{2})\) with \(\alpha_{1} \in E^{i}\) and \(\alpha_{2} \in \left( \bigcup\nolimits_{k = i + 1}^{n - 1} F_k^i\right) \cup F_0^i.\) As \(\alpha_{1} \in E^{i},\) then \(\alpha_{1} = (\alpha_{1}(0), \alpha_{1}(1), \ldots, \alpha_{1}(i))\) with \(\alpha_{1}(j) \in E_{j}^{i}\) for every j, 0 ≤ j ≤ i. As \(\alpha_{2} {\in}\left(\bigcup\nolimits_{k =i+1}^{n-1} F_k^i \right)\cup F_0^i,\) then \(\alpha_{2} \in F^{i}_{k_{0}}\) for any \(k_{0}\) with \(i+1 \, \leq \, k_{0} \, \leq \, n-1\) ó \(\alpha_{2} \in F_{0}^{i}.\) Hence, \(\alpha_{2} = (\alpha_{2}(n-1), \alpha_{2}(i+1), \alpha_{2}(i+2), \ldots, \alpha_{2}(n-2)).\) If \(\alpha_{2} \in F^{i}_{k_{0}}\) for any \(k_{0}\) with \(i+1 \, \leq \, k_{0} \, \leq \, n-1,\) \(\alpha_{2}(n-1)=a_{k_{0}}\) and \(\alpha_{2}(j) \in F_{j}^{k_{0}}\) with \(i+1 \, \leq \, j \, \leq \, n-2.\) If \(\alpha_{2} \in F_{0}^{i},\) \(\alpha_{2}(n-1) = b_{i}\) and \(\alpha_{2}(j) = b_{i}\) with \(i+1 \, \leq \, j \, \leq \, n-2.\) In \(L^{i}\) we define an operation \(\Rightarrow\) as:
Lemma 3.2
Let\({\bf L}^{i} = \langle L^{i}, \wedge, \vee, \Rightarrow, 0, 1\rangle .\)Then\({\bf L}^{i} \in {{\mathcal{SH}}}\)and, consequently, \({\bf L}\)is a subalgebra of\({\bf L}^{i}.\)
Proof
Clearly, \({\bf L}^{i} \models x \Rightarrow x \approx 1.\)
We will see that \({\bf L}^{i} \models x \wedge (x \Rightarrow y) \approx x \wedge y.\) Let \(a , b \in L^{i}\) with \(a < b.\)
If \(a, b \in L,\) then \(a \wedge (a \Rightarrow b) = a \wedge (a \to b) = a \wedge b.\)
Suppose that \(a = b_{i},\) \(b = a_{r}\) with \(i+1 \, \leq \, r \, \leq \, n-1\) and \(\alpha_{2} \in F_{k_{0}}^{i}\) with \(i+1 \, \leq \, k_{0} \, \leq \, n-1.\) Then \(b_{i} \wedge (b_{i} \Rightarrow a_{r}) = b_{i} \wedge \alpha_{2}(r).\) If \(r=n-1,\) \(b_{i} \wedge (b_{i} \Rightarrow 1) = b_{i} \wedge \alpha_{2}(n-1) = b_{i} \wedge a_{k_{0}} = b_{i} = b_{i} \wedge 1.\) If \(r < n-1,\) \(b_i \Rightarrow a_r = \left\{ \begin{array}{lllll} a_{k_0} & \text{if} & k_0 < r\\ x & \text{if} & k_0 \ge r & \text{with} & x \ge a_r \\ \end{array} \right..\)
Consequently, \(b_i \wedge \left({b_i \Rightarrow a_r}\right)= \left\{ \begin{array}{lll} b_i & \text{if} & k_0 < r \\ b_i & \text{if} & k_0 \ge r \\ \end{array} \right. = b_i \wedge a_r\)
Suppose now that \(a = b_{i},\) \(b = a_{r}\) with \(i+1 \, \leq \, r \, \leq \, n-1\) and \(\alpha_{2} \in F_{0}^{i}.\) In this case \(b_{i} \wedge (b_{i} \Rightarrow a_{r}) = b_{i} \wedge \alpha_{2}(r) = b_{i} \wedge b_{i}=b_{i}=b_{i} \wedge a_{r}.\)
Finally, if \(a = b_{i}\) and \(b = a_{r}\) with \(0 \, \leq \, r \, \leq \, i.\) \(a_{r} \wedge (a_{r} \Rightarrow b_{i}) = a_{r} \wedge \alpha_{1}(r).\) Thus,
so,
Hence, \(a_{r} \wedge (a_{r} \Rightarrow b_{i}) = a_{r} \wedge \alpha_{1}(r)=a_{r} = a_{r} \wedge b_{i}.\)
In a similar way, it can be proved that \({\bf L}^i \models x \wedge (y \Rightarrow z) \approx x \wedge [(x \wedge y) \Rightarrow (x \wedge z)].\)
By Lemma 2.3, \({\bf L}^{i}_{\to} \in\ {{\mathcal{SH}}}. \square\)
Remark 3.3
From the previous lemma, it follows that for every n-element semi-Heyting chain L there exists an (n + 1)-element semi-Heyting chain L′ such that L is a subalgebra of L′.
