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:

+2(p1, p1) -> p2
+2(p1, +2(p2, p2)) -> p5
+2(p5, p5) -> p10
+2(+2(x, y), z) -> +2(x, +2(y, z))
+2(p1, +2(p1, x)) -> +2(p2, x)
+2(p1, +2(p2, +2(p2, x))) -> +2(p5, x)
+2(p2, p1) -> +2(p1, p2)
+2(p2, +2(p1, x)) -> +2(p1, +2(p2, x))
+2(p2, +2(p2, p2)) -> +2(p1, p5)
+2(p2, +2(p2, +2(p2, x))) -> +2(p1, +2(p5, x))
+2(p5, p1) -> +2(p1, p5)
+2(p5, +2(p1, x)) -> +2(p1, +2(p5, x))
+2(p5, p2) -> +2(p2, p5)
+2(p5, +2(p2, x)) -> +2(p2, +2(p5, x))
+2(p5, +2(p5, x)) -> +2(p10, x)
+2(p10, p1) -> +2(p1, p10)
+2(p10, +2(p1, x)) -> +2(p1, +2(p10, x))
+2(p10, p2) -> +2(p2, p10)
+2(p10, +2(p2, x)) -> +2(p2, +2(p10, x))
+2(p10, p5) -> +2(p5, p10)
+2(p10, +2(p5, x)) -> +2(p5, +2(p10, x))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

+2(p1, p1) -> p2
+2(p1, +2(p2, p2)) -> p5
+2(p5, p5) -> p10
+2(+2(x, y), z) -> +2(x, +2(y, z))
+2(p1, +2(p1, x)) -> +2(p2, x)
+2(p1, +2(p2, +2(p2, x))) -> +2(p5, x)
+2(p2, p1) -> +2(p1, p2)
+2(p2, +2(p1, x)) -> +2(p1, +2(p2, x))
+2(p2, +2(p2, p2)) -> +2(p1, p5)
+2(p2, +2(p2, +2(p2, x))) -> +2(p1, +2(p5, x))
+2(p5, p1) -> +2(p1, p5)
+2(p5, +2(p1, x)) -> +2(p1, +2(p5, x))
+2(p5, p2) -> +2(p2, p5)
+2(p5, +2(p2, x)) -> +2(p2, +2(p5, x))
+2(p5, +2(p5, x)) -> +2(p10, x)
+2(p10, p1) -> +2(p1, p10)
+2(p10, +2(p1, x)) -> +2(p1, +2(p10, x))
+2(p10, p2) -> +2(p2, p10)
+2(p10, +2(p2, x)) -> +2(p2, +2(p10, x))
+2(p10, p5) -> +2(p5, p10)
+2(p10, +2(p5, x)) -> +2(p5, +2(p10, 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:

+12(p5, p2) -> +12(p2, p5)
+12(p10, +2(p5, x)) -> +12(p10, x)
+12(p5, p1) -> +12(p1, p5)
+12(p10, +2(p1, x)) -> +12(p1, +2(p10, x))
+12(p2, +2(p1, x)) -> +12(p2, x)
+12(p10, +2(p2, x)) -> +12(p2, +2(p10, x))
+12(p10, +2(p2, x)) -> +12(p10, x)
+12(p10, +2(p5, x)) -> +12(p5, +2(p10, x))
+12(p10, p5) -> +12(p5, p10)
+12(p2, p1) -> +12(p1, p2)
+12(p1, +2(p1, x)) -> +12(p2, x)
+12(+2(x, y), z) -> +12(y, z)
+12(p2, +2(p1, x)) -> +12(p1, +2(p2, x))
+12(p2, +2(p2, +2(p2, x))) -> +12(p5, x)
+12(p10, +2(p1, x)) -> +12(p10, x)
+12(p2, +2(p2, +2(p2, x))) -> +12(p1, +2(p5, x))
+12(p5, +2(p2, x)) -> +12(p2, +2(p5, x))
+12(p5, +2(p1, x)) -> +12(p5, x)
+12(p10, p2) -> +12(p2, p10)
+12(p1, +2(p2, +2(p2, x))) -> +12(p5, x)
+12(p5, +2(p5, x)) -> +12(p10, x)
+12(p10, p1) -> +12(p1, p10)
+12(+2(x, y), z) -> +12(x, +2(y, z))
+12(p5, +2(p1, x)) -> +12(p1, +2(p5, x))
+12(p2, +2(p2, p2)) -> +12(p1, p5)
+12(p5, +2(p2, x)) -> +12(p5, x)

