Proof. Let a, b, c ∈Xsuch that T(a, b, c)> θ.
Our main is to prove that T(b, a, c)> θ, for this,
choosing h∈Xwhich is satisfying T(b, a, h)> θ
and proving that h=c.
From our hypotheses, there exists t∈Xsuch
that (a◦a)◦(b◦b)(t)> θ, (18)
and (a◦b)◦(a◦b)(t)> θ. (19)
Taking t1, t2∈Xsuch that T(a, a, t1)> θ and
T(b, b, t2)> θ, thus (18) and (19) give respec-
tively T(t1, t2, t)> θ and T(c, c, t)> θ. Then,
t1◦(b◦b)(t)≥T(b, b, t2)∧T(t1, t2, t)> θ, (20)
and
c◦(a◦b)(t)≥T(a, b, c)∧T(c, c, t)> θ. (21)
Now, let x1, x2, y1, y2∈Xsatisfy T(t1, b, x1)> θ,
T(x1, b, y1)> θ,T(c, a, x2)> θ and T(x2, b, y2)>
θ. Then,
(t1◦b)◦b(y1)> θ and (c◦a)◦b(y2)> θ,
which, because of (20) and (21) together Defini-
tion 3, implies that y1=tand y2=t. It follows
that T(x1, b, t)> θ and T(x2, b, t)> θ, hence
in view of Lemma 1(2) we get x1=x2. Conse-
quently, T(t1, b, x1)> θ and T(c, a, x1)> θ. So
that,
(a◦a)◦b(x1)≥T(a, a, t1)∧T(t1, b, x1)> θ,
(22)
and
(a◦b)◦a(x1)≥T(a, b, c)∧T(c, a, x1)> θ. (23)
Let v, k ∈Xsuch that T(a, c, v)> θ and
T(a, h, k)> θ. Then,
a◦(a◦b)(v)≥T(a, b, c)∧T(a, c, v)> θ
and
a◦(b◦a)(k)≥T(b, a, h)∧T(a, h, k)> θ.
(24)
Combining (22), (23), (24) and using Definition
3, we obtain v=x1and h=x1and therefore,
T(a, c, x1)> θ and T(a, h, x1)> θ. Once again,
using Lemma 1(1) we conclude that c=h. So
that, T(b, a, c)> θ.
Conversely, assuming that T(b, a, c)> θ and us-
ing similar arguments as used above to prove that
T(a, b, c)> θ. Thus, (X, T ) is a commutative
fuzzy group.
Theorem 2. Let (X, T, L)be a fuzzy near-ring.
If there exists z∈ZF(X)∗such that T(z, z, r)>
θ=⇒r∈ZF(X)for all r∈X, then (X, T )is
abelian.
Proof. Suppose that (X, T, L) is a left fuzzy near-
ring and let x, y ∈X. Consider zthe element of
our hypothesis; by Definition 2 there exists t∈X
such that T(z, z, t)> θ which, according to our
hypothesis, implies that t∈ZF(X).
Now, let v, h ∈Xsuch that T(x, y, v)> θ and
L(t, v, h)> θ. Then,
t∗(x◦y)(h)≥T(x, y, v)∧L(t, v, h)> θ. (25)
Taking h1, h2, h0∈Xsuch that L(t, x, h1)> θ,
L(t, y, h2)> θ and T(h1, h2, h0)> θ. Then,
(t∗x)◦(t∗y)(h0)≥
L(t, x, h1)∧L(t, y, h2)∧T(h1, h2, h0).
(26)
From Definition 1 (iii), (25) and (26) give h0=h,
so that
T(h1, h2, h)> θ. (27)
Also, as t∈ZF(X), we have L(x, t, h1)> θ and
L(y, t, h2)> θ, then
x∗(z◦z)(h1)≥T(z, z, t)∧L(x, t, h1)> θ and
y∗(z◦z)(h2)≥T(z, z, t)∧L(y, t, h2)> θ.
Choosing v1, v2, v3, v4∈Xsuch that L(x, z, v1)>
θ,L(y, z, v2)> θ,T(v1, v1, v3)> θ and
T(v2, v2, v4)> θ, we infer that
(x∗z)◦(x∗z)(v3)> θ and (y∗z)◦(y∗z)(v4)> θ.
(28)
Consequently, in virtue of Definition 1 (iii), we
conclude that v3=h1and v4=h2which implies
that T(v1, v1, h1)> θ and L(v2, v2, h2)> θ. Once
again, since z∈ZF(X)∗, we have L(z, x, v1)> θ
and L(z, y, v2)> θ, thus
(z∗x)◦(z∗x)(h1)> θ and (z∗y)◦(z∗y)(h2)> θ.
(29)
Next, choose `1, `2, `3, `4such that T(x, x, `1)>
θ,T(y, y, `2)> θ,L(z, `1, `3)> θ and
L(z, `2, `4)> θ, which yields
z∗(x◦x)(`3)> θ and z∗(y◦y)(`4)> θ. (30)
Invoking Definition 1 (iii), we arrive at `3=h1
and `4=h2. So that, L(z, `1, h1)> θ and
L(z, `2, h2)> θ. Because of (27), we get
(z∗`1)◦(z∗`2)(h)≥
L(z, `1, h1)∧L(z, `2, h2)∧T(h1, h2, h)> θ.
(31)
WSEAS TRANSACTIONS on MATHEMATICS
DOI: 10.37394/23206.2024.23.42
M. Ou-Mha, A. Raji, M. Oukessou, A. Boua