(0) Obligation:

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.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 7 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


+1(p1, +(p1, x)) → +1(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(p5, +(p1, x)) → +1(p5, x)
+1(p2, +(p2, +(p2, x))) → +1(p5, x)
+1(p5, +(p2, x)) → +1(p5, x)
+1(p5, +(p5, x)) → +1(p10, x)
+1(p10, +(p1, x)) → +1(p1, +(p10, x))
+1(p10, +(p1, x)) → +1(p10, x)
+1(p10, +(p2, x)) → +1(p2, +(p10, x))
+1(p10, +(p2, x)) → +1(p10, x)
+1(p10, +(p5, x)) → +1(p5, +(p10, x))
+1(p10, +(p5, x)) → +1(p10, x)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

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

The following usable rules [FROCOS05] were oriented:

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

(7) Obligation:

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

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

(8) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 3 less nodes.

(9) TRUE

(10) Obligation:

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

+1(+(x, y), z) → +1(y, z)
+1(+(x, y), z) → +1(x, +(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.

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


+1(+(x, y), z) → +1(y, z)
+1(+(x, y), z) → +1(x, +(y, z))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

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

The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

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.

(13) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(14) TRUE