The TRS R consists of the following rules:

+2(p1, p1) -> p2
+2(p1, +2(p2, p2)) -> p5
+2(p5, p5) -> p10
+2(+2(x, y), z) -> +2(x, +2(y, z))
+2(p1, +2(p1, x)) -> +2(p2, x)
+2(p1, +2(p2, +2(p2, x))) -> +2(p5, x)
+2(p2, p1) -> +2(p1, p2)
+2(p2, +2(p1, x)) -> +2(p1, +2(p2, x))
+2(p2, +2(p2, p2)) -> +2(p1, p5)
+2(p2, +2(p2, +2(p2, x))) -> +2(p1, +2(p5, x))
+2(p5, p1) -> +2(p1, p5)
+2(p5, +2(p1, x)) -> +2(p1, +2(p5, x))
+2(p5, p2) -> +2(p2, p5)
+2(p5, +2(p2, x)) -> +2(p2, +2(p5, x))
+2(p5, +2(p5, x)) -> +2(p10, x)
+2(p10, p1) -> +2(p1, p10)
+2(p10, +2(p1, x)) -> +2(p1, +2(p10, x))
+2(p10, p2) -> +2(p2, p10)
+2(p10, +2(p2, x)) -> +2(p2, +2(p10, x))
+2(p10, p5) -> +2(p5, p10)
+2(p10, +2(p5, x)) -> +2(p5, +2(p10, x))

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:

+12(p5, p2) -> +12(p2, p5)
+12(p10, +2(p5, x)) -> +12(p10, x)
+12(p5, p1) -> +12(p1, p5)
+12(p10, +2(p1, x)) -> +12(p1, +2(p10, x))
+12(p2, +2(p1, x)) -> +12(p2, x)
+12(p10, +2(p2, x)) -> +12(p2, +2(p10, x))
+12(p10, +2(p2, x)) -> +12(p10, x)
+12(p10, +2(p5, x)) -> +12(p5, +2(p10, x))
+12(p10, p5) -> +12(p5, p10)
+12(p2, p1) -> +12(p1, p2)
+12(p1, +2(p1, x)) -> +12(p2, x)
+12(+2(x, y), z) -> +12(y, z)
+12(p2, +2(p1, x)) -> +12(p1, +2(p2, x))
+12(p2, +2(p2, +2(p2, x))) -> +12(p5, x)
+12(p10, +2(p1, x)) -> +12(p10, x)
+12(p2, +2(p2, +2(p2, x))) -> +12(p1, +2(p5, x))
+12(p5, +2(p2, x)) -> +12(p2, +2(p5, x))
+12(p5, +2(p1, x)) -> +12(p5, x)
+12(p10, p2) -> +12(p2, p10)
+12(p1, +2(p2, +2(p2, x))) -> +12(p5, x)
+12(p5, +2(p5, x)) -> +12(p10, x)
+12(p10, p1) -> +12(p1, p10)
+12(+2(x, y), z) -> +12(x, +2(y, z))
+12(p5, +2(p1, x)) -> +12(p1, +2(p5, x))
+12(p2, +2(p2, p2)) -> +12(p1, p5)
+12(p5, +2(p2, x)) -> +12(p5, x)

The TRS R consists of the following rules:

