Termination w.r.t. Q of the given QTRS could be proven:

(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) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Combined order from the following AFS and order.
+(x1, x2)  =  +(x2)
p2  =  p2
p1  =  p1
p5  =  p5
p10  =  p10

Lexicographic path order with status [LPO].
Quasi-Precedence:

[p2, p1, p5, p10] > +₁

Status:

+₁: [1]
p2: []
p1: []
p5: []
p10: []

AFS:
+(x1, x2)  =  +(x2)
p2  =  p2
p1  =  p1
p5  =  p5
p10  =  p10

From the DPs we obtained the following set of size-change graphs:

• +¹(p1, +(p1, x)) → +¹(p2, x) (allowed arguments on rhs = {2})
  The graph contains the following edges 2 > 2

• +¹(p1, +(p2, +(p2, x))) → +¹(p5, x) (allowed arguments on rhs = {2})
  The graph contains the following edges 2 > 2

• +¹(p2, +(p1, x)) → +¹(p2, x) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 1 >= 1, 2 > 2

• +¹(p2, +(p2, +(p2, x))) → +¹(p5, x) (allowed arguments on rhs = {2})
  The graph contains the following edges 2 > 2

• +¹(p5, +(p2, x)) → +¹(p2, +(p5, x)) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 2 >= 2

• +¹(p10, +(p2, x)) → +¹(p2, +(p10, x)) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 2 >= 2

• +¹(p5, +(p1, x)) → +¹(p1, +(p5, x)) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 2 >= 2

• +¹(p10, +(p1, x)) → +¹(p1, +(p10, x)) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 2 >= 2

• +¹(p5, +(p5, x)) → +¹(p10, x) (allowed arguments on rhs = {2})
  The graph contains the following edges 2 > 2

• +¹(p10, +(p5, x)) → +¹(p5, +(p10, x)) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 2 >= 2

• +¹(p2, +(p1, x)) → +¹(p1, +(p2, x)) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 2 >= 2

• +¹(p2, +(p2, +(p2, x))) → +¹(p1, +(p5, x)) (allowed arguments on rhs = {2})
  The graph contains the following edges 2 > 2

• +¹(p5, +(p1, x)) → +¹(p5, x) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 1 >= 1, 2 > 2

• +¹(p5, +(p2, x)) → +¹(p5, x) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 1 >= 1, 2 > 2

• +¹(p10, +(p1, x)) → +¹(p10, x) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 1 >= 1, 2 > 2

• +¹(p10, +(p2, x)) → +¹(p10, x) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 1 >= 1, 2 > 2

• +¹(p10, +(p5, x)) → +¹(p10, x) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 1 >= 1, 2 > 2

We oriented the following set of usable rules [AAECC05,FROCOS05].

+(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) TRUE

(8) 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.

(9) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
+(x1, x2)  =  +(x1, x2)

From the DPs we obtained the following set of size-change graphs:

• +¹(+(x, y), z) → +¹(y, z) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 1 > 1, 2 >= 2

• +¹(+(x, y), z) → +¹(x, +(y, z)) (allowed arguments on rhs = {1})
  The graph contains the following edges 1 > 1

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(10) TRUE