From the previous results, we can establish the following correspondence:
Theorem 3.4
Let \(S_{i}\) be the set of operations \(\to : L^{i} \times L^{i} \to L^{i}\) such that \({\bf L}^i = \langle L^{i}, \wedge, \vee, \to, 0, 1\rangle \in {{\mathcal{SH}}}\) and L is a subalgebra of \({\bf L}^i.\) Then there exists a bijective correspondence between \(S_{i}\) and \(G^{i}.\)
Proof
We define \(\alpha: S_{i} \to G^{i}\) as \(\alpha(\to) = ((a_{0} \to b_{i}, a_{1} \to b_{i}, \ldots, a_{i} \to b_{i}), (b_{i} \to 1, b_{i} \to a_{i+1}, b_{i} \to a_{i+2}, \ldots, b_{i} \to a_{n-2})).\) By Lemmas 3.1 and 3.2, \(\alpha\) is well defined, and it is onto. The injectivity is left to the reader. \(\square\)
Now we want to determine the cardinal of the set \(G^{i}.\)
Lemma 3.5
Let \(i+1 \, \leq \, k,k' \, \leq \, n-1\) and \(0 \, \leq \, i \, \leq \, n-2.\) Then \(F_{k}^{i} \cap F_{k'}^{i} = \emptyset\) if \(k \not= k'.\)
Proof
Let \(\alpha \in F_{k}^{i} \cap F_{k'}^{i}\) and suppose that \(k < k'.\) As \(\alpha \in F_{k}^{i}, \alpha = (\alpha_{1}, \alpha_{i+1}, \alpha_{i+2}, \ldots, \alpha_{n-2})\) with \(\alpha_{1} = a_{k}\) and \(\alpha_{j} \in F_{j}^{k}\) with \(i+1 \, \leq \, j \, \leq \, n-2.\) As \(\alpha \in F_{k'}^{i}, \alpha = (\alpha_{1}, \alpha_{i+1}, \alpha_{i+2}, \ldots, \alpha_{n-2})\) with \(\alpha_{1} = a_{k'}\) and \(\alpha_{j} \in F_{j}^{k'}\) with \(i+1 \, \leq \, j \, \leq \, n-2.\) Hence, \(a_{k} = a_{k'}.\) Then \(k=k'\) a contradiction. \(\square\)
Lemma 3.6
where
Proof
We have that \(F_{j}^{k} = \left\{ \begin{array}{lll} \left\{a_k\right\} & \text{if} & k < j\\ \left[a_j \right) & \text{if} & k \ge j\\ \end{array} \right.\) and then
Hence, for \(k \not= n-1\) ,
If \(k=n-1,\) \(\prod\nolimits_{j = i + 1}^{n - 2} \left| F_j^k\right| = (n - (i + 1) - 1)!.\)
Besides, as
then \(| E_{j}^{i} | = \left\{ \begin{array}{lll} 1 & \text{if} & a_j \to a_{i + 1} \le a_i \\ |\left[b_i \right)| & \text{if} & a_j \to a_{i + 1} > a_i \\ \end{array} \right.\) with \(|[b_{i})| = n-1-(i+1)+2=n-1-i-1+2=n-i.\)
Let
Then \(G^{i} = E^i \times \left[\left( \bigcup\nolimits_{k = i + 1}^{n - 1}F_k^i \right) \cup F_0^i \right],\) by Lemma 3.5
\(\square\)
From Lemma 3.6 and Theorem 3.4 the next corollary follows.