+2(p1, p1) -> p2
+2(p1, +2(p2, p2)) -> p5
+2(p5, p5) -> p10
+2(+2(x, y), z) -> +2(x, +2(y, z))
+2(p1, +2(p1, x)) -> +2(p2, x)
+2(p1, +2(p2, +2(p2, x))) -> +2(p5, x)
+2(p2, p1) -> +2(p1, p2)
+2(p2, +2(p1, x)) -> +2(p1, +2(p2, x))
+2(p2, +2(p2, p2)) -> +2(p1, p5)
+2(p2, +2(p2, +2(p2, x))) -> +2(p1, +2(p5, x))
+2(p5, p1) -> +2(p1, p5)
+2(p5, +2(p1, x)) -> +2(p1, +2(p5, x))
+2(p5, p2) -> +2(p2, p5)
+2(p5, +2(p2, x)) -> +2(p2, +2(p5, x))
+2(p5, +2(p5, x)) -> +2(p10, x)
+2(p10, p1) -> +2(p1, p10)
+2(p10, +2(p1, x)) -> +2(p1, +2(p10, x))
+2(p10, p2) -> +2(p2, p10)
+2(p10, +2(p2, x)) -> +2(p2, +2(p10, x))
+2(p10, p5) -> +2(p5, p10)
+2(p10, +2(p5, x)) -> +2(p5, +2(p10, 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 9 less nodes.

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

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

+12(p2, +2(p2, +2(p2, x))) -> +12(p5, x)
+12(p10, +2(p5, x)) -> +12(p10, x)
+12(p10, +2(p1, x)) -> +12(p1, +2(p10, x))
+12(p10, +2(p1, x)) -> +12(p10, x)
+12(p2, +2(p1, x)) -> +12(p2, x)
+12(p2, +2(p2, +2(p2, x))) -> +12(p1, +2(p5, x))
+12(p5, +2(p2, x)) -> +12(p2, +2(p5, x))
+12(p10, +2(p2, x)) -> +12(p2, +2(p10, x))
+12(p5, +2(p1, x)) -> +12(p5, x)
+12(p1, +2(p2, +2(p2, x))) -> +12(p5, x)
+12(p10, +2(p2, x)) -> +12(p10, x)
+12(p10, +2(p5, x)) -> +12(p5, +2(p10, x))
+12(p5, +2(p5, x)) -> +12(p10, x)
+12(p1, +2(p1, x)) -> +12(p2, x)
+12(p5, +2(p1, x)) -> +12(p1, +2(p5, x))
+12(p5, +2(p2, x)) -> +12(p5, x)
+12(p2, +2(p1, x)) -> +12(p1, +2(p2, x))

The TRS R consists of the following rules:

+2(p1, p1) -> p2
+2(p1, +2(p2, p2)) -> p5
+2(p5, p5) -> p10
+2(+2(x, y), z) -> +2(x, +2(y, z))
+2(p1, +2(p1, x)) -> +2(p2, x)
+2(p1, +2(p2, +2(p2, x))) -> +2(p5, x)
+2(p2, p1) -> +2(p1, p2)
+2(p2, +2(p1, x)) -> +2(p1, +2(p2, x))
+2(p2, +2(p2, p2)) -> +2(p1, p5)
+2(p2, +2(p2, +2(p2, x))) -> +2(p1, +2(p5, x))
+2(p5, p1) -> +2(p1, p5)
+2(p5, +2(p1, x)) -> +2(p1, +2(p5, x))
+2(p5, p2) -> +2(p2, p5)
+2(p5, +2(p2, x)) -> +2(p2, +2(p5, x))
+2(p5, +2(p5, x)) -> +2(p10, x)
+2(p10, p1) -> +2(p1, p10)
+2(p10, +2(p1, x)) -> +2(p1, +2(p10, x))
+2(p10, p2) -> +2(p2, p10)
+2(p10, +2(p2, x)) -> +2(p2, +2(p10, x))
+2(p10, p5) -> +2(p5, p10)
+2(p10, +2(p5, x)) -> +2(p5, +2(p10, 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.


+12(p2, +2(p2, +2(p2, x))) -> +12(p5, x)
+12(p10, +2(p5, x)) -> +12(p10, x)
+12(p10, +2(p1, x)) -> +12(p10, x)
+12(p2, +2(p1, x)) -> +12(p2, x)
+12(p2, +2(p2, +2(p2, x))) -> +12(p1, +2(p5, x))
+12(p5, +2(p1, x)) -> +12(p5, x)
+12(p1, +2(p2, +2(p2, x))) -> +12(p5, x)
+12(p10, +2(p2, x)) -> +12(p10, x)
+12(p5, +2(p5, x)) -> +12(p10, x)
+12(p1, +2(p1, x)) -> +12(p2, x)
+12(p5, +2(p2, x)) -> +12(p5, x)
The remaining pairs can at least be oriented weakly.

