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(b(x1)) → x1
a(c(x1)) → b(c(c(a(x1))))
b(c(x1)) → a(b(x1))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

A(c(x1)) → A(x1)
A(c(x1)) → B(c(c(a(x1))))
B(c(x1)) → B(x1)
B(c(x1)) → A(b(x1))

The TRS R consists of the following rules:

a(b(x1)) → x1
a(c(x1)) → b(c(c(a(x1))))
b(c(x1)) → a(b(x1))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof

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

A(c(x1)) → A(x1)
A(c(x1)) → B(c(c(a(x1))))
B(c(x1)) → B(x1)
B(c(x1)) → A(b(x1))

The TRS R consists of the following rules:

a(b(x1)) → x1
a(c(x1)) → b(c(c(a(x1))))
b(c(x1)) → a(b(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(c(x1)) → A(x1)
B(c(x1)) → B(x1)
The remaining pairs can at least be oriented weakly.

A(c(x1)) → B(c(c(a(x1))))
B(c(x1)) → A(b(x1))
Used ordering: Polynomial interpretation [25,35]:

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

a(b(x1)) → x1
a(c(x1)) → b(c(c(a(x1))))
b(c(x1)) → a(b(x1))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ QDPOrderProof

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

A(c(x1)) → B(c(c(a(x1))))
B(c(x1)) → A(b(x1))

The TRS R consists of the following rules:

a(b(x1)) → x1
a(c(x1)) → b(c(c(a(x1))))
b(c(x1)) → a(b(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(c(x1)) → B(c(c(a(x1))))
B(c(x1)) → A(b(x1))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(c(x1)) = 1/4 + (4)x_1   
POL(B(x1)) = 1 + (1/4)x_1   
POL(a(x1)) = (4)x_1   
POL(A(x1)) = 1/2 + (4)x_1   
POL(b(x1)) = (1/4)x_1   
The value of delta used in the strict ordering is 3/16.
The following usable rules [17] were oriented:

a(b(x1)) → x1
a(c(x1)) → b(c(c(a(x1))))
b(c(x1)) → a(b(x1))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ PisEmptyProof

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

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