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:

a1(a1(x)) -> b1(b1(x))
b1(b1(a1(x))) -> a1(b1(b1(x)))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

a1(a1(x)) -> b1(b1(x))
b1(b1(a1(x))) -> a1(b1(b1(x)))

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

A1(a1(x)) -> B1(x)
B1(b1(a1(x))) -> A1(b1(b1(x)))
B1(b1(a1(x))) -> B1(x)
B1(b1(a1(x))) -> B1(b1(x))
A1(a1(x)) -> B1(b1(x))

The TRS R consists of the following rules:

a1(a1(x)) -> b1(b1(x))
b1(b1(a1(x))) -> a1(b1(b1(x)))

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:

A1(a1(x)) -> B1(x)
B1(b1(a1(x))) -> A1(b1(b1(x)))
B1(b1(a1(x))) -> B1(x)
B1(b1(a1(x))) -> B1(b1(x))
A1(a1(x)) -> B1(b1(x))

The TRS R consists of the following rules:

a1(a1(x)) -> b1(b1(x))
b1(b1(a1(x))) -> a1(b1(b1(x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


A1(a1(x)) -> B1(x)
A1(a1(x)) -> B1(b1(x))
The remaining pairs can at least be oriented weakly.

B1(b1(a1(x))) -> A1(b1(b1(x)))
B1(b1(a1(x))) -> B1(x)
B1(b1(a1(x))) -> B1(b1(x))
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( A1(x1) ) = max{0, x1 - 2}


POL( a1(x1) ) = x1 + 3


POL( B1(x1) ) = max{0, x1 - 3}


POL( b1(x1) ) = x1



The following usable rules [14] were oriented:

a1(a1(x)) -> b1(b1(x))
b1(b1(a1(x))) -> a1(b1(b1(x)))



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

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

B1(b1(a1(x))) -> A1(b1(b1(x)))
B1(b1(a1(x))) -> B1(x)
B1(b1(a1(x))) -> B1(b1(x))

The TRS R consists of the following rules:

a1(a1(x)) -> b1(b1(x))
b1(b1(a1(x))) -> a1(b1(b1(x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

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

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

B1(b1(a1(x))) -> B1(x)
B1(b1(a1(x))) -> B1(b1(x))

The TRS R consists of the following rules:

a1(a1(x)) -> b1(b1(x))
b1(b1(a1(x))) -> a1(b1(b1(x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


B1(b1(a1(x))) -> B1(x)
B1(b1(a1(x))) -> B1(b1(x))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( B1(x1) ) = max{0, x1 - 3}


POL( b1(x1) ) = x1 + 1


POL( a1(x1) ) = x1 + 3



The following usable rules [14] were oriented:

a1(a1(x)) -> b1(b1(x))
b1(b1(a1(x))) -> a1(b1(b1(x)))



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

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

a1(a1(x)) -> b1(b1(x))
b1(b1(a1(x))) -> a1(b1(b1(x)))

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.