At first, we introduce some notations and definitions that will be used later. The following definition was introduced by Jungck [28].
Definition 2.1. [28] Let (X, d) be a metric space and f, g : X → X. If w = fx = gx, for some x ∈ X, then x is called a coincidence point of f and g, and w is called a point of coincidence of f and g. The pair {f, g} is said to be compatible if and only if , whenever {x_{
n
} } is a sequence in X such that for some t ∈ X.
Let X be a nonempty set and R : X → X be a given mapping. For every x ∈ X, we denote by R^{1}(x) the subset of X defined by
In [19], Nashine and Samet introduced the following concept:
Definition 2.2. [19] Let (X, ≤) be a partially ordered set, and T, S, R : X → X are given mappings, such that TX ⊆ RX and SX ⊆ RX. We say that S and T are weakly increasing with respect to R if for all x ∈ X, we have
and
Remark 2.3. If R : X → X is the identity mapping (Rx = x for all x ∈ X), then S and T are weakly increasing with respect to R implies that S and T are weakly increasing mappings. It is noted that the notion of weakly increasing mappings was introduced in [9] (also see [16, 29]).
Example 2.4. Let X = [0, +∞) endowed with the usual order ≤. Define the mappings T, S, R : X → X by
and
Then, we will show that the mappings S and T are weakly increasing with respect to R.
Let x ∈ X. We distinguish the following two cases.

(i)
Let y ∈ R ^{1}(Tx), that is, Ry = Tx. By the definition of T, we have Tx = 0 and then Ry = 0. By the definition of R, we have y = 0 or y ≥ 1. By the definition of S, in both cases, we have Sy = 0. Then, Tx = 0 = Sy.

(ii)
Let y ∈ R ^{1}(Sx), that is, Ry = Sx. By the definition of S, we have Sx = 0, and then Ry = 0. By the definition of R, we have y = 0 or y ≥ 1. By the definition of T, in both cases, we have Ty = 0. Then, Sx = 0 = Ty.

(i)
Let y ∈ R ^{1}(Tx), that is, Ry = Tx. By the definition of T, we have Tx = x and then Ry = x. By the definition of R, we have Ry = y ^{2}, and then . We have

(ii)
Let y ∈ R ^{1}(Sx), that is, Ry = Sx. By the definition of S, we have , and then . By the definition of R, we have Ry = y ^{2}, and then y = x ^{1/4}. We have
Thus, we proved that S and T are weakly increasing with respect to R.
Example 2.5. Let X = {1, 2, 3} endowed with the partial order ≤ given by
Define the mappings T, S, R : X → X by
We will show that the mappings S and T are weakly increasing with respect to R.
Let x, y ∈ X such that y ∈ R^{1}(Tx). By the definition of S, we have Sy = 1. On the other hand, Tx ∈ {1, 3} and (1, 1), (3, 1) ∈≤. Thus, we have Tx ≤ Sy for all y ∈ R^{1}(Tx).
Let x, y ∈ X such that y ∈ R^{1}(Sx). By the definitions of S and R, we have R^{1}(Sx) = R^{1}(1) = {1}. Then, we have y = 1. On the other hand, 1 = Sx ≤ Ty = T 1 = 1. Then, Sx ≤ Ty for all y ∈ R^{1}(Sx). Thus, we proved that S and T are weakly increasing with respect to R.
Our first result is as follows.
Theorem 2.6. Let (X, ≤) be a partially ordered set, and suppose that there exists a metric d on X such that (X, d) is a complete metric space. Let T, S, R : X → X be given mappings, satisfying for every pair (x, y) ∈ X × X such that Rx and Ry are comparable:
where ψ_{1}and ψ_{2}are generalized altering distance functions, and Φ_{1}(x) = ψ_{1}(x, x, x).
We assume the following hypotheses:

(i)
T, S, and R are continuous.

(ii)
TX ⊆ RX, SX ⊆ RX.

(iii)
T and S are weakly increasing with respect to R.