+12(p10, +2(p1, x)) -> +12(p1, +2(p10, x))
+12(p5, +2(p2, x)) -> +12(p2, +2(p5, x))
+12(p10, +2(p2, x)) -> +12(p2, +2(p10, x))
+12(p10, +2(p5, x)) -> +12(p5, +2(p10, x))
+12(p5, +2(p1, x)) -> +12(p1, +2(p5, x))
+12(p2, +2(p1, x)) -> +12(p1, +2(p2, x))
Used ordering: Polynomial interpretation [21]:

POL(+2(x1, x2)) = 1 + x2   
POL(+12(x1, x2)) = x2   
POL(p1) = 1   
POL(p10) = 0   
POL(p2) = 1   
POL(p5) = 0   

The following usable rules [14] were oriented:

+2(p10, p1) -> +2(p1, p10)
+2(p5, p1) -> +2(p1, p5)
+2(p10, p2) -> +2(p2, p10)
+2(p5, +2(p1, x)) -> +2(p1, +2(p5, x))
+2(p2, +2(p1, x)) -> +2(p1, +2(p2, x))
+2(p10, +2(p1, x)) -> +2(p1, +2(p10, x))
+2(p5, +2(p2, x)) -> +2(p2, +2(p5, x))
+2(p1, +2(p2, +2(p2, x))) -> +2(p5, x)
+2(p1, +2(p1, x)) -> +2(p2, x)
+2(p2, +2(p2, +2(p2, x))) -> +2(p1, +2(p5, x))
+2(p10, +2(p5, x)) -> +2(p5, +2(p10, x))
+2(p10, +2(p2, x)) -> +2(p2, +2(p10, x))
+2(p5, +2(p5, x)) -> +2(p10, x)
+2(p1, +2(p2, p2)) -> p5
+2(p2, +2(p2, p2)) -> +2(p1, p5)
+2(p1, p1) -> p2
+2(p2, p1) -> +2(p1, p2)
+2(p5, p2) -> +2(p2, p5)
+2(p10, p5) -> +2(p5, p10)
+2(p5, p5) -> p10



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

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

+12(p10, +2(p5, x)) -> +12(p5, +2(p10, x))
+12(p10, +2(p1, x)) -> +12(p1, +2(p10, x))
+12(p5, +2(p1, x)) -> +12(p1, +2(p5, x))
+12(p5, +2(p2, x)) -> +12(p2, +2(p5, x))
+12(p10, +2(p2, x)) -> +12(p2, +2(p10, x))
+12(p2, +2(p1, x)) -> +12(p1, +2(p2, x))

The TRS R consists of the following rules:

+2(p1, p1) -> p2
+2(p1, +2(p2, p2)) -> p5
+2(p5, p5) -> p10
+2(+2(x, y), z) -> +2(x, +2(y, z))
+2(p1, +2(p1, x)) -> +2(p2, x)
+2(p1, +2(p2, +2(p2, x))) -> +2(p5, x)
+2(p2, p1) -> +2(p1, p2)
+2(p2, +2(p1, x)) -> +2(p1, +2(p2, x))
+2(p2, +2(p2, p2)) -> +2(p1, p5)
+2(p2, +2(p2, +2(p2, x))) -> +2(p1, +2(p5, x))
+2(p5, p1) -> +2(p1, p5)
+2(p5, +2(p1, x)) -> +2(p1, +2(p5, x))
+2(p5, p2) -> +2(p2, p5)
+2(p5, +2(p2, x)) -> +2(p2, +2(p5, x))
+2(p5, +2(p5, x)) -> +2(p10, x)
+2(p10, p1) -> +2(p1, p10)
+2(p10, +2(p1, x)) -> +2(p1, +2(p10, x))
+2(p10, p2) -> +2(p2, p10)
+2(p10, +2(p2, x)) -> +2(p2, +2(p10, x))
+2(p10, p5) -> +2(p5, p10)
+2(p10, +2(p5, x)) -> +2(p5, +2(p10, x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 6 less nodes.