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:

+(p1, p1) → p2
+(p1, +(p2, p2)) → p5
+(p5, p5) → p10
+(+(x, y), z) → +(x, +(y, z))
+(p1, +(p1, x)) → +(p2, x)
+(p1, +(p2, +(p2, x))) → +(p5, x)
+(p2, p1) → +(p1, p2)
+(p2, +(p1, x)) → +(p1, +(p2, x))
+(p2, +(p2, p2)) → +(p1, p5)
+(p2, +(p2, +(p2, x))) → +(p1, +(p5, x))
+(p5, p1) → +(p1, p5)
+(p5, +(p1, x)) → +(p1, +(p5, x))
+(p5, p2) → +(p2, p5)
+(p5, +(p2, x)) → +(p2, +(p5, x))
+(p5, +(p5, x)) → +(p10, x)
+(p10, p1) → +(p1, p10)
+(p10, +(p1, x)) → +(p1, +(p10, x))
+(p10, p2) → +(p2, p10)
+(p10, +(p2, x)) → +(p2, +(p10, x))
+(p10, p5) → +(p5, p10)
+(p10, +(p5, x)) → +(p5, +(p10, x))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

+(p1, p1) → p2
+(p1, +(p2, p2)) → p5
+(p5, p5) → p10
+(+(x, y), z) → +(x, +(y, z))
+(p1, +(p1, x)) → +(p2, x)
+(p1, +(p2, +(p2, x))) → +(p5, x)
+(p2, p1) → +(p1, p2)
+(p2, +(p1, x)) → +(p1, +(p2, x))
+(p2, +(p2, p2)) → +(p1, p5)
+(p2, +(p2, +(p2, x))) → +(p1, +(p5, x))
+(p5, p1) → +(p1, p5)
+(p5, +(p1, x)) → +(p1, +(p5, x))
+(p5, p2) → +(p2, p5)
+(p5, +(p2, x)) → +(p2, +(p5, x))
+(p5, +(p5, x)) → +(p10, x)
+(p10, p1) → +(p1, p10)
+(p10, +(p1, x)) → +(p1, +(p10, x))
+(p10, p2) → +(p2, p10)
+(p10, +(p2, x)) → +(p2, +(p10, x))
+(p10, p5) → +(p5, p10)
+(p10, +(p5, x)) → +(p5, +(p10, x))

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:

+1(p5, +(p2, x)) → +1(p5, x)
+1(p2, p1) → +1(p1, p2)
+1(p10, p2) → +1(p2, p10)
+1(p5, +(p2, x)) → +1(p2, +(p5, x))
+1(+(x, y), z) → +1(x, +(y, z))
+1(p2, +(p2, p2)) → +1(p1, p5)
+1(p10, p5) → +1(p5, p10)
+1(p10, +(p2, x)) → +1(p10, x)
+1(p10, +(p1, x)) → +1(p1, +(p10, x))
+1(p5, p2) → +1(p2, p5)
+1(p5, p1) → +1(p1, p5)
+1(p10, p1) → +1(p1, p10)
+1(p10, +(p2, x)) → +1(p2, +(p10, x))
+1(p2, +(p2, +(p2, x))) → +1(p5, x)
+1(p5, +(p1, x)) → +1(p5, x)
+1(p10, +(p1, x)) → +1(p10, x)
+1(p1, +(p1, x)) → +1(p2, x)
+1(p10, +(p5, x)) → +1(p5, +(p10, x))
+1(p2, +(p1, x)) → +1(p1, +(p2, x))
+1(p2, +(p1, x)) → +1(p2, x)
+1(p2, +(p2, +(p2, x))) → +1(p1, +(p5, x))
+1(p1, +(p2, +(p2, x))) → +1(p5, x)
+1(p10, +(p5, x)) → +1(p10, x)
+1(p5, +(p5, x)) → +1(p10, x)
+1(p5, +(p1, x)) → +1(p1, +(p5, x))
+1(+(x, y), z) → +1(y, z)

The TRS R consists of the following rules:

+(p1, p1) → p2
+(p1, +(p2, p2)) → p5
+(p5, p5) → p10
+(+(x, y), z) → +(x, +(y, z))
+(p1, +(p1, x)) → +(p2, x)
+(p1, +(p2, +(p2, x))) → +(p5, x)
+(p2, p1) → +(p1, p2)
+(p2, +(p1, x)) → +(p1, +(p2, x))
+(p2, +(p2, p2)) → +(p1, p5)
+(p2, +(p2, +(p2, x))) → +(p1, +(p5, x))
+(p5, p1) → +(p1, p5)
+(p5, +(p1, x)) → +(p1, +(p5, x))
+(p5, p2) → +(p2, p5)
+(p5, +(p2, x)) → +(p2, +(p5, x))
+(p5, +(p5, x)) → +(p10, x)
+(p10, p1) → +(p1, p10)
+(p10, +(p1, x)) → +(p1, +(p10, x))
+(p10, p2) → +(p2, p10)
+(p10, +(p2, x)) → +(p2, +(p10, x))
+(p10, p5) → +(p5, p10)
+(p10, +(p5, x)) → +(p5, +(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:

+1(p5, +(p2, x)) → +1(p5, x)
+1(p2, p1) → +1(p1, p2)
+1(p10, p2) → +1(p2, p10)
+1(p5, +(p2, x)) → +1(p2, +(p5, x))
+1(+(x, y), z) → +1(x, +(y, z))
+1(p2, +(p2, p2)) → +1(p1, p5)
+1(p10, p5) → +1(p5, p10)
+1(p10, +(p2, x)) → +1(p10, x)
+1(p10, +(p1, x)) → +1(p1, +(p10, x))
+1(p5, p2) → +1(p2, p5)
+1(p5, p1) → +1(p1, p5)
+1(p10, p1) → +1(p1, p10)
+1(p10, +(p2, x)) → +1(p2, +(p10, x))
+1(p2, +(p2, +(p2, x))) → +1(p5, x)
+1(p5, +(p1, x)) → +1(p5, x)
+1(p10, +(p1, x)) → +1(p10, x)
+1(p1, +(p1, x)) → +1(p2, x)
+1(p10, +(p5, x)) → +1(p5, +(p10, x))
+1(p2, +(p1, x)) → +1(p1, +(p2, x))
+1(p2, +(p1, x)) → +1(p2, x)
+1(p2, +(p2, +(p2, x))) → +1(p1, +(p5, x))
+1(p1, +(p2, +(p2, x))) → +1(p5, x)
+1(p10, +(p5, x)) → +1(p10, x)
+1(p5, +(p5, x)) → +1(p10, x)
+1(p5, +(p1, x)) → +1(p1, +(p5, x))
+1(+(x, y), z) → +1(y, z)

The TRS R consists of the following rules:

+(p1, p1) → p2
+(p1, +(p2, p2)) → p5
+(p5, p5) → p10
+(+(x, y), z) → +(x, +(y, z))
+(p1, +(p1, x)) → +(p2, x)
+(p1, +(p2, +(p2, x))) → +(p5, x)
+(p2, p1) → +(p1, p2)
+(p2, +(p1, x)) → +(p1, +(p2, x))
+(p2, +(p2, p2)) → +(p1, p5)
+(p2, +(p2, +(p2, x))) → +(p1, +(p5, x))
+(p5, p1) → +(p1, p5)
+(p5, +(p1, x)) → +(p1, +(p5, x))
+(p5, p2) → +(p2, p5)
+(p5, +(p2, x)) → +(p2, +(p5, x))
+(p5, +(p5, x)) → +(p10, x)
+(p10, p1) → +(p1, p10)
+(p10, +(p1, x)) → +(p1, +(p10, x))
+(p10, p2) → +(p2, p10)
+(p10, +(p2, x)) → +(p2, +(p10, x))
+(p10, p5) → +(p5, p10)
+(p10, +(p5, x)) → +(p5, +(p10, x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 9 less nodes.