(iv)
the pairs {T, R} and {S, R} are compatible.
Then, T, S, and R have a coincidence point, that is, there exists u ∈ X such that Ru = Tu = Su.
Proof. Let x_{0} ∈ X be an arbitrary point. Since TX ⊆ RX, there exists x_{1} ∈ X such that Rx_{1} = Tx_{0}. Since SX ⊆ RX, there exists x_{2} ∈ X such that Rx_{2} = Sx_{1}.
Continuing this process, we can construct a sequence {Rx_{
n
} } in X defined by
We claim that
To this aim, we will use the increasing property with respect to R for the mappings T and S. From (2.2), we have
Since Rx_{1} = Tx_{0}, x_{1} ∈ R^{1} (Tx_{0}), and we get
Again,
Since x_{2} ∈ R^{1} (Sx 1), we get
Hence, by induction, (2.3) holds.
Without loss of the generality, we can assume that
Now, we will prove our result on three steps.
Step I. We will prove that
Letting x = x_{2n+1}and y = x_{2n}, from (2.3) and the considered contraction, we have
Suppose that
Using the property of the generalized altering function, this implies that
Hence, we obtain
This implies that
and
Hence, we obtain a contradiction with (2.4). We deduce that
Similarly, letting x = x_{2n+1}and y = x_{2n+2}, from (2.3) and the considered contraction, we have
Suppose that
Then, from (2.9) and (2.10), we obtain
This implies that
and
Hence, we obtain a contradiction with (2.4). We deduce that
Combining (2.8) and (2.11), we obtain
Hence, {d(Rx_{n+1}, Rx_{n+2})} is a decreasing sequence of positive real numbers. This implies that there exists r ≥ 0 such that
Define the function Φ_{2}: [0, +∞) → [0, +∞) by
From (2.6) and (2.12), we obtain
which implies that
Similarly, from (2.9) and (2.12), we obtain
which implies that
Now, combining (2.14) and (2.15), we obtain
This implies that for all , we have
This implies that
Hence,
Now, using (2.13), (2.16), and the continuity of Φ_{2}, we obtain
which implies that r = 0. Hence, (2.5) is proved.
Step II. We claim that {Rx_{
n
} } is a Cauchy sequence.
From (2.5), it will be sufficient to prove that {Rx_{2n}} is a Cauchy sequence. We proceed by negation, and suppose that {Rx_{2n}} is not a Cauchy sequence. Then, there exists ε > 0 for which we can find two sequences of positive integers {m(i)} and {n(i)} such that for all positive integer i,
From (2.17) and using the triangular inequality, we get
Letting i → +∞ in the above inequality, and using (2.5), we obtain
Again, the triangular inequality gives us
Letting i → +∞ in the above inequality, and using (2.5) and (2.18), we get
On the other hand, we have
Then, from (2.5), (2.18), and the continuity of Φ_{1}, and letting i → +∞ in the above inequality, we have
Now, using the considered contractive condition for x = x_{2m(i)1}and y = x_{2n(i)}, we have
Then, from (2.5), (2.19), and the continuity of ψ_{1} and ψ_{2}, and letting i → +∞ in the above inequality, we have
Now, combining (2.20) with the above inequality, we get
which implies that ψ_{2}(ε, 0, 0) = 0, that is a contradiction since ε > 0. We deduce that {Rx_{
n
} } is a Cauchy sequence.
Step III. Existence of a coincidence point.
Since {Rx_{
n
} } is a Cauchy sequence in the complete metric space (X, d), there exists u ∈ X such that
From (2.21) and the continuity of R, we get
By the triangular inequality, we have
On the other hand, we have
Since R and T are compatible mappings, this implies that
Now, from the continuity of T and (2.21), we have
Combining (2.22), (2.24), and (2.25), and letting n → +∞ in (2.23), we obtain
that is,
Again, by the triangular inequality, we have
On the other hand, we have
Since R and S are compatible mappings, this implies that
Now, from the continuity of S and (2.21), we have
Combining (2.22), (2.28), and (2.29), and letting n → + ∞ in (2.27), we obtain
that is,
Finally, from (2.26) and (2.30), we have
that is, u is a coincidence point of T, S, and R. This completes the proof.
In the next theorem, we omit the continuity hypotheses on T, S, and R.
Definition 2.7. Let (X,≤, d) be a partially ordered metric space. We say that X is regular if the following hypothesis holds: if {z_{
n
} } is a nondecreasing sequence in X with respect to ≤ such that z_{
n
} → z ∈ X as n → +∞, then z_{
n
} ≤ z for all .
Now, our second result is the following.
Theorem 2.8. Let (X,≤) be a partially ordered set, and suppose that there exists a metric d on X such that (X, d) is a complete metric space. Let T, S, R : X → X be given mappings satisfying for every pair (x, y) ∈ X × X such that Rx and Ry are comparable,
where ψ_{1}and ψ_{2}are generalized altering distance functions and Φ_{1}(x) = ψ_{1}(x, x, x). We assume the following hypotheses:

(i)
X is regular.

(ii)
T and S are weakly increasing with respect to R.

(iii)
RX is a closed subset of (X, d).

(iv)
TX ⊆ RX, SX ⊆ R X.
Then, T, S, and R have a coincidence point.
Proof. From the proof of Theorem 2.6, we have that {Rx_{
n
} } is a Cauchy sequence in (RX, d) which is complete, since RX is a closed subspace of (X, d). Hence, there exists u = Rv, v ∈ X such that
Since {Rx_{
n
} } is a nondecreasing sequence and X is regular, it follows from (2.31) that Rx_{
n
} ≤ Rv for all . Hence, we can apply the considered contractive condition. Then, for x = v and y = x_{2n}, we obtain
Letting n → +∞ in the above inequality, and using (2.5), (2.31), and the properties of ψ_{1} and ψ_{2}, then we have
This implies that ψ_{2}(0, d(Rv, Sv), 0) = 0, which gives us that d(Rv, Sv) = 0, i.e.,
Similarly, for x = x_{2n+1}and y = v, we obtain
Letting n → +∞ in the above inequality, we get
This implies that ψ_{2}(0, 0, d(Rv, Tv)) = 0 and then,
Now, combining (2.32) and (2.33), we obtain
Hence, v is a coincidence point of T, S, and R. This completes the proof.
Now, we present an example to illustrate the obtained result given by the previous theorem. Moreover, in this example, we will show that Theorem 1.6 of Choudhury cannot be applied.
Example 2.9. Let X = {4, 5, 6} endowed with the usual metric d(x, y) = x  y for all x, y ∈ X, and ≤:= {(4, 4), (5, 5), (6, 6), (6, 4)}. Clearly, ≤ is a partial order on X. Consider the mappings T, S, R : X → X defined by
We will show that T and S are weakly increasing with respect to R. In the case under study, we have to check that Tx ≤ T(Tx) for all x ∈ X.
For x = 4, we have
For x = 5, we have
For x = 6, we have
Thus, we have proved that T and S are weakly increasing with respect to R.
Now, we will show that (X, ≤, d) is regular.
Let {z_{
n
} } be a nondecreasing sequence in X with respect to ≤ such that z_{
n
}→ z ∈ X as n → +∞. Then, we have z_{
n
}≤ z_{n+1}, for all .

If z_{0} = 4, then z_{0} = 4 ≤ z_{1}. From the definition of ≤, we have z_{1} = 4. By induction, we get z_{
n
}= 4 for all and z = 4. Then, z_{
n
}≤ z for all .

If z_{0} = 5, then z_{0} = 5 ≤ z_{1}. From the definition of ≤, we have z_{1} = 5. By induction, we get z_{
n
}= 5 for all and z = 5. Then, z_{
n
}≤ z for all .

