**Proposition 5.1** *The class of* *d*-*algebras and the class of pre*-*idempotent groupoids are Smarandache disjoint*.

*Proof* Let (X,\ast ,0) be both a *d*-algebra and a pre-idempotent groupoid. Then a\ast b=(a\ast b)\ast (a\ast b)=0 and b\ast a=(b\ast a)\ast (b\ast a)=0, by pre-idempotence and (D2), for any a,b\in X. By (D3), it follows that a=b, which proves that |X|=1. □

**Proposition 5.2** *The class of groups and the class of pre*-*idempotent groupoids are Smarandache disjoint*.

*Proof* Let (X,\ast ,0) be both a group and a pre-idempotent groupoid. Then, for any x\in X, we have x=x\ast e=(x\ast e)\ast (x\ast e)=x\ast x. It follows that e=x\ast {x}^{-1}=(x\ast x)\ast {x}^{-1}=x\ast (x\ast {x}^{-1})=x\ast e=x, proving that |X|=1. □

A groupoid (X,\ast ) is said to be an {L}^{4}-*groupoid* if, for all a,b\in X,

(L1) ((b\ast a)\ast b)\ast (b\ast a)=b\ast a,

(L2) (b\ast a)\ast ((b\ast a)\ast b)=(b\ast a)\ast b.

**Example 5.3** Let X:=\mathbf{R} be the set of all real numbers. Define a map \phi :X\to X by \phi (x):=\lceil x\rceil -\frac{1}{2}, where \lceil x\rceil is the ceiling function. Then \phi (\phi (x))=\lceil \phi (x)\rceil -\frac{1}{2}=\lceil \lceil x\rceil -\frac{1}{2}\rceil -\frac{1}{2}=\lceil \phi (x)\rceil -\frac{1}{2}=\phi (x). Define a binary operation ‘∗’ on *X* by x\ast y:=\phi (y) for all x,y\in X. Then (X,\ast ) is an {L}^{4}-groupoid. In fact, for all a,b\in X, ((b\ast a)\ast b)\ast (b\ast a)=\phi (b)\ast \phi (a)=\phi (\phi (a))=\phi (a)=b\ast a. Moreover, (b\ast a)\ast ((b\ast a)\ast b)=\phi (a)\ast \phi (b)=\phi (\phi (b))=\phi (b)=(b\ast a)\ast b.

**Proposition 5.4** *Let* (X,\ast ) *be a groupoid and let* p(x)\in {Z}_{2}[[x]] *such that* p(x)={x}^{4}q(x) *for some* q(x)\in {Z}_{2}[[x]]. *If there exist* u,v\in X *such that* {[a,b]}_{p(x)}=\{a,b,u,v,u,v,{\alpha}_{1},{\alpha}_{2},\dots \}, *where* {\alpha}_{i}\in {Z}_{2} *for any* a,b\in X, *then* (X,\ast ) *is an* {L}^{4}-*groupoid*.

*Proof* Since p(x)={x}^{4}q(x), we have {\bigwedge}_{p(x)}=\{L,L,L,L,{\alpha}_{1},{\alpha}_{2},\dots \}, where {\alpha}_{i}\in \{L,R\}. It follows that {[a,b]}_{p(x)}=\{a,b,b\ast a,(b\ast a)\ast b,((b\ast a)\ast b)\ast (b\ast a),(((b\ast a)\ast b)\ast (b\ast a))\ast ((b\ast a)\ast b),\dots \}. This shows that u=b\ast a, v=(b\ast a)\ast b, and hence ((b\ast a)\ast b)\ast (b\ast a)=b\ast a and (b\ast a)\ast ((b\ast a)\ast b)=(b\ast a)\ast b, proving the proposition. □

**Proposition 5.5** *Every* {L}^{4}-*groupoid is pre*-*idempotent*.

*Proof* Given a,b\in X, we have (b\ast a)\ast (b\ast a)=((b\ast a)\ast b)\ast (b\ast a)=b\ast a, proving the proposition. □

**Proposition 5.6** *The class of* {L}^{4}-*groupoids and the class of groups are Smarandache disjoint*.

*Proof* Let (X,\ast ) be both an {L}^{4}-groupoid and a group with identity *e*. Then ((b\ast a)\ast b)\ast (b\ast a)=b\ast a for all a,b\in X. Since any group has the cancellation laws, we obtain (b\ast a)\ast b=e. If we apply this to (L2), then we have (b\ast a)\ast e=e. This means that b\ast a=e. It follows that e=b\ast a=((b\ast a)\ast b)\ast (b\ast a)=(e\ast b)\ast e=b, proving that |X|=1. □

**Proposition 5.7** *The class of* {L}^{4}-*groupoids and the class of* *BCK*-*algebras are Smarandache disjoint*.

*Proof* Let (X,\ast ) be both an {L}^{4}-groupoid and a *BCK*-algebra with a special element 0\in X. Given a,b\in X, we have 0=0\ast a=(b\ast b)\ast a=(b\ast a)\ast b=(b\ast a)\ast ((b\ast a)\ast b)=(b\ast a)\ast 0=b\ast a. Similarly, a\ast b=0. Since *X* is a *BCK*-algebra, a=b for all a,b\in X, proving that X=\{0\}. □

Let p(x)\in {Z}_{2}[[x]]. A groupoid (X,\ast ) is said to be a *Fibonacci semi-lattice* if for any a,b\in X, there exists u=u(a,b,p(x)) in *X* depending on *a*, *b*, p(x) such that {[a,b]}_{p(x)}=\{a,b,u,u,\dots \}.