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

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

+1(p10, +(p2, x)) → +1(p2, +(p10, x))
+1(p5, +(p2, x)) → +1(p5, x)
+1(p2, +(p2, +(p2, x))) → +1(p5, x)
+1(p5, +(p1, x)) → +1(p5, x)
+1(p10, +(p1, x)) → +1(p10, x)
+1(p1, +(p1, x)) → +1(p2, x)
+1(p5, +(p2, x)) → +1(p2, +(p5, x))
+1(p10, +(p5, x)) → +1(p5, +(p10, x))
+1(p10, +(p2, x)) → +1(p10, x)
+1(p2, +(p1, x)) → +1(p1, +(p2, x))
+1(p10, +(p1, x)) → +1(p1, +(p10, x))
+1(p2, +(p1, x)) → +1(p2, x)
+1(p2, +(p2, +(p2, x))) → +1(p1, +(p5, x))
+1(p1, +(p2, +(p2, x))) → +1(p5, x)
+1(p10, +(p5, x)) → +1(p10, x)
+1(p5, +(p5, x)) → +1(p10, x)
+1(p5, +(p1, x)) → +1(p1, +(p5, x))

The TRS R consists of the following rules:

+(p1, p1) → p2
+(p1, +(p2, p2)) → p5
+(p5, p5) → p10
+(+(x, y), z) → +(x, +(y, z))
+(p1, +(p1, x)) → +(p2, x)
+(p1, +(p2, +(p2, x))) → +(p5, x)
+(p2, p1) → +(p1, p2)
+(p2, +(p1, x)) → +(p1, +(p2, x))
+(p2, +(p2, p2)) → +(p1, p5)
+(p2, +(p2, +(p2, x))) → +(p1, +(p5, x))
+(p5, p1) → +(p1, p5)
+(p5, +(p1, x)) → +(p1, +(p5, x))
+(p5, p2) → +(p2, p5)
+(p5, +(p2, x)) → +(p2, +(p5, x))
+(p5, +(p5, x)) → +(p10, x)
+(p10, p1) → +(p1, p10)
+(p10, +(p1, x)) → +(p1, +(p10, x))
+(p10, p2) → +(p2, p10)
+(p10, +(p2, x)) → +(p2, +(p10, x))
+(p10, p5) → +(p5, p10)
+(p10, +(p5, x)) → +(p5, +(p10, x))

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.


+1(p10, +(p2, x)) → +1(p2, +(p10, x))
+1(p5, +(p2, x)) → +1(p5, x)
+1(p2, +(p2, +(p2, x))) → +1(p5, x)
+1(p5, +(p1, x)) → +1(p5, x)
+1(p10, +(p1, x)) → +1(p10, x)
+1(p1, +(p1, x)) → +1(p2, x)
+1(p5, +(p2, x)) → +1(p2, +(p5, x))
+1(p10, +(p2, x)) → +1(p10, x)
+1(p2, +(p1, x)) → +1(p1, +(p2, x))
+1(p10, +(p1, x)) → +1(p1, +(p10, x))
+1(p2, +(p1, x)) → +1(p2, x)
+1(p2, +(p2, +(p2, x))) → +1(p1, +(p5, x))
+1(p1, +(p2, +(p2, x))) → +1(p5, x)
+1(p5, +(p1, x)) → +1(p1, +(p5, x))
The remaining pairs can at least be oriented weakly.

+1(p10, +(p5, x)) → +1(p5, +(p10, x))
+1(p10, +(p5, x)) → +1(p10, x)
+1(p5, +(p5, x)) → +1(p10, x)
Used ordering: Polynomial interpretation [25,35]:

POL(p1) = 1   
POL(p10) = 0   
POL(p5) = 0   
POL(+1(x1, x2)) = (4)x_2   
POL(p2) = 1/4   
POL(+(x1, x2)) = (1/4)x_1 + (4)x_2   
The value of delta used in the strict ordering is 1/4.
The following usable rules [17] were oriented:

+(p2, +(p2, p2)) → +(p1, p5)
+(p2, p1) → +(p1, p2)
+(p5, p2) → +(p2, p5)
+(p5, p1) → +(p1, p5)
+(p10, p1) → +(p1, p10)
+(p10, p5) → +(p5, p10)
+(p10, p2) → +(p2, p10)
+(p1, p1) → p2
+(p5, p5) → p10
+(p1, +(p2, p2)) → p5
+(p1, +(p2, +(p2, x))) → +(p5, x)
+(p10, +(p1, x)) → +(p1, +(p10, x))
+(p1, +(p1, x)) → +(p2, x)
+(p2, +(p1, x)) → +(p1, +(p2, x))
+(p10, +(p2, x)) → +(p2, +(p10, x))
+(p10, +(p5, x)) → +(p5, +(p10, x))
+(p5, +(p1, x)) → +(p1, +(p5, x))
+(p5, +(p2, x)) → +(p2, +(p5, x))
+(p2, +(p2, +(p2, x))) → +(p1, +(p5, x))
+(p5, +(p5, x)) → +(p10, x)



