Termination w.r.t. Q of the following Term Rewriting System could be proven:

Q restricted rewrite system:
The TRS R consists of the following rules:

f(0(x1)) → s(0(x1))
d(0(x1)) → 0(x1)
d(s(x1)) → s(s(d(p(s(x1)))))
f(s(x1)) → d(f(p(s(x1))))
p(s(x1)) → x1

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

f(0(x1)) → s(0(x1))
d(0(x1)) → 0(x1)
d(s(x1)) → s(s(d(p(s(x1)))))
f(s(x1)) → d(f(p(s(x1))))
p(s(x1)) → x1

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

F(s(x1)) → D(f(p(s(x1))))
F(s(x1)) → F(p(s(x1)))
F(s(x1)) → P(s(x1))
D(s(x1)) → P(s(x1))
D(s(x1)) → D(p(s(x1)))

The TRS R consists of the following rules:

f(0(x1)) → s(0(x1))
d(0(x1)) → 0(x1)
d(s(x1)) → s(s(d(p(s(x1)))))
f(s(x1)) → d(f(p(s(x1))))
p(s(x1)) → x1

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

F(s(x1)) → D(f(p(s(x1))))
F(s(x1)) → F(p(s(x1)))
F(s(x1)) → P(s(x1))
D(s(x1)) → P(s(x1))
D(s(x1)) → D(p(s(x1)))

The TRS R consists of the following rules:

f(0(x1)) → s(0(x1))
d(0(x1)) → 0(x1)
d(s(x1)) → s(s(d(p(s(x1)))))
f(s(x1)) → d(f(p(s(x1))))
p(s(x1)) → x1

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

D(s(x1)) → D(p(s(x1)))

The TRS R consists of the following rules:

f(0(x1)) → s(0(x1))
d(0(x1)) → 0(x1)
d(s(x1)) → s(s(d(p(s(x1)))))
f(s(x1)) → d(f(p(s(x1))))
p(s(x1)) → x1

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


D(s(x1)) → D(p(s(x1)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(D(x1)) = (9/4)x_1   
POL(s(x1)) = 1/2 + (9/4)x_1   
POL(p(x1)) = (1/2)x_1   
The value of delta used in the strict ordering is 9/16.
The following usable rules [17] were oriented:

p(s(x1)) → x1



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

f(0(x1)) → s(0(x1))
d(0(x1)) → 0(x1)
d(s(x1)) → s(s(d(p(s(x1)))))
f(s(x1)) → d(f(p(s(x1))))
p(s(x1)) → x1

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof

Q DP problem:
The TRS P consists of the following rules:

F(s(x1)) → F(p(s(x1)))

The TRS R consists of the following rules:

f(0(x1)) → s(0(x1))
d(0(x1)) → 0(x1)
d(s(x1)) → s(s(d(p(s(x1)))))
f(s(x1)) → d(f(p(s(x1))))
p(s(x1)) → x1

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


F(s(x1)) → F(p(s(x1)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(s(x1)) = 4 + (4)x_1   
POL(p(x1)) = 7/4 + (1/2)x_1   
POL(F(x1)) = (4)x_1   
The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented:

p(s(x1)) → x1



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

f(0(x1)) → s(0(x1))
d(0(x1)) → 0(x1)
d(s(x1)) → s(s(d(p(s(x1)))))
f(s(x1)) → d(f(p(s(x1))))
p(s(x1)) → x1

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.