Definition 1 (see [17].)A function d : ℨ × ℨ⟶[0, ∞) is b-metric if, for every $$,

• (a)

  $$

• (b)

  $$

• (c)

  $$

The ordered pair (ℨ, d) is a b-MS. It is key to note that b-MS are not metrizable, especially b-metric might not be a continuous function of its variable.

Definition 2 (see [27].)A binary operation in such a way $$ is said to be a continuous triangular norm (t-norm) if

• (a)

  $$ and $$

• (b)

  $$ is continuous

• (c)

  $$

• (d)

  $$ whenever $$

Definition 3 (see [28].)Let $$ be a t-norm. Then, $$, is defined by

Let

Then, t-norm $$ is of H-type with the assuming the functions family $$ which is equicontinuous at k = 1, where

The t-norm $$ is an example of H-type.

Each t-norm $$ and t-conorm $$ can be drawn out to an n-ary operation taking $$ (see [27]):

Definition 4. $$, and $$ can be extended in n-ary in the following ways:

Definition 5 (see [29], t-conorm.)“A binary operation $$ is continuous t-conorm if

• (a)

  $$

• (b)

  $$ is continuous

• (c)

  $$

• (d)

  $$ whenever a ≤ c & q ≤ s, ∀a, q, c, s ∈ [0,1]

Definition 6. A t-conorm, $$ is defined by

Let

Then, t-conorm $$ is of H-type with the assuming the functions’ family $$ which is equicontinuous at k = 0, where

The t-conorm $$ is an example of H-type.

The t-conorm can be stretched out by associativity in an n-ary operation taking, for $$, the values

Example 1. The extensions of the t-conorm in n-ary are $$ are the following:

For any $$ in [0,1], the t-norm and t-conorm can be stretched out to a countable illimitable operation, that is,

The sequence $$ is increasing, and $$ is decreasing and has an upper bound and lower bound, respectively, so the limit exists.

In this fixed-point theory, see [30, 31], it is very interesting to discuss the classes of t-norms $$ and the classes of t-conorm $$ and $$ in [0, 1] such that $$ and

Proposition 1. Let $$ in [0, 1] be $$ and $$ be a t-norm of H-type. Then,

Proposition 2. Let $$ in [0, 1] ne $$ and $$ be a t-conorm of H-type. Then,

Definition 7 (see [23].)A 3-tuple $$ is called a b-FMS. If ∀ ð, v, z ∈ ℨ, s, k > 0,

• (a)

  ℳ(ð, v, k) > 0

• (b)

  ℳ(ð, v, k) = 1,  ∀ k > 0 iff ð = v

• (c)

  ℳ(ð, v, k) = ℳ(v, ð, k)

• (d)

  $$

• (e)

  ℳ(ð, v, ⋅) : [0, ∞)⟶[0,1] (left continuous)

where ℨ is a nonempty set, $$ is continuous t-norm, and ℳ is a FS on ℨ² × (0, ∞).

Example 2 (see [9].)Let $$, where p > 1. Then, ℳ is a b-metric with b = 2^p−1.

Definition 8. (see [7, 8]). A 5-tuple $$ is called intuitionistic fuzzy rectangular metric space. If, for every ð, v, z ∈ ℨ and s, k > 0,

• (a)

  $$

• (b)

  ℳ(ð, v, 0) = 0

• (c)

  ℳ(ð, v, k) = 1, ∀k > 0 iff ð = v

• (d)

  ℳ(ð, v, k) = ℳ(v, ð, k)

• (e)

  $$

• (f)

  ℳ(ð, v, ⋅) : [0, ∞)⟶[0,1] (left continuous)

• (g)

  lim_k⟶∞ℳ(ð, v, k) = 1, ∀ð, v ∈ ℨ

• (h)

  $$

• (i)

  $$ iff ð = v

• (j)

  $$

• (k)

  $$ (right continuous)

• (l)

  $$

where ℨ is a random set, $$ is continuous t-norm, $$ is continuous t-conorm, and $$ are FSs on X² × (0, ∞).

