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 QDPSizeChangeProof (⇔)
        ↳7 TRUE
    ↳8 QDP
        ↳9 QDPSizeChangeProof (⇔)
        ↳10 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) 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

Recursive path order with status [RPO].
Quasi-Precedence:

[+1, p2, p1, p5] > p10

Status:
+1: [1]
p2: multiset
p1: multiset
p5: multiset
p10: multiset

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