International Mathematical Forum, 5, 2010, no. 12, 557-571 Triregular Leftmost without Loop and Reverse Arc Graph Varieties of Graph Algebra of Type (2,0) Montri Thongmoon and Tiang Poomsa-ard Department of Mathematics, Faculty of Science Mahasarakham University, Mahasarakham 44150, Thailand montri.t@msu.ac.th, tiang@kku.ac.th Abstract Graph algebras establish a connection between directed graphs without multiple edges and special universal algebras of type (2,0). We say that a graph G satisfies a term equation s t if the corresponding graph algebra A(G) satisfies s t. A class of graph algebras V is called a graph variety if V = Mod g Σ where Σ is a subset of T (X) T (X). A graph variety V = Mod g Σ is called a triregular leftmost without loop and reverse arc graph variety if Σ is a set of triregular leftmost without loop and reverse arc term equations. In this paper we characterize all triregular leftmost without loop and reverse arc graph varieties. For varieties and other basic concept of universal algebra see e.g. [2] Mathematics Subject Classification: 05B25, 08B15 Keywords: varieties, triregular leftmost without loop and reverse arc graph varieties, term, binary algebra, graph algebra 1 Introduction. Graph algebras have been invented in [12] to obtain examples of nonfinitely based finite algebras. To recall this concept, let G = (V, E) be a (directed) graph with the vertex set V and the set of edges E V V. Define the graph algebra A(G) corresponding to G with the underlying set V { }, where is a symbol outside V, and with two basic operations, namely a nullary operation pointing to and a binary one denoted by juxtaposition, given for
558 M. Thongmoon and T. Poomsa-ard u, v V { } by uv = { u, if (u, v) E,, otherwise. In [11] graph varieties had been investigated for finite undirected graphs in order to get graph theoretic results (structure theorems) from universal algebra via graph algebras. In [10] these investigations are extended to arbitrary (finite) directed graphs where the authors ask for a graph theoretic characterization of graph varieties, i.e., of classes of graphs which can be defined by term equations for their corresponding graph algebras. The answer is a theorem of Birkhoff-type, which uses graph theoretic closure operations. A class of finite directed graphs is equational (i.e., a graph variety) if and only if it is closed with respect to finite restricted pointed subproducts and isomorphic copies. In [6] T. Poomsa-ard characterized hyperidentitis in the class of associative graph algebras. In [7], [8] T. Poomsa-ard, J. Wetweerapong and C. Samartkoon characterized hyperidentitis in the class of idempotent graph algebras and the class of transitive graph algebras respectively. In [4] J. Khampakdee and T. Poomsa-ard characterized hyperidentities in the class of (xy)x x(yy). In [3] M. Krapeedaeng and T. Poomsa-ard characterized all biregular leftmost graph varieties. In [1] A. Anantpinitwatna and T. Poomsa-ard characterized hyperidentities in biregular leftmost graph varieties. We say that a graph variety V = Mod g Σ is called a triregular without loop and reverse arc graph variety if Σ is a set of triregular without loop and reverse arc term equations. In this paper we characterize all triregular without loop and reverse arc graph varieties. 2 Terms and graph varieties In [9] R. Pöschel was introduced terms for graph algebras, the underlying formal language has to contain a binary operation symbol (juxtaposition) and a symbol for the constant (denoted by, too). Definition 2.1. The set T (X) of all terms over the alphabet is defined inductively as follows: X = {x 1, x 2, x 3,...} (i) every variable x i, i = 1, 2, 3,..., and are terms; (ii) if t 1 and t 2 are terms, then t 1 t 2 is a term;