If z_{0} = 6, then z_{0} = 6 ≤ z_{1}. From the definition of ≤, we have z_{1} ∈ {6, 4}. By induction, we get z_{
n
}∈ {6, 4} for all . Suppose that there exists p ≥ 1 such that z_{
p
}= 4. From the definition of ≤, we get z_{
n
}= z_{
p
}= 4 for all n ≥ p. Thus, we have z = 4 and z_{
n
}≤ z for all . Now, suppose that z_{
n
}= 6 for all . In this case, we get z = 6, and z_{
n
}≤ z for all . Thus, we proved that in all the cases considered, we have z_{
n
}≤ z for all . Then, (X, ≤, d) is regular.
Now, define the functions ψ_{1}, ψ_{2} : [0, +∞) × [0, +∞) × [0, +∞) → [0, +∞) by
and
Clearly, ψ_{1} and ψ_{2} are the generalized altering distance functions, and for every x, y ∈ X such that Rx ≤ Ry, inequality (2.1) is satisfied.
Now, we can apply Theorem 2.8 to deduce that T, S, and R have a coincidence point u = 4. Note that u is also a fixed point of T since S = T, and R is the identity mapping.
On the other hand, taking x = 4 and y = 5, we get
Thus, Inequality (1.2) is not satisfied for x = 4 and y = 5. Then, Theorem 1.6 of Choudhury [5] cannot be applied in this case.
If R : X → X is the identity mapping, we can deduce easily the following common fixed point results.
The next result is an immediate consequence of Theorem 2.6.
Corollary 2.10. Let (X, ≤) be a partially ordered set, and suppose that there exists a metric d on X such that (X, d) is a complete metric space. Let T, S : X → X be given mappings satisfying for every pair (x, y) ∈ X × X such that x and y are comparable. Then,
where ψ_{1}and ψ_{2}are generalized altering distance functions and Φ_{1}(x) = ψ_{1}(x, x, x). We assume the following hypotheses:

(i)
T and S are continuous.

(ii)
T and S are weakly increasing.
Then, T and S have a common fixed point, that is, there exists u ∈ X such that u = Tu = Su.
The following result is an immediate consequence of Theorem 2.8.
Corollary 2.11. Let (X, ≤) be a partially ordered set and suppose that there exists a metric d on X such that (X, d) is a complete metric space. Let T, S : X → X be given mappings satisfying for every pair (x, y) ∈ X × X such that x and y are comparable. Then,
where ψ_{1}and ψ_{2}are generalised altering distance functions and Φ_{1}(x) = ψ_{1}(x, x, x). We assume the following hypotheses:

(i)
X is regular.

(ii)
T and S are weakly increasing.
Then, T and S have a common fixed point.
A number of fixed point results may be obtained by assuming different forms for the functions ψ_{1} and ψ_{2}. In particular, fixed point results under various contractive conditions may be derived from the above theorems. For example, if we consider
where s > 0 and 0 < k = k_{1} + k_{2} + k_{3}< 1, then we obtain the following results.
The next result is an immediate consequence of Corollary 2.10.
Corollary 2.12. Let (X, ≤) be a partially ordered set, and suppose that there exists a metric d on X such that (X, d) is a complete metric space. Let T, S : X → X be given mappings satisfying for every pair (x, y) ∈ X × X such that x and y are comparable. Then,
where s > 0 and 0 < k = k_{1} + k_{2} + k_{3}< 1. We assume the following hypotheses:

(i)
T and S are continuous.

(ii)
T and S are weakly increasing.
Then, T and S have a common fixed point, that is, there exists u ∈ X such that u = Tu = Su.
The next result is an immediate consequence of Corollary 2.11.
Corollary 2.13. Let (X, ≤) be a partially ordered set, and suppose that there exists a metric d on X such that (X, d) is a complete metric space. Let T, S : X → X be given mappings satisfying for every pair (x, y) ∈ X × X such that x and y are comparable. Then,
where s > 0 and 0 < k = k_{1} + k_{2} + k_{3}< 1. We assume the following hypotheses:

(i)
X is regular.

(ii)
T and S are weakly increasing.
Then, T and S have a common fixed point.
Remark 2.14. Other fixed point results may also be obtained under specific choices of ψ_{1} and ψ_{2}.