Example 3. Let $$, where p > 1. Then, ℳ and $$ are b-metric with b = 2^p−1.

Definition 9 (see [9].)A function ζ : ℜ⟶ℜ is called b-nondecreasing if ð > bv implies ζ(ð) ≥ ζ(ð), ∀ð, v ∈ ℜ.

Definition 10. A function ζ : ℜ⟶ℜ is called b-nonincreasing if ð < bv implies ζ(ð) ≤ ζ(v), ∀ð, v ∈ ℜ.

Lemma 1 (see [23], [32].)Let $$ be a b-FMS. Then, ℳ(ð, v, k) is b-increasing with respect to t, ∀ð, v ∈ ℨ.

Lemma 2. Let $$ be an b-IFMS. Then, $$ is b-decreasing with respect to t, ∀ð, v ∈ ℨ.

Definition 11 (see [23], [32].)Let $$ be an b-IFMS. For k > 0, the open ball ℬ(v, r, k) with centre ð ∈ ℨ and radius 0 < r < 1 are defined as

A sequence $$

• (a)

  Converges to ð if $$ as n⟶∞, for each k > 0. In this case, we write $$.

• (b)

  Is a Cauchy sequence if $$ in such a way that $$.

• (c)

  Converges to ð if $$ as $$, for each k > 0. In this case, we write $$.

• (d)

  Is a Cauchy sequence if $$ in such a way that $$.

Definition 12. The intuitionistic fuzzy b-MS $$ is complete if and only if every Cauchy sequence is a convergent sequence.

Lemma 3. In an b-IFMS $$, we have a sequence $$ in ℨ converges to ð; then, it is definitely a Cauchy sequence and ð is unique.

Proposition 3. Let $$ be a b-IFMS and $$ converges to x. Then,

Remark 1. A b-IFM is not continuous generally.

Example 4. Let ℨ = [0, ∞), ℳ(ð, v, k) = e^{−d(ð, v)/k}, and $$, and

Then, $$ be a b-IFMS with b = 6.

Lemma 4. Let $$ in a b-IFMS $$ be

and there exist ð₀, ð₁ ∈ ℨ and v ∈ (0,1) such that

Then, $$ is a Cauchy sequence.

Proof. Let ω ∈ (0,1). Then, the sum $$ is convergent, and then, there exists $$ such that $$, for every $$. Let $$. Since ℳ is b-increasing by Definition 8(e) and $$ is b-decreasing by Definition 8(j), for every k > 0, we obtain

By (19), it implies $$:

By (19), it implies $$.

Since n > m and b > 1, we have

where v = bμ/ω is v ∈ (0,1). By equation (20), it implies $$ is a Cauchy sequence.

Corollary 1. Let $$ be a b-IFMS $$, where $$ and $$ are H-type. If there exists μ ∈ (0, 1/b) such that

holds, then $$ is a Cauchy sequence.

Lemma 5. If there exists μ ∈ (0,1) and ð, v ∈ ℨ such that

holds, then ð = v.

Proof. Condition (25) implies that

By definition of b-IFMS, we have ð = v.

Theorem 1. Let $$ be a b-IFMS, which is complete and Q : ℨ⟶ℨ. Suppose there exists μ ∈ (0, 1/b) such that

holds. Furthermore, there exists ð₀ ∈ ℨ and v ∈ (0,1) such that holds. Then, Q has a unique fixed point in ℨ.

Proof. Let ð₀ ∈ ℨ and $$. If we take $$ and $$ in (27), then we have

By Lemma 4, it implies $$ is Cauchy sequence. Since $$ is complete, then $$. Therefore,

and

By (27), as n⟶∞, we have

Let us assume that x and y are fixed points for Q. By (24), we obtain

Using Lemma 5, we have ð = v.

Example 5. Assume ℨ = [0,1]. Using Example 3 for p = 2 implies $$ is a b-IFMS with b = 2 and b-IFM. We define

Let $$. Then,

for 1/b > μ > k²; hence, equation (27) will be satisfied, and we can say that Q possess a fixed point in ℨ which is unique.