Triregular leftmost without loop and reverse arc graph varieties 559 (iii) T (X) is the set of all terms which can be obtained from (i) and (ii) in finitely many steps. Terms built up from the two-element set X 2 = {x 1, x 2 } of variables are thus binary terms. We denote the set of all binary terms by T (X 2 ). The leftmost variable of a term t is denoted by L(t) and rightmost variable of a term t is denoted by R(t). A term, in which the symbol occurs is called a trivial term. Definition 2.2. For each non-trivial term t of type τ = (2, 0) one can define a directed graph G(t) = (V (t), E(t)), where the vertex set V (t) is the set of all variables occurring in t and the edge set E(t) is defined inductively by E(t) = φ if t is a variable and E(t 1 t 2 ) = E(t 1 ) E(t 2 ) {(L(t 1 ), L(t 2 ))} where t = t 1 t 2 is a compound term. L(t) is called the root of the graph G(t), and the pair (G(t), L(t)) is the rooted graph corresponding to t. Formally, we assign the empty graph φ to every trivial term t. Definition 2.3. A non-trivial term t of type τ = (2, 0) is called a term without loop and reverse arc if and only if for any x V (t), (x, x) / E(t) and for any x, y V (t) with x y if (x, y) E(t), then (y, x) / E(t). A term equation s t of type τ = (2, 0) is called triregular leftmost without loop and reverse arc term equation if and only if V (s) = V (t), V (t) = 3, L(s) = L(t) and s, t are terms without loop and reverse arc. Definition 2.4. We say that a graph G = (V, E) satisfies a term equation s t if the corresponding graph algebra A(G) satisfies s t (i.e., we have s = t for every assignment V (s) V (t) V { }), and in this case, we write G = s t. Given a class G of graphs and a set Σ of term equations (i.e., Σ T (X) T (X)) we introduce the following notation: G = Σ if G = s t for all s t Σ, G = s t if G = s t for all G G, G = Σ if G = Σ for all G G, IdG = {s t s, t T (X), G = s t}, Mod g Σ = {G G is a graph and G = Σ}, V g (G) = Mod g IdG. V g (G) is called the graph variety generated by G and G is called graph variety if V g (G) = G. G is called equational if there exists a set Σ of term equations such that G = Mod g Σ. Obviously V g (G) = G if and only if G is an equational class. In [9] it was shown that any non-trivial term t over the class of graph algebras has a uniquely determined normal form term NF (t) and there is an
560 M. Thongmoon and T. Poomsa-ard algorithm to construct the normal form term to a given term t. Now, we want to describe how to construct the normal form term. Let t be a non-trivial term. The normal form term of t is the term NF (t) constructed by the following algorithm: (i) Construct G(t) = (V (t), E(t)). (ii) Construct for every x V (t) the list l x = (x i1,..., x ik(x) ) of all outneighbors (i.e., (x, x ij ) E(t), 1 j k(x)) ordered by increasing indices i 1... i k(x) and let s x be the term (...((xx i1 )x i2 )...x ik(x) ). (iii) Starting with x := L(t), Z := V (t), s := L(t), choose the variable x i Z V (s) with the least index i, substitute the first occurrence of x i by the term s xi, denote the resulting term again by s and put Z := Z \ {x i }. Continue this procedure while Z φ. The resulting term is the normal form NF (t). The algorithm stops after a finite number of steps, since G(t) is a rooted graph. Without difficulties one shows G(NF (t)) = G(t), L(NF (t)) = L(t). Definition 2.5. Let G = (V, E) and G = (V, E ) be graphs. A homomorphism h from G into G is a mapping h : V V carrying edges to edges,that is, for which (u, v) E implies (h(u), h(v)) E. Further in [5] it was proved the following proposition: Proposition 2.1. Let G = (V, E) be a graph and let h : X { } V { } be an evaluation of the variables such that h( ) =. Consider the canonical extension of h to the set of all terms. Then there holds: if t is a trivial term then h(t) =. Otherwise, if h : G(t) G is a homomorphism of graphs, then h(t) = h(l(t)), and if h is not a homomorphism of graphs, then h(t) =. By the property of the normal form term and Proposition 2.1, we have the following proposition: Proposition 2.2. Let G = (V, E) be a graph s and t be non-trivial terms. Then G = s t if and only if G = NF (s) NF (t). 3 Triregular leftmost without loop and reverse arcs graph varieties We see that if s and t are terms such that V (s) V (t) or L(s) L(t), then there exists a complete graph (a complete graph which has more than one vertices) which does not belong to the graph variety Mod g {s t}. In this study we will investigate only the graph varieties which contain all complete graphs. By Proposition 2.2, we see that for any Σ T (X) T (X) and Σ is the set of term equations NF (s) NF (t), if s t Σ. Then Mod g Σ