Corollary 3.7
The number of non-isomorphic algebras\({\bf L}^{i}\)is given by the number:
where
Corollary 3.8
The number of non-isomorphic\((n+1)\)-element semi-Heyting chains\({\bf L}^{i}\)such that\({\bf L}\)is a subalgebra of\({\bf L}^{i}\)is given by
where
It remains to consider the case \(n=2.\)
Let \({\bf 2}\) and \({{\mathbf{\overline{2}}}}\) denote the two-element semi-Heyting chains whose operation \(\to\) satisfies \(0 \to 1 = 1\) and \(0 \to 1 = 0,\) respectively. It is easy to see that \({\bf 2}\) is a Heyting algebra (which is also a Boolean algebra), while \({{\mathbf{\overline{2}}}}\) is not.
Let \({\bf L}\) be either \({\bf 2}\) or \({{\mathbf{\overline{2}}}}.\) Let \({{\mathbf{L}}}'=\langle L', \wedge, \vee, 0, 1 \rangle\) be a chain with \(L': 0 < b < 1\) and \(b \notin L.\) If \({\bf L}\) is a subalgebra of \({\bf L}',\) then we have just to determine the elements \(0 \to b\) and \(b \to 1.\) From Lemma 2.5, if \(0 \to 1 = 0\) then \(0 \to b = 0\) and if \(0 \to 1 = 1\) then \(0 \to b \, \geq \, b.\) In addition, \(b \to 1 \, \geq \, b.\)
So we have
Corollary 3.9
There are two non-isomorphic\(3\)-element semi-Heyting chains that contain\({\bf 2}\)as a subalgebra and\(\text{four}\)non-isomorphic\(3\)-element semi-Heyting chains that contain\({{\mathbf{\overline{2}}}}\)as a subalgebra.
Theorem 3.10
Let\(n \, \geq \, 3.\)There is an\((n+1)\)-element semi-Heyting chain that does not have an n-element subalgebra.
Proof
Let \(L': b_{0} < b_{1} < \cdots < b_{n-1} < b_{n}.\) We define \(\to: L' \times L' \to L'\) by means of
It is easy to prove that \(\to\) is a semi-Heyting implication.
Suppose that there exists a subalgebra \({\bf L}\) of \({\bf L}'\) and \(|{\bf L}| = n.\) Then there exists \(0 < k < n\) such that \(L = L' \setminus \{b_{k}\}.\) Now, \(b_{k} = b_{k-1} \to b_{k+1},\) and then \(b_k \in L,\) a contradiction. \(\square\)
4 Equational basis for \({{\mathcal{CSH}}}\)
In this section, we give equational bases for the variety \({{\mathcal{CSH}}}\) and some of its subvarieties.
Lemma 4.1
Let L be a semi-Heyting chain. Then L satisfies the following identity
Proof
Let \(a, b \in L.\) As L is a chain, \(a \, \leq \, b\) or \(b < a.\) If \(a \, \leq \, b,\) by Lemma 1.2, \(a \, \leq \, a \to b.\) Hence, \(a \vee (a \to b)= a \to b,\) so \((a \vee (a \to b)) \to (a \to b) = (a \to b) \to (a \to b) = 1.\) If \(b < a,\) \(b = b \wedge a,\) and then \(b \to (a \wedge b) = b \to b = 1.\) \(\square\)
Lemma 4.2
Let\({\mathbb{V}}\)be the subvariety of\({{\mathcal{SH}}}\)defined by (Ch). If\({\bf L} \in {\mathbb{V}}\)is subdirectly irreducible, then\({\bf L}\)is a chain.
Proof
Let \({\bf L} \in {\mathbb{V}}\) subdirectly irreducible and let \(a, b \in L.\) Since L satisfies (Ch), \(((a \vee (a \rightarrow b)) \rightarrow (a \rightarrow b)) \vee (b \rightarrow (a \wedge b)) = 1.\) As L is subdirectly irreducible, 1 is \(\vee\)-irreducible. Thus, \((a \vee (a \rightarrow b)) \rightarrow (a \rightarrow b) = 1\) or \(b \rightarrow (a \wedge b) = 1.\)
Suppose that \((a \vee (a \rightarrow b)) \rightarrow (a \rightarrow b) = 1.\) Then \(a \vee (a \rightarrow b) \, \leq \, a \to b,\) and thus, \(a \, \leq \, a \vee (a \rightarrow b) \, \leq \, a \to b.\) Hence, \(a \wedge b = a \wedge (a \to b) = a,\) so \(a \, \leq \, b.\)
If \(b \rightarrow (a \wedge b) = 1,\) then \(b \, \leq \, a\wedge b,\) and thus, \(b \, \leq \, a.\)
Hence, \({\bf L}\) is a chain. \(\square\)
From Lemmas 4.1 and 4.2, we have the following:
Theorem 4.3
An equational basis for \({{\mathcal{CSH}}}\) relative to \({{\mathcal{SH}}}\) is given by
Let \({{\mathcal{C}}}_{n}\) denote the subvariety of \({{\mathcal{SH}}}\) generated by all the \(n\)-element chains, \(n \, \geq \, 2.\)
It is easily seen that the following theorem holds.
