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:

a(s(x1)) → s(s(s(p(s(b(p(p(s(s(x1))))))))))
b(s(x1)) → s(s(s(p(p(s(s(c(p(s(p(s(x1))))))))))))
c(s(x1)) → p(s(p(s(a(p(s(p(s(x1)))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

a(s(x1)) → s(s(s(p(s(b(p(p(s(s(x1))))))))))
b(s(x1)) → s(s(s(p(p(s(s(c(p(s(p(s(x1))))))))))))
c(s(x1)) → p(s(p(s(a(p(s(p(s(x1)))))))))
p(p(s(x1))) → p(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:

P(p(s(x1))) → P(x1)
B(s(x1)) → P(s(p(s(x1))))
B(s(x1)) → P(s(s(c(p(s(p(s(x1))))))))
C(s(x1)) → P(s(x1))
A(s(x1)) → P(s(s(x1)))
B(s(x1)) → P(s(x1))
B(s(x1)) → P(p(s(s(c(p(s(p(s(x1)))))))))
C(s(x1)) → P(s(a(p(s(p(s(x1)))))))
A(s(x1)) → B(p(p(s(s(x1)))))
A(s(x1)) → P(p(s(s(x1))))
A(s(x1)) → P(s(b(p(p(s(s(x1)))))))
B(s(x1)) → C(p(s(p(s(x1)))))
C(s(x1)) → A(p(s(p(s(x1)))))
C(s(x1)) → P(s(p(s(x1))))
C(s(x1)) → P(s(p(s(a(p(s(p(s(x1)))))))))

The TRS R consists of the following rules:

a(s(x1)) → s(s(s(p(s(b(p(p(s(s(x1))))))))))
b(s(x1)) → s(s(s(p(p(s(s(c(p(s(p(s(x1))))))))))))
c(s(x1)) → p(s(p(s(a(p(s(p(s(x1)))))))))
p(p(s(x1))) → p(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:

P(p(s(x1))) → P(x1)
B(s(x1)) → P(s(p(s(x1))))
B(s(x1)) → P(s(s(c(p(s(p(s(x1))))))))
C(s(x1)) → P(s(x1))
A(s(x1)) → P(s(s(x1)))
B(s(x1)) → P(s(x1))
B(s(x1)) → P(p(s(s(c(p(s(p(s(x1)))))))))
C(s(x1)) → P(s(a(p(s(p(s(x1)))))))
A(s(x1)) → B(p(p(s(s(x1)))))
A(s(x1)) → P(p(s(s(x1))))
A(s(x1)) → P(s(b(p(p(s(s(x1)))))))
B(s(x1)) → C(p(s(p(s(x1)))))
C(s(x1)) → A(p(s(p(s(x1)))))
C(s(x1)) → P(s(p(s(x1))))
C(s(x1)) → P(s(p(s(a(p(s(p(s(x1)))))))))

The TRS R consists of the following rules:

a(s(x1)) → s(s(s(p(s(b(p(p(s(s(x1))))))))))
b(s(x1)) → s(s(s(p(p(s(s(c(p(s(p(s(x1))))))))))))
c(s(x1)) → p(s(p(s(a(p(s(p(s(x1)))))))))
p(p(s(x1))) → p(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 11 less nodes.

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

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

P(p(s(x1))) → P(x1)

The TRS R consists of the following rules:

a(s(x1)) → s(s(s(p(s(b(p(p(s(s(x1))))))))))
b(s(x1)) → s(s(s(p(p(s(s(c(p(s(p(s(x1))))))))))))
c(s(x1)) → p(s(p(s(a(p(s(p(s(x1)))))))))
p(p(s(x1))) → p(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.


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

POL(P(x1)) = (4)x_1   
POL(p(x1)) = 1/2 + (1/4)x_1   
POL(s(x1)) = (4)x_1   
The value of delta used in the strict ordering is 2.
The following usable rules [17] were oriented: none



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

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

a(s(x1)) → s(s(s(p(s(b(p(p(s(s(x1))))))))))
b(s(x1)) → s(s(s(p(p(s(s(c(p(s(p(s(x1))))))))))))
c(s(x1)) → p(s(p(s(a(p(s(p(s(x1)))))))))
p(p(s(x1))) → p(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:

B(s(x1)) → C(p(s(p(s(x1)))))
C(s(x1)) → A(p(s(p(s(x1)))))
A(s(x1)) → B(p(p(s(s(x1)))))

The TRS R consists of the following rules:

a(s(x1)) → s(s(s(p(s(b(p(p(s(s(x1))))))))))
b(s(x1)) → s(s(s(p(p(s(s(c(p(s(p(s(x1))))))))))))
c(s(x1)) → p(s(p(s(a(p(s(p(s(x1)))))))))
p(p(s(x1))) → p(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.


A(s(x1)) → B(p(p(s(s(x1)))))
The remaining pairs can at least be oriented weakly.

B(s(x1)) → C(p(s(p(s(x1)))))
C(s(x1)) → A(p(s(p(s(x1)))))
Used ordering: Polynomial interpretation [25,35]:

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

p(p(s(x1))) → p(x1)
p(s(x1)) → x1



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

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

B(s(x1)) → C(p(s(p(s(x1)))))
C(s(x1)) → A(p(s(p(s(x1)))))

The TRS R consists of the following rules:

a(s(x1)) → s(s(s(p(s(b(p(p(s(s(x1))))))))))
b(s(x1)) → s(s(s(p(p(s(s(c(p(s(p(s(x1))))))))))))
c(s(x1)) → p(s(p(s(a(p(s(p(s(x1)))))))))
p(p(s(x1))) → p(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 0 SCCs with 2 less nodes.