Triregular leftmost without loop and reverse arc graph varieties 561 and Mod g Σ are the same graph variety. Hence if we want to find all triregular leftmost without loop and reverse arcs graph varieties, then it is enough to find all graph varieties Mod g Σ such that Σ is any subset of T T, where T = {s 1, s 2, s 3, s 4, s 5, s 6, s 7 } and s 1 = x(yz), s 2 = x(zy), s 3 = (xy)z, s 4 = x(y(zx)), s 5 = x(z(yx)), s 6 = (x(yz))z, s 7 = (xy)(zy)}. Clearly, for each i = 1, 2, 3,..., 7, K 0 = Mod g {s i s i } is the set of all graph algebras. In [5] it was proved: Proposition 3.1. Let s and t be non-trivial terms from T (X) with variables V (s) = V (t) = {x 0, x 1,..., x n } and L(s) = L(t). Then a graph G = (V, E) satisfies s t if and only if the graph algebra A(G) has the following property: A mapping h : V (s) V is a homomorphism from G(s) into G iff it is a homomorphism from G(t) into G. Proposition 3.1 gives a method to check whether a graph G = (V, E) satisfies the term equation s t. The following are all graphs with at most two vertices which satisfy at least one term equation s i s j, i = 1, 2,..., 7, j = 1, 2,..., 7, i j. G 1 G 2 G 3 G 4 G 5 G 6 G 7 G 8 G 9 The following are all connected graphs with three vertices which each edges are above graphs. G 10 G 11 G 12 G 13 G 14 G 15 G 16 G 17 G 18 G 19 G 20 G 21 G 22 G 23
562 M. Thongmoon and T. Poomsa-ard G 27 G 24 G 25 G 26 G 28 G 29 G 30 G 31 G 32 G 33 G 34 G 35 G 36 G 37 G 38 G 39 G 40 G 41 G 42 G 43 G 44 G 45 G 46 G 47 G 48 G 49 G 50 G 51 G 52 G 53 G 54 G 55 G 56 G 57 G 58 G 59 G 60 G 61 G 62 G 63 G 64 G 65 G 66 G 67 G 68 G 69 G 70 G 71 G 72
Triregular leftmost without loop and reverse arc graph varieties 563 G 73 G 74 G 75 G 76 G 77 G 78 G 79 G 80 G 81 G 82 G 83 G 84 G 85 G 86 G 87 G 88 G 89 G 90 G 91 G 92 G 93 G 94 G 95 Then all triregular leftmost without loop and reverse arc graph varieties are characterized by the following theorems: Theorem 3.1. Let G = (V, E) be a graph and K 1 = Mod g {x(yz) x(zy)}. Then G K 1 if and only if for any a, b, c V, (a, b), (b, c) E if and only if (a, c), (c, b) E. Proof. Let G = (V, E) be a graph. Suppose that G K 1 and for any a, b, c V, suppose that (a, b), (b, c) E. Let s and t be non-trivial terms such that s = x(yz) and t = x(zy) and let h : V (s) V be a function such that h(x) = a, h(y) = b and h(z) = c. We see that h is a homomorphism from G(s) into G. By Proposition 3.1, we have h is a homomorphism from G(t) into G. Since (x, z), (z, y) E(t), we have (h(x), h(z)) = (a, c) E and (h(z), h(y)) = (c, b) E. In the same way, if (a, c), (c, b) E, then we can prove that (a, b), (b, c) E. Conversely, suppose that G = (V, E) be a graph which has property that, for any a, b, c V, (a, b), (b, c) E if and only if (a, c), (c, b) E. Let s and t be non-trivial terms such that s = x(yz) and t = x(zy) and let h : V (s) V be a function. Suppose that h is a homomorphism from G(s) into G. Since
564 M. Thongmoon and T. Poomsa-ard (x, y), (y, z) E(s), we have (h(x), h(y)), (h(y), h(z)) E. By assumption we get (h(x), h(z)), (h(z), h(y)) E. Hence h is a homomorphism from G(t) into G. In the same way we can prove that, if h is a homomorphism from G(t) into G, then it is a homomorphism from G(s) into G. Then, by Proposition 3.1 we get A(G) satisfies s t. By Theorem 3.1, we see that the graphs which belong to K 1 are the graphs G i A, A = { G 1,G 2,G 3,G 4,G 9,G 10,G 11,G 14,G 24,G 26,G 64,G 95 } and all graphs such that every induced subgraph with at most three vertices of each component is one of these graphs. Theorem 3.2. Let G = (V, E) be a graph and K 2 = Mod g {x(yz) (xy)z}. Then G K 2 if and only if for any a, b, c V, if (a, b) E, then (b, c) E if and only if (a, c) E. Proof. Let G = (V, E) be a graph. Suppose that G K 2 and for any a, b, c V, suppose that (a, b), (b, c) E. Let s and t be non-trivial terms such