Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

a2(0, b2(0, x)) -> b2(0, a2(0, x))
a2(0, x) -> b2(0, b2(0, x))
a2(0, a2(1, a2(x, y))) -> a2(1, a2(0, a2(x, y)))
b2(0, a2(1, a2(x, y))) -> b2(1, a2(0, a2(x, y)))
a2(0, a2(x, y)) -> a2(1, a2(1, a2(x, y)))

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

a2(0, b2(0, x)) -> b2(0, a2(0, x))
a2(0, x) -> b2(0, b2(0, x))
a2(0, a2(1, a2(x, y))) -> a2(1, a2(0, a2(x, y)))
b2(0, a2(1, a2(x, y))) -> b2(1, a2(0, a2(x, y)))
a2(0, a2(x, y)) -> a2(1, a2(1, a2(x, y)))

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:

A2(0, x) -> B2(0, b2(0, x))
B2(0, a2(1, a2(x, y))) -> A2(0, a2(x, y))
B2(0, a2(1, a2(x, y))) -> B2(1, a2(0, a2(x, y)))
A2(0, b2(0, x)) -> B2(0, a2(0, x))
A2(0, a2(1, a2(x, y))) -> A2(1, a2(0, a2(x, y)))
A2(0, b2(0, x)) -> A2(0, x)
A2(0, a2(x, y)) -> A2(1, a2(1, a2(x, y)))
A2(0, x) -> B2(0, x)
A2(0, a2(x, y)) -> A2(1, a2(x, y))
A2(0, a2(1, a2(x, y))) -> A2(0, a2(x, y))

The TRS R consists of the following rules:

a2(0, b2(0, x)) -> b2(0, a2(0, x))
a2(0, x) -> b2(0, b2(0, x))
a2(0, a2(1, a2(x, y))) -> a2(1, a2(0, a2(x, y)))
b2(0, a2(1, a2(x, y))) -> b2(1, a2(0, a2(x, y)))
a2(0, a2(x, y)) -> a2(1, a2(1, a2(x, y)))

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:

A2(0, x) -> B2(0, b2(0, x))
B2(0, a2(1, a2(x, y))) -> A2(0, a2(x, y))
B2(0, a2(1, a2(x, y))) -> B2(1, a2(0, a2(x, y)))
A2(0, b2(0, x)) -> B2(0, a2(0, x))
A2(0, a2(1, a2(x, y))) -> A2(1, a2(0, a2(x, y)))
A2(0, b2(0, x)) -> A2(0, x)
A2(0, a2(x, y)) -> A2(1, a2(1, a2(x, y)))
A2(0, x) -> B2(0, x)
A2(0, a2(x, y)) -> A2(1, a2(x, y))
A2(0, a2(1, a2(x, y))) -> A2(0, a2(x, y))

The TRS R consists of the following rules:

a2(0, b2(0, x)) -> b2(0, a2(0, x))
a2(0, x) -> b2(0, b2(0, x))
a2(0, a2(1, a2(x, y))) -> a2(1, a2(0, a2(x, y)))
b2(0, a2(1, a2(x, y))) -> b2(1, a2(0, a2(x, y)))
a2(0, a2(x, y)) -> a2(1, a2(1, a2(x, y)))

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 4 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
QDP
          ↳ QDPOrderProof

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

A2(0, x) -> B2(0, b2(0, x))
A2(0, b2(0, x)) -> B2(0, a2(0, x))
B2(0, a2(1, a2(x, y))) -> A2(0, a2(x, y))
A2(0, b2(0, x)) -> A2(0, x)
A2(0, x) -> B2(0, x)
A2(0, a2(1, a2(x, y))) -> A2(0, a2(x, y))

The TRS R consists of the following rules:

a2(0, b2(0, x)) -> b2(0, a2(0, x))
a2(0, x) -> b2(0, b2(0, x))
a2(0, a2(1, a2(x, y))) -> a2(1, a2(0, a2(x, y)))
b2(0, a2(1, a2(x, y))) -> b2(1, a2(0, a2(x, y)))
a2(0, a2(x, y)) -> a2(1, a2(1, a2(x, y)))

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.


A2(0, b2(0, x)) -> A2(0, x)
A2(0, x) -> B2(0, x)
A2(0, a2(1, a2(x, y))) -> A2(0, a2(x, y))
The remaining pairs can at least be oriented weakly.

A2(0, x) -> B2(0, b2(0, x))
A2(0, b2(0, x)) -> B2(0, a2(0, x))
B2(0, a2(1, a2(x, y))) -> A2(0, a2(x, y))
Used ordering: Polynomial interpretation [21]:

POL(0) = 2   
POL(1) = 0   
POL(A2(x1, x2)) = 2·x1 + 2·x2   
POL(B2(x1, x2)) = 2·x2   
POL(a2(x1, x2)) = 2 + x1 + x2   
POL(b2(x1, x2)) = x1 + x2   

The following usable rules [14] were oriented:

a2(0, x) -> b2(0, b2(0, x))
a2(0, a2(x, y)) -> a2(1, a2(1, a2(x, y)))
b2(0, a2(1, a2(x, y))) -> b2(1, a2(0, a2(x, y)))
a2(0, b2(0, x)) -> b2(0, a2(0, x))
a2(0, a2(1, a2(x, y))) -> a2(1, a2(0, a2(x, y)))



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

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

A2(0, x) -> B2(0, b2(0, x))
B2(0, a2(1, a2(x, y))) -> A2(0, a2(x, y))
A2(0, b2(0, x)) -> B2(0, a2(0, x))

The TRS R consists of the following rules:

a2(0, b2(0, x)) -> b2(0, a2(0, x))
a2(0, x) -> b2(0, b2(0, x))
a2(0, a2(1, a2(x, y))) -> a2(1, a2(0, a2(x, y)))
b2(0, a2(1, a2(x, y))) -> b2(1, a2(0, a2(x, y)))
a2(0, a2(x, y)) -> a2(1, a2(1, a2(x, y)))

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.


A2(0, x) -> B2(0, b2(0, x))
The remaining pairs can at least be oriented weakly.

B2(0, a2(1, a2(x, y))) -> A2(0, a2(x, y))
A2(0, b2(0, x)) -> B2(0, a2(0, x))
Used ordering: Polynomial interpretation [21]:

POL(0) = 2   
POL(1) = 2   
POL(A2(x1, x2)) = 2 + 3·x1   
POL(B2(x1, x2)) = 2·x2   
POL(a2(x1, x2)) = 2·x1   
POL(b2(x1, x2)) = 0   

The following usable rules [14] were oriented:

a2(0, x) -> b2(0, b2(0, x))
a2(0, a2(x, y)) -> a2(1, a2(1, a2(x, y)))
b2(0, a2(1, a2(x, y))) -> b2(1, a2(0, a2(x, y)))
a2(0, b2(0, x)) -> b2(0, a2(0, x))
a2(0, a2(1, a2(x, y))) -> a2(1, a2(0, a2(x, y)))



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

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

A2(0, b2(0, x)) -> B2(0, a2(0, x))
B2(0, a2(1, a2(x, y))) -> A2(0, a2(x, y))

The TRS R consists of the following rules:

a2(0, b2(0, x)) -> b2(0, a2(0, x))
a2(0, x) -> b2(0, b2(0, x))
a2(0, a2(1, a2(x, y))) -> a2(1, a2(0, a2(x, y)))
b2(0, a2(1, a2(x, y))) -> b2(1, a2(0, a2(x, y)))
a2(0, a2(x, y)) -> a2(1, a2(1, a2(x, y)))

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