Theorem 2. Let $$ be a b-IFMS and Q : ℨ⟶ℨ. Assume that μ ∈ (0, 1/b) such that

and there exists ð₀ ∈ ℨ,  v ∈ (0,1) such that

Then, Q has a unique fixed point in ℨ.

Proof. Let $$. By (36), with $$ and $$, for every $$, we have

If $$, then

Lemma 5 implies $$. Therefore,

and by lemma, we have $$ is a Cauchy sequence. Hence, $$:

Let us prove ð is a fixed point for Q. Let ω₁ ∈ (μb, 1) and ω₂ = 1 − ω₁. By (36), we have

Taking n⟶∞ and using (41), we have

where v = bμ/ω₁ ∈ (0,1). Therefore, we obtain

Lemma 5 implies Qð = ð.

Suppose that ð and v are fixed points for Q (i.e., Qð = ð &Qv = v). By (36), we obtain

By Lemma 5, it follows that Qð = Qv, which implies that ð = v.

Example 6. Let $$, and $$. Then, $$ is b-IFMS which is complete with b = 2. Let

•  

  Case (i): if ð, v ∈ [1,2), then $$, and conditions (36) will be trivially satisfied.

•  

  Case (ii): if ð ∈ [1,2), v ∈ (0,1), then μ ∈ (1/4, 1/2); we have

  $$ (47)

•  

  Case (iii): for μ ∈ (1/4, 1/2), we have

  $$ (48)

•  

  Case (iv): if ð, v ∈ (0,1), then μ ∈ (1/4, 1/2); we have

So, condition (36) is satisfied ∀ ð, v ∈ ð, k > 0, and by Theorem 2, Q possess a unique fixed point.

Theorem 3. Let $$ be a complete b-IFMS and Q : X⟶X. If

for μ(0, 1/b³), then Q possess a unique fixed point in v.

Proof. Let ð₀ ∈ ℨ and $$. By (37), with $$ and $$, using Definition 8(d) and $$, we have

Proceeding as in proof of Theorem 2, Lemma 5, and Corollary 1, it follows that

and $$ is a Cauchy sequence. So, there exists ð ∈ ℨ such that $$ and

Let ω₁ ∈ (μb³, 1) and ω₂ = 1 − ω₁. By (50) and Definition 8(e), for $$, we have

for all $$ and k > 0. Taking n⟶∞ and using (41), we obtain and by Lemma 5, with v = b²μ/ω₁ ∈ (0,1), it follows that Qð = ð.

By condition (50), ð = Qð and v = Qv; we have

and by Lemma 3, it follows that ð = v.

Theorem 4. Let $$, be a b-IFMS which is complete and Q : ℨ⟶ℨ, for some μ ∈ (0, 1/b²), such that

and there exists ð₀ ∈ ℨ,  v ∈ (0,1); we have

Proof. Let $$. Taking $$ and $$ in condition (57), and $$, we have

Since $$ is b-nondecreasing and $$; $$ is b-nonincreasing and $$.

Therefore,

for $$. By lemma, and $$ is a Cauchy sequence. Since $$ is a complete b-IFMS, then $$ and ω₁ ∈ (b³μ, 1) and ω₁ ∈ (b³μ, 1). Also and for $$, we have for every $$ and k > 0. Taking n⟶∞ and using and by Lemma 5 with v = b³μ/ω₁ ∈ (0,1), we have Qð = ð and

Thus, by Lemma 5, we have ð = v.

Example 7. Let $$, or $$, $$. Then, $$ is a complete b-IFM with b = 2.

Define Q : ℨ⟶ℨ as f(0) = f(1) = 0, f(3) = 0.

We observe that if ð = v or ð, v ∈ {0,1}, then $$, and (57) is satisfied.

If x = 1 and y = 3, then μ ∈ (1/9, 1/4) and

In the same way, if we take ð = 3 and v = 1 and for μ ∈ (1/9, 1/4), condition (57) is satisfied, for all ð, v ∈ ℨ, k > 0, and Q possess a unique fixed in ℨ.