that s = x(yz) and t = (xy)z and let h : V (s) V be a function such that h(x) = a, h(y) = b and h(z) = c. We see that h is a homomorphism from G(s) into G. By Proposition 3.1, we have h is a homomorphism from G(t) into G. Since (x, z) E(t), we have (h(x), h(z)) = (a, c) E. In the same way, if (a, b), (a, c) E, then we can prove that (b, c) E. Conversely, suppose that G = (V, E) be a graph which has property that, for any a, b, c V, if (a, b) E, then (b, c) E if and only if (a, c) E. Let s and t be non-trivial terms such that s = x(yz) and t = (xy)z and let h : V (s) V be a function. Suppose that h is a homomorphism from G(s) into G. Since (x, y), (y, z) E(s), we have (h(x), h(y)), (h(y), h(z)) E. By assumption we get (h(x), h(y)), (h(x), h(z)) E. Hence h is a homomorphism from G(t) into G. In the same way we can prove that, if h is a homomorphism from G(t) into G, then it is a homomorphism from G(s) into G. Then, by Proposition 3.1 we get A(G) satisfies s t. By Theorem 3.2, we see that the graphs which belong to K 2 are the graphs G i B, B = {G 1,G 2,G 4,G 9,G 14,G 26,G 64,G 95 } and all graphs such that every induced subgraph with at most three vertices of each component is one of these graphs. Theorem 3.3. Let G = (V, E) be a graph and K 3 = Mod g {x(yz) x(y(zx))}. Then G K 3 if and only if for any a, b, c V if (a, b), (b, c) E, then (c, a) E. Proof. Let G = (V, E) be a graph. Suppose that G K 3 and for any a, b, c V, suppose that (a, b), (b, c) E. Let s and t be non-trivial terms such that s = x(yz) and t = x(y(zx)) and let h : V (s) V be a function such that h(x) = a, h(y) = b and h(z) = c. We see that h is a homomorphism from G(s)
Triregular leftmost without loop and reverse arc graph varieties 565 into G. By Proposition 3.1, we have h is a homomorphism from G(t) into G. Since (z, x) E(t), we have (h(z), h(x)) = (c, a) E. Conversely, suppose that G = (V, E) be a graph which has property that, for any a, b, c V, if (a, b), (b, c) E, then (c, a) E. Let s and t be nontrivial terms such that s = x(yz) and t = x(y(zx)) and let h : V (s) V be a function. Suppose that h is a homomorphism from G(s) into G. Since (x, y), (y, z) E(s), we have (h(x), h(y)), (h(y), h(z)) E. By assumption we get (h(z), h(x)) E. Hence h is a homomorphism from G(t) into G. Clearly that, if h is a homomorphism from G(t) into G, then it is a homomorphism from G(s) into G. Then, by Proposition 3.1 we get A(G) satisfies s t. By Theorem 3.3, we see that the graphs which belong to K 3 are the graphs G i C, C = { G 1,G 2 G 3,G 9,G 10,G 24,G 66,G 95 } and all graphs such that every induced subgraph with at most three vertices of each component is one of these graphs. Theorem 3.4. Let G = (V, E) be a graph and K 4 = Mod g {x(yz) x(z(yx))}. Then G K 4 if and only if for any a, b, c V, (a, b), (b, c) E if and only if (a, c), (c, b), (b, a) E. Proof. The proof is similar as the proof of Theorem 3.1. By Theorem 3.4, we see that the graphs which belong to K 4 are the graphs G i D, D = {G 1,G 2,G 3,G 9,G 10,G 24,G 95 } and all graphs such that every induced subgraph with at most three vertices of each component is one of these graphs. Theorem 3.5. Let G = (V, E) be a graph and K 5 = Mod g {x(yz) (x(yz))z}. Then G K 5 if and only if for any a, b, c V if (a, b), (b, c) E, then (a, c) E. Proof. The proof is similar as the proof of Theorem 3.3. By Theorem 3.5, we see that the graphs which belong to K 5 are the graphs G i E, E = {G 1,G 2,...,G 6,G 9,...,G 15,G 24,...,G 29,G 52,...,G 59,G 64,G 65,G 82,G 83, G 95 } and all graphs such that every induced subgraph with at most three vertices of each component is one of these graphs. Theorem 3.6. Let G = (V, E) be a graph and K 6 = Mod g {(xy)z x(y(zx))}. Then G K 6 if and only if for any a, b, c V, if (a, b) E, then (a, c) E if and only if (b, c), (c, a) E. Proof. The proof is similar as the proof of Theorem 3.2. By Theorem 3.6, we see that the graphs which belong to K 6 are the graphs G i F, F = {G 1,G 2,G 9,G 95 } and all graphs such that every induced subgraph with at most three vertices of each component is one of these graphs.