Theorem 4.4
An equational basis for \({{\mathcal{C}}}_{2}\) is given by
and
Observe that \({{\mathcal{C}}}_{2}\) is the subvariety generated by the algebras 2 and \({{\mathbf{\overline{2}}}},\) which have the 2-element chain as their lattice reduct and whose operation \(\to\) satisfies \(0 \to 1 = 1\) and \(0 \to 1 = 0,\) respectively.
Theorem 4.5
An equational basis for \({{\mathcal{C}}}_{n}\) with \(n \, \geq \, 3\) is given by the identities
and
Proof
Let \({\bf L}\) be an \(n\)-element semi-Heyting chain, \(L: 0 < a_{1} < a_{2} < \cdots < a_{n-2} < 1.\) By Theorem 4.3, \({\bf L}\) satisfies (Ch).
Let us prove that L satisfies \({\rm (H_n)}.\) Let \(z_{1}, z_{2}, \ldots, z_{n-1} \in L.\) If \(z_{k} \in \{0,1\}\) for some k, \(1 \, \leq \, k \, \leq \, n-1,\) then \(z_{k} \vee z_{k}^{*} = 1.\) Suppose that \(z_{k} \not\in \{0,1\}\) for every \(k,\) \(1 \, \leq \, k \, \leq \, n-1,\) i.e., \(z_{k} \in \{a_{1}, a_{2}, \ldots, a_{n-2}\}\) for every \(k.\) Then there exists \(j < i\) such that \(z_{i} = z_{j},\) and so \(z_{i} \to z_{j} = 1.\) Hence, L satisfies \({\rm (H_n)}.\)
Let \({\mathbb{V}}\) be the subvariety of \({{\mathcal{SH}}}\) defined by \({\rm (Ch)}\) and \({\rm (H_n)}\) and consider a subdirectly irreducible algebra \({{\mathbf{L}}} \in {\mathbb{V}}.\) By Theorem 4.3, \({\bf L} \in {{\mathcal{CSH}}},\) and by Lemma 4.2, \(L\) is a chain. Suppose that exists \(a_{1}, a_{2}, \ldots, a_{n-2}, a_{n-1} \in L\) such that \(0 < a_{1} < a_{2} < \cdots < a_{n-2} < a_{n-1} < 1.\) Then \(\bigvee_{i=1}^{n-1} \left( a_{i} \vee a_{i}^{*} \right) = \bigvee_{i=1}^{n-1} \left( a_{i} \vee 0 \right) = a_{n-1}.\) By hypothesis, \(a_{n-1} \vee \bigvee_{i,j=1;j < i}^{n-1} \left( a_{i} \to a_{j} \right) = 1,\) and as \(1\) is \(\vee\)-irreducible, \(a_{i} \to a_{j} = 1\) for some \(j < i.\) Hence, \(a_{i} \, \leq \, a_{j},\) a contradiction. Thus, \(|L| \, \leq \, n,\) and consequently \({\bf L} \in {{\mathcal{C}}}_{n}.\) \(\square\)
As an immediate corollary of Theorem 4.3 we can determine equational bases for the following subvarieties of \({{\mathcal{CSH}}}\) introduced in [5]: \({{\mathcal{CFTT}}} (0 \rightarrow 1 \approx 1), {{\mathcal{CFTD}}} ((0 \rightarrow 1)^{*} \approx 0),\) quasi-Heyting chains \((y \, \leq \, x \rightarrow y),\) Heyting chains \( ((x \wedge y) \rightarrow x \approx 1), {{\mathcal{CFTF}}} (0 \rightarrow 1 \approx 0)\) and conmutative semi-Heyting chains \((x \rightarrow y \approx y \rightarrow x).\)
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Acknowledgments
The support of CONICET is grateful acknowledged by J. M. Cornejo and J. P. Díaz Varela.
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J. M. Cornejo dedicates this work to his wife, Carina Foresi.
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Abad, M., Cornejo, J.M. & Díaz Varela, J.P. The variety generated by semi-Heyting chains. Soft Comput 15, 721–728 (2010). https://doi.org/10.1007/s00500-010-0604-0
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DOI: https://doi.org/10.1007/s00500-010-0604-0