On BCK Identities

On BCK Identities
Andrzej Wro´n ski
November8,2006
Abstract
We discuss the problem of axiomatizing BCK identities and present arguments supporting a certain conjecture.
The question how to axiomatize identities of BCK-algebras arose when it became clear that BCK-algebras form a proper quasivariety.
By all means there is no need to recall the definition of BCK-algebras here but–just tofix the notation–let us say that they are structures of the form A= A,→,1 such that the following conditions hold for every a,b,c∈A:
(B)(a→b)→((c→a)→(c→b))=1,
(C)(a→(b→c))→(b→(a→c))=1,
(K)a→(b→a)=1,
(!)if a→b=1=b→a then a=b.
Let the symbol BCK denotes the quasivariety determined by the con-ditions above.Note that BCK verifies formal identities from the following list:
(1)1→x≈x,
(2)x→(y→z)≈((x→(y→z))→z)→z.
(3)x→(y→z)≈y→(x→z).
Moreover,we can generalize(2)and(3)in the following manner:
1
(4)x0→(x1→(···(x n→z)···))≈
((x0→(x1→(···(x n→z)···)))→z)→z,
(5)x0→(x1→(···(x n→z)···))≈x f
0→(x f
1
→(···(x f
n
→z)···)),
for any permutation f of{0,...,n}.
The identity(5)allows us to treat antecedents of the operation→as a multiset and thus,the expression of the formΓ→b can be safely used to abbreviate any expression of the form a0→(a1→(···(a n→b)···))where Γis a multiset of all antecedents a0,...,a n.Now,the identity(4)can be written simply as:
Γ→z≈((Γ→z)→z)→z
.
Structures closely related to BCK-algebras are so called pocrims(partially ordered commutative integral monoids)having the form A= A,·,→,1 where A= A,→,1 is a BCK-algebra and for every a,b,c∈A,
a→(b→c)≈(a·b)→c
It is known that every pocrim is a pocrim subreduct of so called involutive pocrim understood as a structure of the form A= A,·,→,1,0 where A= A,·,→,1 is a pocrim and for every a∈A:
(a→0)→0≈a
Involutive pocrims are term equivalent to structures resembling Boolean algebras.Putting:
¬a:=a→0,
a+b:=¬(¬a·¬b)
one gets from an involutive pocrim A a term equivalent structure A,·,+,¬,1 obeying a lot of familiar Boolean identities.For example,de Morgan identi-ties hold and the operation of product can be defined in terms of→and¬by putting a·b:=¬(a→¬b).However,the product and sum operations of involutive pocrims are not idempotent in general.
Now we will take a closer look at the free BCK-algebra F(A)= F,→,1 with the set A of free generators.First,let us note that F(A)can be decom-posed into isomorphic subalgebras of the form F a,→,1 corresponding to members of A,with universes defined as follows:
F a:={Γ→a:Γis afinite multiset of members of F}
2
The result of M.Nagayama from SL(1994)227-234[1]yields that:
Fact1For every a,b∈A,if a=b then F a∩F b={1}.
It is worth noticing that every component of the form F a can be equipped with the structure of an involutive pocrim.Indeed,putting:
0:=a,
(Γ→a) (∆→a):=((Γ→a)→((∆→a)→a))→a
and applying identities(4)and(5)one easily verifies that the structure:
F a(A):= F a, ,→,1,a
is an involutive pocrim.
Now,for every doubleton{a,b}⊆A,the cycle(a,b)gives rise to a BCK-isomorphismσ(a,b):F a(A)−→F b(A).In consequence,the whole of F(A) can be recovered from each single component of the form F a(A),a∈A –the whole information needed is there.Note also,that the subalgebra of the involutive pocrim F a(A)generated by set{c→a:c∈A,c=a}is a free member of the quasivariety of involutive pocrims with|A−{a}|free generators.
Let us consider a certain variety H(to be further thought of as a possible candidate for H(BCK)).We define H as the variety determined by the following set of formal identities:
(i)x→1≈1,
(ii)(x→y)→((z→x)→(z→y))≈1,
(iii)1→x≈x,
(iv)x→(y→z)≈((x→(y→z))→z)→z,
(v)x→(y→z)≈y→(x→z).
First,note that an easy induction proves for H the important generalized BCK-identities i.e.
(4)x0→(x1→(···(x n→z)···))≈
((x0→(x1→(···(x n→z)···)))→z)→z,
3
(5)x0→(x1→(···(x n→z)···))≈x f
0→(x f
1
→(···(x f
n
→z)···)),
for any permutation f of{0,...,n}.
Note that H⊇H(BCK)and next:
Fact2H and BCK verify the same identities of the form:ϕ≈1.
Indeed,let v be a one-to-one mapping of variables occurring in the term ϕinto the set of free generators of an algebra F= F,→F,1F being a free member of H.Define a relationη:={ a,b :a→F b=1F=b→F a}and observe thatη∈Cg(F),F/η∈BCK,1F/η={1F}.Now,supposing that F/η|=ϕ≈1one gets that v(ϕ)/η=1F/ηandfinally v(ϕ)=1F which clearly means that H|=ϕ≈1.
The fact noted above implies that the Nagayama’s result for BCK-identities carries over to the variety H i.e.we have:
Fact3If H|=Γ→x≈∆→y where x,y are distinct variables then H|=Γ→x≈1≈∆→y.
Thus,all what is needed for showing that H=H(BCK)is a verification of the following:
Conjecture4If H|=(Γ→x)→(∆→x)≈1≈(∆→x)→(Γ→x) then H|=Γ→x≈∆→x.
References
[1]M.Nagayama,On a Property of BCK-Identities,Studia Logica,53
(1994),pp.227–234.
4。