566 M. Thongmoon and T. Poomsa-ard Theorem 3.7. Let G = (V, E) be a graph and K 7 = Mod g {(xy)z (x(yz))z}. Then G K 7 if and only if for any a, b, c V if (a, b), (a, c) E, then (b, c) E. Proof. The proof is similar as the proof of Theorem 3.3. By Theorem 3.7, we see that the graphs which belong to K 7 are the graphs G i H, H = {G 1,G 2,G 4,G 9,G 14,G 26,G 34,G 51,G 64,G 95 } and all graphs such that every induced subgraph with at most three vertices of each component is one of these graphs. Theorem 3.8. Let G = (V, E) be a graph and K 8 = Mod g {x(y(zx)) x(z(yx))}. Then G K 8 if and only if for any a, b, c V, (a, b), (b, c), (c, a) E if and only if (a, c), (c, b), (b, a) E. Proof. The proof is similar as the proof of Theorem 3.1. By Theorem 3.8, we see that the graphs which belong to K 8 are the graphs G i I, I = {G 1,G 2,...,G 65,G 78,...;G 83,G 92,...,G 95 } and all graphs such that every induced subgraph with at most three vertices of each component is one of these graphs. Theorem 3.9. Let G = (V, E) be a graph and K 9 = Mod g {x(y(zx)) (x(yz))z}. Then G K 9 if and only if for any a, b, c V, if (a, b), (b, c) E, then (c, a) E if and only if (a, c) E. Proof. The proof is similar as the proof of Theorem 3.2. By Theorem 3.9, we see that the graphs which belong to K 9 are the graphs G i J, J = {G 1,G 2,G 3,G 7,G 8,G 9,G 10,G 16,G 24,G 30,G 31,G 38,G 39,G 46,...,G 51, G 92,...,G 95 } and all graphs such that every induced subgraph with at most three vertices of each component is one of these graphs. Theorem 3.10. Let G = (V, E) be a graph and K 10 = Mod g {(x(yz))z (xy)(zy)}. Then G K 10 if and only if for any a, b, c V, if (a, c), (a, b) E, then (b, c) E if and only if (c, b) E. Proof. The proof is similar as the proof of Theorem 3.2. By Theorem 3.10, we see that the graphs which belong to K 10 are the graphs G i L, L = {G 1,...,G 4,G 7,...,G 11,G 14,G 16,G 19, G 24,G 26,G 30,G 31,G 32,G 34,G 38, G 39,G 41,G 43,G 46,...,G 52,G 55,G 60, G 61,G 64,G 66,G 70,G 73,G 78,G 80,G 89,G 92,...G 95 } and all graphs such that every induced subgraph with at most three vertices of each component is one of these graphs. By checking the graphs in K 1 and the graphs in Mod g {x(yz) (xy)(zy)}, we found that they are the same graph variety. After that we use the Proposition 3.1 recheck again as the following theorem:
Triregular leftmost without loop and reverse arc graph varieties 567 Theorem 3.11. Let K = Mod g {x(yz) (xy)(zy)}. Then K = K 1. Proof. Let G = (V, E) be a graph. Suppose that G K 1. Let s, t and t are terms such that s = x(yz), t = (xy)(zy) and t = x(zy) and let h : V (s) V be a function. Suppose that h is a homomorphism from G(s) into G. Since (x, y), (y, z) E(s), we have (h(x), h(y)), (h(y), h(z)) E. Since G K 1. By Theorem 3.1 we have (h(x), h(z)), (h(z), h(y)) E. So h is a homomorphism from G(t ) into G. From (h(x), h(y)), (h(x), h(z)), (h(z), h(y)) E, we have that h is a homomorphism from G(t) into G. Suppose that h is a homomorphism from G(t) into G. Since (x, y), (z, y), (x, z) E(t), we have (h(x), h(y)), (h(z), h(y)), (h(x), h(z)) E. Since