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(x, y) -> b2(x, b2(0, c1(y)))
c1(b2(y, c1(x))) -> c1(c1(b2(a2(0, 0), y)))
b2(y, 0) -> y

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

a2(x, y) -> b2(x, b2(0, c1(y)))
c1(b2(y, c1(x))) -> c1(c1(b2(a2(0, 0), y)))
b2(y, 0) -> 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:

C1(b2(y, c1(x))) -> C1(b2(a2(0, 0), y))
A2(x, y) -> B2(0, c1(y))
C1(b2(y, c1(x))) -> C1(c1(b2(a2(0, 0), y)))
C1(b2(y, c1(x))) -> A2(0, 0)
C1(b2(y, c1(x))) -> B2(a2(0, 0), y)
A2(x, y) -> C1(y)
A2(x, y) -> B2(x, b2(0, c1(y)))

The TRS R consists of the following rules:

a2(x, y) -> b2(x, b2(0, c1(y)))
c1(b2(y, c1(x))) -> c1(c1(b2(a2(0, 0), y)))
b2(y, 0) -> 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:

C1(b2(y, c1(x))) -> C1(b2(a2(0, 0), y))
A2(x, y) -> B2(0, c1(y))
C1(b2(y, c1(x))) -> C1(c1(b2(a2(0, 0), y)))
C1(b2(y, c1(x))) -> A2(0, 0)
C1(b2(y, c1(x))) -> B2(a2(0, 0), y)
A2(x, y) -> C1(y)
A2(x, y) -> B2(x, b2(0, c1(y)))

The TRS R consists of the following rules:

a2(x, y) -> b2(x, b2(0, c1(y)))
c1(b2(y, c1(x))) -> c1(c1(b2(a2(0, 0), y)))
b2(y, 0) -> 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 3 less nodes.

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

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

C1(b2(y, c1(x))) -> C1(b2(a2(0, 0), y))
C1(b2(y, c1(x))) -> C1(c1(b2(a2(0, 0), y)))
C1(b2(y, c1(x))) -> A2(0, 0)
A2(x, y) -> C1(y)

The TRS R consists of the following rules:

a2(x, y) -> b2(x, b2(0, c1(y)))
c1(b2(y, c1(x))) -> c1(c1(b2(a2(0, 0), y)))
b2(y, 0) -> 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(x, y) -> C1(y)
The remaining pairs can at least be oriented weakly.

C1(b2(y, c1(x))) -> C1(b2(a2(0, 0), y))
C1(b2(y, c1(x))) -> C1(c1(b2(a2(0, 0), y)))
C1(b2(y, c1(x))) -> A2(0, 0)
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( c1(x1) ) = 2


POL( 0 ) = 0


POL( C1(x1) ) = max{0, x1 - 1}


POL( A2(x1, x2) ) = x2 + 1


POL( a2(x1, x2) ) = x1 + 2


POL( b2(x1, x2) ) = x1 + x2



The following usable rules [14] were oriented:

b2(y, 0) -> y
c1(b2(y, c1(x))) -> c1(c1(b2(a2(0, 0), y)))
a2(x, y) -> b2(x, b2(0, c1(y)))



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

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

C1(b2(y, c1(x))) -> C1(b2(a2(0, 0), y))
C1(b2(y, c1(x))) -> C1(c1(b2(a2(0, 0), y)))
C1(b2(y, c1(x))) -> A2(0, 0)

The TRS R consists of the following rules:

a2(x, y) -> b2(x, b2(0, c1(y)))
c1(b2(y, c1(x))) -> c1(c1(b2(a2(0, 0), y)))
b2(y, 0) -> 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 1 less node.

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

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

C1(b2(y, c1(x))) -> C1(b2(a2(0, 0), y))
C1(b2(y, c1(x))) -> C1(c1(b2(a2(0, 0), y)))

The TRS R consists of the following rules:

a2(x, y) -> b2(x, b2(0, c1(y)))
c1(b2(y, c1(x))) -> c1(c1(b2(a2(0, 0), y)))
b2(y, 0) -> y

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