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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
  ↳5 QDP
    ↳6 QDPOrderProof (⇔)
    ↳7 QDP
    ↳8 DependencyGraphProof (⇔)
    ↳9 TRUE
  ↳10 QDP
    ↳11 QDPOrderProof (⇔)
    ↳12 QDP
    ↳13 PisEmptyProof (⇔)
    ↳14 TRUE

(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(p10, x)
+1(p10, +(p2, x)) → +1(p10, x)
+1(p10, +(p5, x)) → +1(p10, x)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
+1(x1, x2)  =  x2
+(x1, x2)  =  +(x2)
p1  =  p1
p2  =  p2
p5  =  p5
p10  =  p10

Lexicographic Path Order [LPO].
Precedence:
+1 > p2 > p5 > p10
p1 > p2 > p5 > p10

The following usable rules [FROCOS05] were oriented:

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

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

(8) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 6 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: Combined order from the following AFS and order.
+1(x1, x2)  =  x1
+(x1, x2)  =  +(x1, x2)

Lexicographic Path Order [LPO].
Precedence:
trivial

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