G K 1 and (h(x), h(z)), (h(z), h(y)) E. By Theorem 3.1 we have (h(x), h(y)), (h(y), h(z)) E. So we have that h is a homomorphism from G(s) into G. Then, by Proposition 3.1 we get G K. Let G K. Suppose that h is a homomorphism from G(s) into G. Since (x, y), (y, z) E(s), we have (h(x), h(y)), (h(y), h(z)) E. Since G K. By Proposition 3.1 we have h is a homomorphism from G(t) into G. Since (z, y), (x, z) E(t), we get (h(z), h(y)), (h(x), h(z)) E. Hence h is a homomorphism from G(t ) into G. Suppose that h is a homomorphism from G(t ) into G. Since (x, z), (z, y), E(t ), we have (h(x), h(z)), (h(z), h(y)) E. Let h : V (t ) V be a function such that h (x) = h(x), h (y) = h(z) and h (z) = h(y). Hence h is a homomorphism from G(s) into G. Since G K, we get that h is a homomorphism from G(t) into G. Since (x, z), (z, y) E(t) we have (h (x), h (z)) = (h(x), h(y)) E and (h (z), h (y)) = (h(y), h(z)) E. Hence h is a homomorphism from G(s) into G. Then, by Proposition 3.1 we get G K 1. Hence K = K 1. In the similar way, we have the following K 1 = Mod g {x(yz) x(zy)} = Mod g {x(yz) (xy)(zy)} = Mod g {x(zy) (x(yz))z}; K 2 = Mod g {x(yz) (xy)z} = Mod g {x(zy) (xy)z}; K 3 = Mod g {x(yz) x(y(zx))} = Mod g {x(zy) x(z(yx))}; K 4 = Mod g {x(yz) x(z(yx))} = Mod g {x(zy) x(y(zx))}; K 5 = Mod g {x(yz) (x(yz))z} = Mod g {x(zy) (xy)(zy)}; K 6 = Mod g {(xy)z x(y(zx))} = Mod g {(xy)z x(z(yx))};
568 M. Thongmoon and T. Poomsa-ard K 7 = Mod g {(xy)z (x(yz))z} = Mod g {(xy)z (xy)(zy)}; K 9 = Mod g {x(y(zx)) (x(yz))z} = Mod g {x(y(zx)) (xy)(zy)} = Mod g {x(z(yx)) (x(yz))z} = Mod g {x(z(yx)) (xy)(zy)}. Hence there are only ten triregular leftmost without loop and reverse arc graph varieties Mod g {s i s j } with i j for all i = 1, 2, 3,..., 7, j = 1, 2, 3,..., 7. Since for any Σ T T, we have the triregular leftmost without loop and reverse arc graph varieties Mod g Σ = s t Σ Mod g{s t}. Therefore we can find all triregular leftmost without loop and reverse arc graph varieties in the following way: At first let A = {K 0, K 1,..., K 10 }. Find the intersection between two elements of A. If there are some of them do not belong to A, then add them into A to get the set A 1. Do the same manner for A 1 to get A 2. Continuing this process until to get the set A n such that all intersection between two elements of A n belong to itself. Then, A n close under inclusion. Let K 11 = K 5 K 10. Then by Theorem 3.5 and Theorem 3.10 we see that the graphs which belong to K 11 are the graphs G i M, M = {G 1, G 2,G 3,G 4,G 9, G 10,G 11,G 14,G 24,G 26,G 52,G 55,G 64,G 95 } and all graphs such that every induced subgraph with at most three vertices of each component is one of these graphs. Let K 12 = K 7 K 9. Then by Theorem 3.7 and Theorem 3.9 we see that the graphs which belong to K 12 are the graphs G i N, N = {G 1, G 2,G 9,G 51,G 95 } and all graphs such that every induced subgraph with at most three vertices of each component is one of these graphs. Let K 13 = K 8 K 10. Then by Theorem 3.8 and Theorem 3.10 we see that the graphs which belong to K 