Note that every Fibonacci semi-lattice is a pre-idempotent groupoid satisfying one of the conditions (b\ast a)\ast b=a\ast b, b\ast (b\ast a)=b\ast a, b\ast (a\ast b)=a\ast b, (a\ast b)\ast b=a\ast b separately (and simultaneously).

**Proposition 5.8** *Let* (X,\ast ) *be a groupoid and let* p(x)\in {Z}_{2}[[x]] *such that* p(x)={x}^{3}q(x) *for some* q(x)\in {Z}_{2}[[x]] *with* {q}_{0}=1. *Then* (X,\ast ) *is a Fibonacci semi*-*lattice*.

*Proof* Since p(x)={x}^{3}q(x), where q(x)\in {Z}_{2}[[x]] with {q}_{0}=1, we have {\bigwedge}_{p(x)}=\{L,L,L,R,{\alpha}_{1},{\alpha}_{2},\dots \}, where {\alpha}_{i}\in \{L,R\}. If we let u:=b\ast a, v:=(b\ast a)\ast b, then {[a,b]}_{p(x)}=\{a,b,b\ast a=u,(b\ast a)\ast b=v,((b\ast a)\ast b)\ast (b\ast a)=v\ast u=u,\dots \}. It follows that b\ast a=u=v\ast u=v=(b\ast a)\ast b. Hence {[a,b]}_{p(x)}=\{a,b,b\ast a,b\ast a,\dots \}, proving the proposition. □

A groupoid (X,\ast ) is said to be an {\mathit{LRL}}^{2}-*groupoid* if for all a,b\in X,

({\mathit{LRL}}^{2}1) (b\ast (b\ast a))\ast (b\ast a)=b\ast a,

({\mathit{LRL}}^{2}2) (b\ast a)\ast (b\ast (b\ast a))=b\ast (b\ast a).

**Proposition 5.9** *Let* (X,\ast ) *be a groupoid and let* p(x)\in {Z}_{2}[[x]] *such that* {p}_{0}={p}_{2}={p}_{3}=0, {p}_{1}=1. *Then* (X,\ast ) *is an* {\mathit{LRL}}^{2}-*groupoid*.

*Proof* Since p(x)\in {Z}_{2}[[x]] such that {p}_{0}={p}_{2}={p}_{3}=0, {p}_{1}=1, we have {\bigwedge}_{p(x)}=\{L,R,L,L,{\alpha}_{1},{\alpha}_{2},\dots \}, where {\alpha}_{i}\in \{L,R\}. If we let u:=b\ast a, v:=b\ast (b\ast a), then {[a,b]}_{p(x)}=\{a,b,b\ast a=u,b\ast (b\ast a)=v,(b\ast (b\ast a))\ast (b\ast a)=u,[(b\ast (b\ast a))\ast (b\ast a)]\ast [b\ast (b\ast a)],\dots \}. It follows that (b\ast (b\ast a))\ast (b\ast a)=b\ast a and (b\ast a)\ast (b\ast (b\ast a))=b\ast (b\ast a), proving the proposition. □

**Proposition 5.10** *The class of* {\mathit{LRL}}^{2}-*groupoids and the class of groups are Smarandache disjoint*.

*Proof* Let (X,\ast ) be both an {\mathit{LRL}}^{2}-groupoid and a group with a special element e\in X. Given a,b\in X, we have (b\ast (b\ast a))\ast (b\ast a)=b\ast a. Since every group has cancellation laws, we obtain b\ast (b\ast a)=e. It follows that b\ast a=(b\ast a)\ast e=(b\ast a)\ast (b\ast (b\ast a))=b\ast (b\ast a)=e, and hence e=b\ast (b\ast a)=b\ast e=b. This proves that |X|=1, proving the proposition. □

A groupoid (X,\ast ) is said to be an {R}^{4}-*groupoid* if for all a,b\in X,

({R}^{4}1) (a\ast b)\ast (b\ast (a\ast b))=a\ast b,

({R}^{4}2) (b\ast (a\ast b))\ast (a\ast b)=b\ast (a\ast b).

**Proposition 5.11** *Let* (X,\ast ) *be a groupoid and let* p(x)\in {Z}_{2}[[x]] *such that* {p}_{0}={p}_{1}={p}_{2}={p}_{3}=1. *Then* (X,\ast ) *is an* {R}^{4}-*groupoid*.