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

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

+1(p10, +(p5, x)) → +1(p10, x)
+1(p5, +(p5, x)) → +1(p10, x)
+1(p10, +(p5, x)) → +1(p5, +(p10, x))

The TRS R consists of the following rules:

+(p1, p1) → p2
+(p1, +(p2, p2)) → p5
+(p5, p5) → p10
+(+(x, y), z) → +(x, +(y, z))
+(p1, +(p1, x)) → +(p2, x)
+(p1, +(p2, +(p2, x))) → +(p5, x)
+(p2, p1) → +(p1, p2)
+(p2, +(p1, x)) → +(p1, +(p2, x))
+(p2, +(p2, p2)) → +(p1, p5)
+(p2, +(p2, +(p2, x))) → +(p1, +(p5, x))
+(p5, p1) → +(p1, p5)
+(p5, +(p1, x)) → +(p1, +(p5, x))
+(p5, p2) → +(p2, p5)
+(p5, +(p2, x)) → +(p2, +(p5, x))
+(p5, +(p5, x)) → +(p10, x)
+(p10, p1) → +(p1, p10)
+(p10, +(p1, x)) → +(p1, +(p10, x))
+(p10, p2) → +(p2, p10)
+(p10, +(p2, x)) → +(p2, +(p10, x))
+(p10, p5) → +(p5, p10)
+(p10, +(p5, x)) → +(p5, +(p10, x))

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.


+1(p10, +(p5, x)) → +1(p10, x)
+1(p5, +(p5, x)) → +1(p10, x)
+1(p10, +(p5, x)) → +1(p5, +(p10, x))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(p1) = 4   
POL(p10) = 13/4   
POL(p5) = 4   
POL(+1(x1, x2)) = (2)x_1 + (4)x_2   
POL(p2) = 4   
POL(+(x1, x2)) = (4)x_1 + (4)x_2   
The value of delta used in the strict ordering is 21/2.
The following usable rules [17] were oriented:

+(p2, +(p2, p2)) → +(p1, p5)
+(p2, p1) → +(p1, p2)
+(p5, p2) → +(p2, p5)
+(p5, p1) → +(p1, p5)
+(p10, p1) → +(p1, p10)
+(p1, p1) → p2
+(p5, p5) → p10
+(p10, p5) → +(p5, p10)
+(p1, +(p2, p2)) → p5
+(p10, p2) → +(p2, p10)
+(p1, +(p2, +(p2, x))) → +(p5, x)
+(p10, +(p1, x)) → +(p1, +(p10, x))
+(p1, +(p1, x)) → +(p2, x)
+(p2, +(p1, x)) → +(p1, +(p2, x))
+(p10, +(p2, x)) → +(p2, +(p10, x))
+(p10, +(p5, x)) → +(p5, +(p10, x))
+(p5, +(p1, x)) → +(p1, +(p5, x))
+(p5, +(p2, x)) → +(p2, +(p5, x))
+(p2, +(p2, +(p2, x))) → +(p1, +(p5, x))
+(p5, +(p5, x)) → +(p10, x)



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

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

+(p1, p1) → p2
+(p1, +(p2, p2)) → p5
+(p5, p5) → p10
+(+(x, y), z) → +(x, +(y, z))
+(p1, +(p1, x)) → +(p2, x)
+(p1, +(p2, +(p2, x))) → +(p5, x)
+(p2, p1) → +(p1, p2)
+(p2, +(p1, x)) → +(p1, +(p2, x))
+(p2, +(p2, p2)) → +(p1, p5)
+(p2, +(p2, +(p2, x))) → +(p1, +(p5, x))
+(p5, p1) → +(p1, p5)
+(p5, +(p1, x)) → +(p1, +(p5, x))
+(p5, p2) → +(p2, p5)
+(p5, +(p2, x)) → +(p2, +(p5, x))
+(p5, +(p5, x)) → +(p10, x)
+(p10, p1) → +(p1, p10)
+(p10, +(p1, x)) → +(p1, +(p10, x))
+(p10, p2) → +(p2, p10)
+(p10, +(p2, x)) → +(p2, +(p10, x))
+(p10, p5) → +(p5, p10)
+(p10, +(p5, x)) → +(p5, +(p10, 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.