13 are the graphs G i O, O = {G 1, G 2,G 3,G 4,G 7,G 8, G 9,G 10,G 11,G 14, G 16,G 19,G 24,G 26,G 30,G 31,G 32,G 34,G 38, G 39,G 41,G 43,G 46,G 47, G 48,G 49,G 50,G 51, G 52,G 55,G 60,G 61,G 64,G 78,G 80,G 92,G 93,G 94,G 95 } and all graphs such that every induced subgraph with at most three vertices of each component is one of these graphs. Further we prove that K 3 K 9 = K 4 and K 3 K 10 as the following theorems. Theorem 3.12. K 3 K 9 = K 4. Proof. Let G = (V, E) be a graph. Suppose that G K 4. For any a, b, c V, suppose that (a, b), (b, c) E. Since G K 4, we have (a, c), (c, b), (b, a) E. From (a, c), (c, b) E and G K 4 we have (c, a) E. Hence G K 3. For any a, b, c V, suppose (a, b), (b, c), (c, a) E. From (b, c), (c, a) E and G K 4, we have (a, c) E. Suppose (a, b), (b, c), (a, c) E. From (a, b), (b, c) E and G K 4, we have (c, b), (b, a) E. From (a, c), (c, b) E and G K 4 we have (c, a) E. Hence G K 9. So G K 3 K 9. Therefore
Triregular leftmost without loop and reverse arc graph varieties 569 K 4 K 3 K 9. Suppose that G K 3 K 9. For any a, b, c V, suppose that (a, b), (b, c) E. Since G K 3, we have (c, a) E. From (a, b), (b, c), (c, a) E and G K 9, we have (a, c) E. From (c, a), (a, b), (b, c) E and G K 9, we have (c, b) E. From (b, c), (c, a), (a, b) E and G K 9 we have (b, a) E. For any a, b, c V, suppose that (a, c), (c, b), (b, a) E. Since G K 9, we have (a, b) E. From (b, a), (a, c), (c, b) E and G K 9, we have (b, c) E. Then, we have G K 4. Hence K 3 K 9 K 4. Therefore K 3 K 9 = K 4. Theorem 3.13. K 3 K 10. Proof. Let G = (V, E) be a graph. Let G K 3 and for any a, b, c V, suppose that (a, c), (a, b), (b, c) E. From (a, b), (b, c) E and G K 3, we have (c, a) E. From (c, a), (a, c) E and G K 3, we have (c, c) E. From (b, c), (c, c) E and G K 3, we have (c, b) E. Suppose that (a, c), (a, b), (c, b) E. From (a, c), (c, b) E and G K 3, we have (b, a) E. From (b, a), (a, b) E and G K 3, we have (b, b) E. From (c, b), (b, b) E and G K 3, we have (b, c) E. Then G K 10. Therefore K 3 K 10. In the similar way, we have K 8 K 10 = K 13, K 5 K 10 = K 11, K 7 K 9 = K 12, K 1 K 7 = K 2, K 2 K 4 = K 6, K 12 K 2 = K 6, K 12 K 4 = K 6, K 1 K 9 = K 4, K 1 K 3 = K 4, K 9 K 13, K 5 K 8, K 1 K 11, K 2 K 1, K 11 K 13 and K 7 K 13. We found that the set K = {K 0, K 1, K 3,..., K 13 } close under inclusion. Hence K is the set of all triregular leftmost without loop and reverse arc graph varieties. Since the intersection of any family of triregular leftmost without loop and reverse arc graph varieties is again a triregular leftmost without loop and reverse arc graph variety, K forms a poset under inclusion, in which two elements have a greatest lower bound (their intersection) and least upper bound (the intersection of all triregular leftmost without loop and reverse arc graph varieties which contain both of them), Thus we have a lattice (K;, ), called the triregular leftmost without loop and reverse arc lattice of graph varieties. The following is the Hasse diagram of this lattice.
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