*Proof* Since p(x)\in {Z}_{2}[[x]] such that {p}_{0}={p}_{1}={p}_{2}={p}_{3}=1, we have {\bigwedge}_{p(x)}=\{R,R,R,R,{\alpha}_{1},{\alpha}_{2},\dots \}, where {\alpha}_{i}\in \{L,R\}. If we let u:=a\ast b, v:=b\ast (a\ast b), then {[a,b]}_{p(x)}=\{a,b,a\ast b=u,b\ast (a\ast b)=v,(a\ast b)\ast (b\ast (a\ast b))=u,[b\ast (a\ast b)]\ast [(a\ast b)\ast (b\ast (a\ast b))],\dots \}. It follows that a\ast b=u=u\ast v=(a\ast b)\ast (b\ast (a\ast b)) and b\ast (a\ast b)=v=v\ast u=(b\ast (a\ast b))\ast (a\ast b), proving the proposition. □

**Theorem 5.12** *The class of* {R}^{4}-*groupoids and the class of groups are Smarandache disjoint*.

*Proof* Let (X,\ast ) be both an {R}^{4}-groupoid and a group with a special element e\in X. Given a,b\in X, we have (a\ast b)\ast (b\ast (a\ast b))=a\ast b. Since every group has cancellation laws, we obtain b\ast (a\ast b)=e. By applying ({R}^{4}2), we have e\ast (a\ast b)=e, *i.e.*, a\ast b=e. By ({R}^{4}1), b=e\ast (b\ast e)=e. This proves that |X|=1, proving the proposition. □

**Proposition 5.13** *Every implicative* *BCK*-*algebra is an* {R}^{4}-*groupoid*.

*Proof* If (X,\ast ,0) is an implicative *BCK*-algebra, then a\ast (b\ast a)=a and (a\ast b)\ast b=a\ast b for any a,b\in X. It follows immediately that (X,\ast ) is an {R}^{4}-groupoid. □

**Remark** The condition, *implicativity*, is important for a *BCK*-algebra to be an {R}^{4}-groupoid.

**Example 5.14** Let X:=\{0,1,2,3\} be a set with the following table:

Then (X,\ast ,0) is a *BCK*-algebra, but not implicative, since 1\ast (3\ast 1)=0\ne 1. Moreover, it is not a {R}^{4}-groupoid since 3\ast 2=1\ne 0=(3\ast 2)\ast [2\ast (3\ast 2)].

**Theorem 5.15** *Every* *BCK*-*algebra* (X,\ast ,0) *inherited from a poset* (X,\le ) *is an* {R}^{4}-*groupoid*.

*Proof* If (X,\ast ,0) is a *BCK*-algebra (X,\ast ,0) inherited from a poset (X,\le ), then the operation ‘∗’ is defined by

x\ast y:=\{\begin{array}{cc}0\hfill & \text{if}x\le y,\hfill \\ x\hfill & \text{otherwise}.\hfill \end{array}

Then the condition ({R}^{4}1) holds. In fact, given x,y\in X, if x\le y, then x\ast y=0 and y\ast x=y. It follows that (x\ast y)\ast (y\ast (x\ast y))=0\ast (y\ast 0)=0=x\ast y. If y\le x, then y\ast x=0 and x\ast y=x. It follows that (x\ast y)\ast (y\ast (x\ast y))=x\ast (y\ast x)=x\ast 0=x=x\ast y. If *x* and *y* are incomparable, then x\ast y=x and y\ast x=y. It follows that (x\ast y)\ast (y\ast (x\ast y))=x\ast (y\ast x)=x\ast y.

We claim that ({R}^{4}2) holds. Given x,y\in X, if x\le y, then x\ast y=0, y\ast x=y. It follows that (y\ast (x\ast y))\ast (x\ast y)=(y\ast 0)\ast 0=y=y\ast 0=y\ast (x\ast y). If y\le x, then y\ast x=0, x\ast y=x. It follows that (y\ast (x\ast y))\ast (x\ast y)=(y\ast x)\ast x=0\ast x=0=y\ast x=y\ast (x\ast y). If *x* and *y* are incomparable, then x\ast y=x and y\ast x=y. It follows that (y\ast (x\ast y))\ast (x\ast y)=(y\ast x)\ast x=y\ast x=y\ast (x\ast y). This proves that the *BCK*-algebra (X,\ast ,0) inherited from a poset (X,\le ) is an {R}^{4}-groupoid. □

Note that the *BCK*-algebra (X,\ast ,0) inherited from a poset (X,\le ) need not be an implicative *BCK*-algebra unless the poset (X,\le ) is an antichain [[14], Corollary 9].

**Example 5.16** Let (X,\ast ) be a left-zero semigroup, *i.e.*, x\ast y=x for all x,y\in X. Then (X,\ast ) is an {R}^{4}-groupoid, but not a *BCK*-algebra.